HA150.pdf

HA150 SQL Basics for HANA

Material Number: 50119271

SQL Basics For HANA – Agenda Motivation And Basic Concepts Reading Data From A Table Or View

Exercises 1 - 19

Aggregating Data

Exercises 20 - 29

Reading Data From Multiple Tables Part I

Exercises 30 - 40

Reading Data From Multiple Tables Part II

Exercises 41 - 45

Understanding NULL Values Changing Data Stored in Tables

Exercises 46 - 53

Defining How Data Is Stored

Exercises 54 - 58

Using Views For Data Access

Exercises 59 - 60

Defining Data Access Database Transactions ©

2013 SAP AG. All rights reserved.

2

SQL Basics For HANA Course

The SQL Basics course is designed to ... 

Explain basic concepts in the database world and the Relational Database Model



Refresh and deepen SQL knowledge, especially for developing on HANA database systems

The SQL Basics course is not designed to ... 

Explain implementation details of SAP HANA



Supersede official SAP HANA documentation



Solve specific development problems



Serve as a basis for decision-making for development projects

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3

Unit 1 Motivation & Basic Concepts

4

Unit 1: Motivation And Basic Concepts Learning Objectives After completing this unit, you will be able to: • List some database models • Understand the motivation for and foundation of the relational model • Understand why learning SQL is important when dealing with HANA • Understand how the relation model and SQL are related • List the three sub-languages of SQL • Understand the sample database used throughout the course

© 2013 SAP AG. All rights reserved.

5

Basic Concepts

What are the terms •

Database

•

Database System

•

Database Management System ?

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6

DBS = DB + DBMS



The Database Management System (DBMS) manages the database  

Every access to the database (create, read, insert, update, delete) goes exclusively through the DBMS The DBMS exercises complete control over the database



Database (DB) = structured collection of "records"



Database System (DBS) = specific database + DBMS

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DBS = DB + DBMS

User / Application

DBMS DBS

DB ©


8

DBS = DB + DBMS



„What Database do you use?“ 



„Which Database Management System do you use?“ 



„We are using HANA as Database.“ 



„We are using SAP HANA as Database Management System.“ 



In everyday life, these terms are usually used incorrectly …

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9

Basics Concepts

3 Level Schema Architecture

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10


Objective: Changes at a lower level should not affect a higher level (if possible)

Application/User

External Schema 1

Application/User …

External Schema n

Conceptual Schema

Internal Schema

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External Level

Conceptual Level

Internal Level

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The address book should not display salary data  External Level (VIEW) An Employee has a D-Number, name, and salary and is assigned to a department  Conceptual Level (TABLE) The board mostly accesses employee data according to descending order of salary (which has to be very fast)  Internal Level (INDEX)

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12

Basic Concepts

Which Database Models are available?

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13

Database Models



Hierarchical Database Model



Network Database Model



Relational Database Model



Object-relational Database Model



Object-oriented Database Model



XML-based Database Model



…

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14

Basic Concepts


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15


Invented in the late 1960s by Edgar F. Codd (1923-2003)  IBM Almaden Research Lab in San Jose (California)

First prototype in the mid-1970s • "System R", IBM San Jose

Relational DBMS products (selection) • • • •

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Oracle (since 1979) IBM DB2 (since 1983) SQL Server (since 1989) SAP HANA (since 2011)


16


Initial Development Goals (1960s):  Simple but mathematically profound database model – Easy to understand but still mathematically precise

 Simple but mathematically-based database language – Easy to learn, but semantics still described with mathematical precision – Provable equivalence of two queries

 Integrity monitoring largely by the DBMS  Storage & retrieval of data is the responsibility of the DBMS (and not of the user) – Descriptive language rather than navigation (optimizer selects optimal execution strategy)

 Clean separation between conceptual and internal schema

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17

Basic Concepts

Why is the Relational Database Model actually called Relational Database Model?

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18

Relations

A relation (in the sense of mathematics) is the subset of the Cartesian product of sets

R⊆AxBxC

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19

Comparative Relations (=,≠,<,≤,>,≥)

N = {1, 2, 3, 4, 5} N × N = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (2,5), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5)}

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Comparative Relations (=,≠,<,≤,>,≥) ≤ ⊆ N × N ≤ ⊆ {(1,1), (2,1), (3,1), (4,1), (5,1),

(1,2), (2,2), (3,2), (4,2), (5,2),

(1,3), (2,3), (3,3), (4,3), (5,3),

(1,4), (2,4), (3,4), (4,4), (5,4),

(1,5), (2,5), (3,5), (4,5), (5,5)}

≤ = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,2), (2,3), (2,4), (2,5), (3,3), (3,4), (3,5), (4,4), (4,5), (5,5)} (2,3) ϵ ≤ 2 ≤ 3

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Relations PersNumber = {P036407, P040824, P052867, …} Name = {Ben, Paul, Raja, …} HiringYear = {2001, 2008, 2010, …} PersNumber × Name × HiringYear= {(P036407,Ben,2001), (P036407,Ben,2008), (P036407,Paul,2001), (P036407,Paul,2008), (P035607,Raja,2001), (P036407,Raja,2008), (P040824,Ben,2001), (P040824,Ben,2008), (P040824,Paul,2001), (P040824,Paul,2008), (P040824,Raja,2001), (P040824,Raja,2008), (P052867,Ben,2001), (P052867,Jim,2008), (P052867,Paul,2001), (P052867,Vishal,2008), (P052867,Raja,2001), (P052867,Werner,2008),

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(P036407,Ben,2010), (P036407,Paul,2010), (P036407,Raja,2010), (P040824,Ben,2010), (P040824,Paul,2010), (P040824,Raja,2010), (P052867,Ben,2010), (P052867,Paul,2010), (P052867,Raja,2010), …}

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Relations Employee ⊆ PersNumber × Name × HiringYear Employee ⊆ {(P036407,Ben,2001), (P036407,Paul,2001), (P035607,Raja,2001), (P040824,Ben,2001), (P040824,Paul,2001), (P040824,Raja,2001), (P052867,Ben,2001), (P052867,Paul,2001), (P052867,Raja,2001),

(P036407,Ben,2008), (P036407,Paul,2008), (P036407,Raja,2008), (P040824,Ben,2008), (P040824,Paul,2008), (P040824,Raja,2008), (P052867,Jim,2008), (P052867,Vishal,2008), (P052867,Werner,2008),

(P036407,Ben,2010), (P036407,Paul,2010), (P036407,Raja,2010), (P040824,Ben,2010), (P040824,Paul,2010), (P040824,Raja,2010), (P052867,Ben,2010), (P052867,Paul,2010), (P052867,Raja,2010), …}

Employee= {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}

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Relations & Tables

A relation can be represented as a table Employee= {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}

Employee

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PersNumber

Name

HiringYear

P036407 P040824 P052867

Paul Ben Raja

2001 2008 2010

24

Basic Concepts

Which languages are available for the Relational Database Model?

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25

Relational Languages

Relational Algebra  

Formal basis for DBMS internal query optimization 6 basic operations (selection, projection, cross join, union, difference, rename)

Relational Calculus 

Tuple variables and quantifiers

SQL 

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Standardized and used in practice


26

Why SQL?

Why SQL is important?

Why it is worth working with SQL

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Why SQL? → Reasons #1 and #2 In almost every business application scenario, the data is managed using database systems. The most significant are database systems based on the relational data model and using SQL (Structured Query Language) as a database language.

SQL is a widely-established, powerful, standardized database language many application programmers have experience in. There is (so far) no other database language that has all the advantages mentioned.

SAP HANA is a relational data base management system and SAP HANA supports SQL ©


28

Why SQL?

Different application architectures and development models are possible with SAP HANA Data Marts with SAP HANA

Standalone HANA Apps

BI Tools, MS Excel, Web Browser …

Standalone Application SAP HANA clients

Traditional DBMS

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HANA Content / Database Tables

HANA Content / Database Tables

SAP HANA

Schema replication

SAP HANA

BI 4.0 (optional) Semantic Layer

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Why SQL?

Different application architectures and development models are possible with SAP HANA HANA as primary Database HANA as Accelerator (secondary DB) (for AS ABAP) SAP GUI, Browser-based GUI

SAP GUI, Browser-based GUI

SAP Application Server

SAP Application Server

replication

Data

SAP HANA

ABAP Schema

ABAP Schema / Database Tables

SAP HANA

E.g. CRM on HANA – NW 7.40

Aggregation Levels

Traditional DBMS

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Why SQL? → Reason #3

HANA supports an extended version of SQL

No matter which HANA-based architecture or development model you work with, it is likely that using HANA SQL will be beneficial.

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31

SQL

SQL Features

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SQL Features

 SQL is standardized – No uncontrolled growth with respect to syntax and semantics

 SQL is descriptive (rather than procedural) – The "what" and not "how" is described

 SQL execution is optimized – SQL statement is first parsed and optimized, and then executed – Optimizer determines optimal execution plan (at least in theory)

 SQL is multi-set oriented (and not single record-based) – Using a single SQL statement multiple table rows can be read, modified or deleted in one go

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33

SQL Language Elements

SQL language elements can be divided into three categories: 

DML = Data Manipulation Language SELECT, INSERT, UPDATE, DELETE



DDL = Data Definition Language CREATE, ALTER, DROP, RENAME



DCL = Data Control Language GRANT, REVOKE

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34

SQL

SQL is standardized!

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35

SQL Standard: History

1970s

SEQUEL (Structured English Query Language) as a language for "System R" • "System R" was the first relational prototype developed by IBM in San Jose

End of 1970

Renaming of SEQUEL into SQL (Structured Query Language)

1982

American National Standards Institute (ANSI) begins with standardization of SQL

1986

First version of the SQL standard is adopted (as ANSI) SQL-86 ("SQL0“)

1987

International Organization for Standardization (ISO) does SQL standard (ISO standard)

1989

SQL-89 ("SQL1")

1992

SQL-92 ("SQL2"), 3 attenuation levels: Entry Level, Intermediate Level, Full Level

1999

SQL:1999 ("SQL3")

2003

SQL:2003 ("SQL4"), "SQL/XML:2006" 2006 official standard part of SQL:2003

2008

SQL:2008 ("SQL5"), officially: "ISO/IEC 9075:2008" and "DIN 9075" (> 3,000 pages)

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SQL

SQL ≠ relational

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37

SQL

SQL is the most important database language for relational database model. Although SQL is based on the relational database model, it deviates in important points from the "purely relational" model: 

No (primary) key required 



NULL values are allowed 



3-valued logic required ("A = A" need not be TRUE)

The language is not closed   

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Duplicates (identical rows) allowed (multi-sets instead of sets)

Order of the result rows may be relevant (ORDER BY) Anonymous result columns allowed Duplicate result column names allowed


38

Database Objects

What can be found in the database?

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39

Database Objects

The table is the primary database object - but not the only one. Apart from tables a database usually contains:     

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Views to simplify and limit data access Indexes to speed up (certain) read accesses Constraints to ensure data consistency Stored procedures for more complex tasks Triggers to selectively respond to particular events


40

Basic Concepts

What is a table?

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41

Tables

A (database) table represents a relation (but: multi-set of rows instead of set of tuples)

Employee PersNumber

Name

HiringYear

P036407

Paul

2001

P040824

Ben

2008

P052867

Raja

2010

Employee = {(P036407,Paul,2001), (P040824,Ben,2008), (P052867,Raja,2010)}

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42

Tables

A table consists of rows and columns 

A table can therefore be represented two-dimensionally

A table represents a relation (with multi-sets property) 

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A table represents a (unordered) multi-set of points in n-dimensional space:  Each point of the multi-set corresponds to a table row  Each of the n dimensions is equivalent to a table column


43

Table rows as points in space

Name

Paul HiringYear Raja

(P040824, Ben, 2008) 2010 2008

Ben

2001 PersNumber

P030407

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P040824

P052867

44

Projection on the plane

Projection list

Name

SELECT PersNumber, Name FROM Employee WHERE PersNumber = 'P040824';

Paul

HiringYear Raja

(P030824, Ben, 2008) 2010 2008

Ben

(P040824, Ben)

2001 PersNumber

P036407

©


P040824

P052867

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Projection on an axis

Projection list

Name

SELECT Name FROM Employee WHERE PersNumber = 'P040824';

Paul

HiringYear Raja

(P040824, Ben, 2008) 2010 2008

Ben

2001 PersNumber

P036407

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P040824

P052867

46

Basic Concepts

What are the components of a (database) table?

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47

Tables

Table-Name

Primary Key

Column Name

Employee PersNumber Name P036407 Paul Table Row P040824 Ben P052867 Raja

HiringYear 2001 2008 2010 (Table Column) value

(Table) Column

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48

Basic Concepts

What is a key?

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49

Key

Key = is a set of columns which serves to uniquely identify any row in the table. The ability to uniquely identify rows must apply in principle (and not only for the rows existing at a certain point in time).

Employee PersNumber Name P036407 Paul P040824 Ben P052867 Raja

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HiringYear 2001 2008 2010

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Multi-column Keys

A key can consist of multiple columns

Employee Name Country City BuildingNr Block Floor Room SeatNr Mr A

DE

WDF

03

A

5

29

1

Ms B

DE

WDF

03

A

5

29

2

Ms C

DE

WDF

03

A

5

29

3

…

…

…

…

…

…

…

…

 In this example the key consists of 7 columns  1 Key (not 7 keys)!

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Multiple Keys

There may be more than one key

Employee PNumber Name TaxID

Passport ID

Plate Number

ChasisNr

CarLicense ID

SWIFT

IBAN

P000815

…

…

…

HD-MM 815

…

…

…

…

P004711

…

…

…

HD-ML 4711

…

…

…

…

P012345

…

…

…

HD-OI 2345

…

…

…

…

…

…

…

…

…

…

…

…

…

 Here are 7 keys (provided there are no joint accounts)  Just 1 key is selected as Primary Key

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Basic Concepts

What is a Foreign-Key?

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53

Foreign-Keys

Foreign-Key = set of columns, which is a (primary) key in an(other) table •

The foreign-key can refer to its own table

•

It is not necessary to share the same name(s)

•

The foreign-key can contain only those values that occur as a (primary) key value in another table (in addition and if applicable, NULL values are allowed)

•

The foreign-key is usually not a key!

Employee

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DNumber D010000 D010001 D010002 D010003 D010004 D010005


Name Mr A Ms B Mr C Ms D Mr E Ms F

DepartmentNR A01 A01 A02 A02 A02 A03

Foreign-key Relationship

Department DNr Function A01 HANA-Development A02 HANA-Sales A03 HANA-Training A04 ABAP-Development

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Foreign-Keys

A Foreign-Key can consist of multiple columns

Publication PaperID Topic P001 In-Memory P002 Column Store P003 Row Store P004 SQL & XML P005 XML & SQL P006 DB Recovery

Uni HPI HPI

StudentID 12345 12345

HPI FSU FSU FSU

77777 12345 12345 77777

Student

Uni StudentID Name FirstName HPI 12345 A B HPI 77777 C D FSU 12345 E F FSU 77777 G H

 This example shows 1 Foreign-Key built on 2 columns

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55

Foreign-Keys

There can be multiple Foreign-Keys

Vendor

VID L1 L2 L3 …

©

Name Heidelberg Paper Wiesloch Paper Walldorf Paper …


Delivery VID PNr Year Quantity

L1 L1 L1 L1 L2 L2 L3 …

A1 A1 A1 A2 A1 A2 A1 …

2010 2011 2012 2012 2011 2011 2012 …

320 570 925 102 577 100 999 …

Product PNr Description A1 Sticky Notes A2 Printing Paper A3 Envelopes … …

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Foreign-Keys

The foreign-key can refer to its own table

Employee

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DNumber D010000 D010001 D010002 D010003 D010004 D010005 D010006 D010007

Name Mr A Ms B Mr C Ms D Mr E Ms F Mr G Ms H

DNumberManager D010007 D010007 D010004 D010002 D010002 D010005 D010005

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Example Scenario

Fictional vehicle registration office as an example scenario

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58

Example: Registration Office

 The database of a fictional registration office will serve as the basis for further explanations.  The tables in this database have been specifically tailored to the SQL course and are not an example of good database design.    

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The officials working in the fictional registration office have a manager. Each vehicle is registered for exactly one owner (or is unregistered). There is a list of vehicles that have been reported stolen. Owners, who have at least three vehicles registered, are assigned to one or multiple contacts.


59

Example: Registration Office

Official(PNr, Name, Overtime, Salary, Manager)

Contact(PersNumber, OwnerID) Owner(OwnerID, Name, Birthday, City)

Car(CarID, PlateNumber, Brand, Color, HP, Owner) Stolen(PlateNumber, reported_at)

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60

Officials

Official

©

PNr

Name

P01

Mr A

P02

Salary

Manager

10

A09

P04

Mr B

10

A10

P04

P03

Ms C

20

A09

P04

P04

Ms D

NULL

A12

P09

P05

Mr E

10

A08

P08

P06

Mr F

18

A09

P08

P07

Ms G

22

A11

P08

P08

Ms H

NULL

A13

P09

P09

Mr I

NULL

A14

NULL


Overtime

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Owner

Owner

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OwnerID

Name

Birthday

City

H01

Ms T

20.06.1934

Wiesloch

H02

Ms U

11.05.1966

Hockenheim

H03

SAP AG

NULL

Walldorf

H04

HDM AG

NULL

Heidelberg

H05

Mr V

21.04.1952

Leimen

H06

Ms W

01.06.1957

Wiesloch

H07

IKEA

NULL

Walldorf

H08

Mr X

30.08.1986

Walldorf

H09

Ms Y

10.02.1986

Sinsheim

H10

Mr Z

03.02.1986

Ladenburg

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Contact

Contact

©


PersNumber

OwnerID

P01

H03

P01

H04

P01

H07

P04

H03

P04

H04

P08

H04

P08

H07

P09

H03

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Cars Car

©

CarID

PlateNumber

Brand

Color

F01 F02 F03 F04 F05 F06 F07 F08 F09 F10

HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507

Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW

red black blue white black yellow blue black red black

75 120 184 136 170 260 116 160 105 140

H06 H03 H03 H07 H03 H05 H03 H07 H02 H04

F11 F12 F13 F14 F15 F16 F17 F18 F19 F20

HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 NULL HD-Y 333 HD-MQ 2006 HD-VW 2012 NULL

BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi

green red red white black green orange red black green

184 105 136 170 136 120 184 90 125 184

H02 H04 H07 H03 H03 NULL H09 H03 H01 NULL


HP

Owner

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Stolen Cars Stolen

©

PlateNumber

reported_at

HD-VW 1999

20.06.2012

HD-V 106

01.06.2012

HD-Y 333

21.05.2012


65

Registration Office & EU

 A new (fictional) European Union directive requires that the information about which vehicle is registered to which owner, has to be stored in a central transnational database.  The vehicle identification number (CarID) is unique across the EU, but not the Owner ID. Owner_EU(Country, OwnerID, Name, Birthday, City) Car_EU(CarID, PlateNumber, Brand, Color, HP, Country, Owner)

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66

Owner (EU-wide) Owner_EU

©

Country

OwnerID

Name

Birthday

City

D

H01

Ms T

20.06.1934

Wiesloch

D

H02

Ms U

11.05.1966

Hockenheim

D

H03

SAP AG

NULL

Walldorf

D

H04

HDM AG

NULL

Heidelberg

D

H05

Mr V

21.04.1952

Leimen

D

H06

Ms W

01.06.1957

Wiesloch

D

H07

IKEA

NULL

Walldorf

D

H08

Mr X

30.08.1986

Walldorf

D

H09

Ms Y

10.02.1986

Sinsheim

D

H10

Mr Z

03.02.1986

Ladenburg

A

H01

Ms O

21.05.1977

Wien

A

H02

Mr P

02.08.1977

Salzburg

E

H01

Señor Q

18.02.1925

Madrid

E

H02

Señora R

27.02.1927

Barcelona


67

Cars (EU-wide) Car_EU

©

CarID F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 F21 F22 F23 F24


PlateNumber HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 NULL HD-Y 333 HD-MQ 2006 HD-VW 2012 NULL W-302 ML S-215 MM 1409 EMM 3206 MLM

Brand Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi Mercedes VW Audi VW

Color red black blue white black yellow blue black red black green red red white black green orange red black green black silvern blue black

HP 75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184 170 184 116 170

Country D D D D D D D D D D D D D D D NULL D D D NULL A A E E

Owner H06 H03 H03 H07 H03 H05 H03 H07 H02 H04 H02 H04 H07 H03 H03 NULL H09 H03 H01 NULL H01 H02 H01 H02

68

Unit 1: Motivation And Basic Concepts Summary

You should now be able to: • List some database models

• • • • •

Understand the motivation for and foundation of the relational model Understand why learning SQL is important when dealing with HANA Understand how the relation model and SQL are related List the three sub-languages of SQL Understand the sample database used throughout the course


Internal

6969

Unit 2 Reading data from a table or view

70

Unit 2: Reading Data From A Table Or View

After completing this unit, you will be able to: • Write simple database queries using SQL‘s SELECT statement • Project columns in and out of queries using the SELECT clause • Avoid duplicates in SELECT statement result sets • Include columns based on conditions • Use built-in functions in column lists and WHERE clauses • Limit results sets to the first N rows • Ensure a specific order in SELECT statement result sets • Restrict the result set using the WHERE clause


Internal

7171

Reading Database Access

Which SQL statement can be used for reading data from a table or view?

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72


SELECT … FROM … WHERE …  It

is used to read from a table or view



The SELECT statement is the central construct for read access to data (database) with SQL



The SELECT statement can include the following (optional) clauses:

WHERE, GROUP BY, HAVING, ORDER BY

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73


The SELECT statement

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74

SELECT Statement

SELECT FROM WHERE GROUP BY HAVING ORDER BY

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Column, Column, COUNT(*) Table Condition Column, Column Group_Condition Column ASC, Column DESC;

75

SELECT

SELECT clause

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76

SELECT Clause

You can specify a single column in the projection list:

SELECT Name FROM Official;

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NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I

77

SELECT Clause

You can specify multiple columns in the projection list:

SELECT PNr, Name, Salary FROM Official;

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PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09


Salary -----A09 A10 A09 A12 A08 A09 A11 A13 A14

78

SELECT Clause

The sequence of the columns in the projection list is relevant:

SELECT PNr, Name FROM Official; PNR --P01 P02 P03 P04 …

©

NAME ---Mr A Mr B Ms C Ms D …


SELECT Name, PNr FROM Official; NAME ---Mr A Mr B Ms C Ms D …

PNR --P01 P02 P03 P04 …

79

SELECT Clause

The same column can appear several times in the projection list: We do not recommend using this option

SELECT PNr, PNr, PNr FROM Official;

©


PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

80

SELECT Clause

An asterisk (*) in the projection list represents "all columns“:

SELECT * FROM Official; PNR --P01 P02 P03 P04 …

©



OVERTIME -------10 10 20 ? …

SALARY -----A09 A10 A09 A12 …

MANAGER ------P04 P04 P04 P09 …

81

SELECT Clause Columns can be specified in the projection list in addition to the asterisk (*): We do not recommend using this option

SELECT PNr, *, Name FROM Official; PNR --P01 P02 P03 P04 …

©

PNR --P01 P02 P03 P04 …



OVERTIME -------10 10 20 ? …

SALARY -----A09 A10 A09 A12 …

MANAGER ------P04 P04 P04 P09 …


82

SELECT Clause

The asterisk (*) can be used multiple times in the projection list: We do not recommend using this option

SELECT *, *, * FROM Official; PNR

NAME

OVERTIME

SALARY

MANAGER

PNR

NAME

OVERTIME

SALARY

MANAGER

PNR

NAME

OVERTIME

SALARY

MANAGER

---

----

--------

------

-------

---

----

--------

------

-------

---

----

--------

------

-------

P01

Mr A

10

A09

P04

P01

Mr A

10

A09

P04

P01

Mr A

10

A09

P04

P02

Mr B

10

A10

P04

P02

Mr B

10

A10

P04

P02

Mr B

10

A10

P04

P03

Ms C

20

A09

P04

P03

Ms C

20

A09

P04

P03

Ms C

20

A09

P04

P04

Ms D

?

A12

P09

P04

Ms D

?

A12

P09

P04

Ms D

?

A12

P09

P05

Mr E

10

A08

P08

P05

Mr E

10

A08

P08

P05

Mr E

10

A08

P08

P06

Mr F

18

A09

P08

P06

Mr F

18

A09

P08

P06

Mr F

18

A09

P08

P07

Ms G

22

A11

P08

P07

Ms G

22

A11

P08

P07

Ms G

22

A11

P08

P08

Ms H

?

A13

P09

P08

Ms H

?

A13

P09

P08

Ms H

?

A13

P09

P09

Mr I

?

A14

?

P09

Mr I

?

A14

?

P09

Mr I

?

A14

?

©


83

SELECT Clause

You can generate additional “artificial” result columns:

SELECT 'Today', Name, 'has been assigned to salary group', Salary FROM Official; 'Today' ------Today Today Today Today Today …

©

NAME ---Mr A Mr B Ms C Ms D Mr E …


'has been assigned to salary group' ----------------------------------has been assigned to salary group has been assigned to salary group has been assigned to salary group has been assigned to salary group has been assigned to salary group …

SALARY -----A09 A10 A09 A12 A08 …

84

SELECT Clause

The additional artificial result column can have a numeric type:

SELECT 'The working week of', Name, 'amounts to', 40, 'hours.' FROM Official; 'The working week of' --------------------The working week of The working week of The working week of The working week of The working week of The working week of The working week of The working week of The working week of

©



'amounts to' -----------amounts to amounts to amounts to amounts to amounts to amounts to amounts to amounts to amounts to

40 -40 40 40 40 40 40 40 40 40

'hours'. -------hours. hours. hours. hours. hours. hours. hours. hours. hours.

85

SELECT Clause

The projection list may only contain artificial result columns:

SELECT 'Good Morning!' FROM Official;

©


'Good Morning!' --------------Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning! Good Morning!

86

SELECT Clause

If the projection list only contains artificial result columns, you can use the special table “Dummy” as reference:

SELECT 'Good Morning!' FROM Dummy;

'Good Morning!' --------------Good Morning!

The “DUMMY“ table contains one column and one row:

SELECT * FROM Dummy;

©


DUMMY ----X

87

SELECT Clause

SELECT a, 'b', "c", 1, '2', "3" FROM "4";

a

©

A … … …

'b' --b b b

c … … …

1 1 1 1

'2' --2 2 2

3 … … …

Existing column named “A“ (“A“ as capital letter)

'b'

Artificial result column with string “b“ as value in each row

"c"

Existing column named “c“ (“c“ as lower case letter)

1

Artificial result column with 1 as numeric value in each row

'2'

Artificial result column with string “2“ as value in each row

"3"

Existing column named “3“

"4"

Existing table named “4“


88

SELECT Clause

The projection list can have computed columns. If a NULL value is used in an arithmetic operation, the result is also a NULL value:

SELECT Name, Overtime * 60 FROM Official;

©



OVERTIME*60 ----------600 600 1200 ? 600 1080 1320 ? ? 89

SELECT Clause

0 * NULL results in NULL (and not 0):

SELECT Name, Overtime, Overtime * 0 FROM Official; NAME ---Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I ©


OVERTIME -------10 10 20 ? 10 18 22 ? ?

OVERTIME*0 ---------0 0 0 ? 0 0 0 ? ? 90

SELECT Clause

You can explicitly name computed result columns:

SELECT Name, Overtime * 60 AS Overminutes FROM Official; NAME ---Mr A Mr B Ms C Ms D …

©


OVERMINUTES ----------600 600 1200 ? …

91

SELECT Clause

Quotation marks are required, if the result column name should include lowercase letters:

SELECT Name, Overtime * 60 AS "Overminutes" FROM Official; NAME ---Mr A Mr B Ms C Ms D …

©


Overminutes ----------600 600 1200 ? …

92

SELECT Clause

Quotation marks are also required, if the result column name should contain blanks:

SELECT Name, Overtime * 60 AS "overtime minutes" FROM Official; NAME ---Mr A Mr B Ms C Ms D …

©


overtime minutes ---------------600 600 1200 ? …

93

SELECT Clause

You can also rename non-calculated result columns:

SELECT PNr AS PersNumber, Salary AS "Salary Group" FROM Official; PERSNUMBER ---------P01 P02 P03 P04 …

©


Salary Group -----------A09 A10 A09 A12 …

94

SELECT Clause

In the (re)naming of result columns the keyword "AS“ can be omitted:

SELECT PNr PersNumber, Salary "Salary Group" FROM Official; PERSNUMBER ---------P01 P02 P03 P04 …

©


Salary Group -----------A09 A10 A09 A12 …

95

SELECT Clause

You can use existing column names for (re)naming of result columns:

SELECT PNr Salary, Salary PNr FROM Official;

©


SALARY -----P01 P02 P03 P04 P05 P06 P07 P08 P09

PNR --A09 A10 A09 A12 A08 A09 A11 A13 A14

96

SELECT Clause

Multiple result columns can have the same name: We do not recommend using this option

SELECT PNr AS xyz, Name AS xyz, Salary AS xyz FROM Official; XYZ --P01 P02 P03 P04 …

©


XYZ ---Mr A Mr B Ms C Ms D …

XYZ --A09 A10 A09 A12 …

97

SELECT Clause

 

You can use functions in the projection list. The corresponding result columns can be named explicitly.

 Which owner is born in which year?

SELECT Name, YEAR(Birthday) AS "Year of Birth" FROM Owner;

©


NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z

Year of Birth ------------1934 1966 ? ? 1952 1957 ? 1986 1986 1986

98

SELECT Clause

 When were vehicle owners first allowed to drive a car in Germany?

SELECT Name, ADD_YEARS(Birthday, 18) AS "18th Birthday" FROM Owner; NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z

©


18th Birthday ------------1952-06-20 1984-05-11 ? ? 1970-04-21 1975-06-01 ? 2004-08-30 2004-02-10 2004-02-03

99

SELECT Clause

Function calls can be nested  On which day is the owner’s 18th birthday?

SELECT Name, DAYNAME ( ADD_YEARS (Birthday, ROUND(ABS(-18.2)) ) ) AS Weekday FROM Owner;

©



WEEKDAY ------FRIDAY FRIDAY ? ? TUESDAY SUNDAY ? MONDAY TUESDAY TUESDAY 100

Functions

Which functions are provided by SAP HANA?

©


101

Functions

SAP HANA provides following functions (selection):

©

Function

Explanation

YEAR(Date)

year

ADD_YEARS(Date, n)

n years later

DAYNAME(Date)

weekday (English)

CURRENT_DATE

current date

ABS(Number)

absolute value

ROUND(Number)

rounding

SQRT(Number)

square root

UPPER(String)

convert to upper case

SUBSTR(String, Start, Length)

cut out of a string (substring)

LENGTH(String)

length of a string


102

SELECT Clause

You can qualify the column name by adding the table name:

SELECT Official.Name FROM Official;

©



103

SELECT Clause

The qualification with the table name is also allowed, if the output table has to contain all columns:

SELECT Official.* FROM Official; PNR --P01 P02 P03 P04 …

©



OVERTIME -------10 10 20 ? …

SALARY -----A09 A10 A09 A12 …

MANAGER ------P04 P04 P04 P09 …

104

SELECT Clause

The projection list can include a mix of qualified and unqualified column names:

SELECT Name, Official.PNr FROM Official;

©



PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

105

Tuple Variables Or Table Aliases

Using Tuple Variables, also known as Table Aliases

©


106

Tuple Variables aka Table Aliases

You can use tuple variables in the projection list. Tuple variables are defined in the FROM clause and also referred to as table aliases in SAP HANA (and most other DBMSs).

SELECT o.Name, o.PNr FROM Official o; You can use the (optional) key word AS in the definition of a tuple variable:

SELECT o.Name, o.PNr FROM Official AS o;

©



PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

107


Despite having defined table aliases the projection list may contain unqualified column names:

SELECT Name, PNr FROM Official o; A mix of qualified and unqualified column names is possible:

SELECT o.Name, PNr FROM Official o;

©



PNR --P01 P02 P03 P04 P05 P06 P07 P08 P09

108


If a table alias is defined in the FROM clause, you are not allowed to use the corresponding table name for qualification of a column name:

SELECT Official.Name FROM Official o;

SELECT Official.Name FROM Official;

SELECT o.Name FROM Official o;

SELECT o.Name FROM Official;

SELECT Name FROM Official o;

SELECT Name FROM Official;

©


109

Case Differentiation

Can the projection list contain columns that are based on a case differentiation?

©


110


The projection list can contain columns that are based on a case differentiation These columns can be named explicitly:

SELECT *, CASE WHEN WHEN ELSE END AS FROM Car;

©


HP < 120 THEN 'low' HP >= 120 AND HP < 180 THEN 'medium' 'high' Rating

111

Case Differentiation CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20

©

PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 ? HD-Y 333 HD-MQ 2006 HD-VW 2012 ?


BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi

COLOR -----red black blue white black yellow blue black red black green red red white black green orange red black green

HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184

OWNER ----H06 H03 H03 H07 H03 H05 H03 H07 H02 H04 H02 H04 H07 H03 H03 ? H09 H03 H01 ?

RATING -----low medium high medium medium high low medium low medium high low medium medium medium medium high low medium high

112


If a case differentiation does not correspond to any of the listed cases, a NULL value is returned:

SELECT CarID, CASE Color WHEN 'red' WHEN 'green' WHEN 'yellow' WHEN 'blue' WHEN 'white' WHEN 'black' END AS Color FROM Car;

©


THEN THEN THEN THEN THEN THEN

'FF0000' '00FF00' 'FFFF00' '0000FF' 'FFFFFF' '000000'

CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20

COLOR -----FF0000 000000 0000FF FFFFFF 000000 FFFF00 0000FF 000000 FF0000 000000 00FF00 FF0000 FF0000 FFFFFF 000000 00FF00 ? FF0000 000000 00FF00

113


Both the THEN and the ELSE branch can contain references to table columns:

SELECT CarID, CASE WHEN PlateNumber IS NOT NULL THEN PlateNumber ELSE 'The Car is not registered!' END AS PlateNumber, CASE Brand WHEN 'Mercedes' THEN 'Mercedes-Benz' WHEN 'VW' THEN 'Volkswagen' ELSE Brand END AS Brand, Color, HP FROM Car;

©


114

Case Differentiation CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20

©

PLATENUMBER -------------------------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-VW 1999 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 The Car is not registered! HD-Y 333 HD-MQ 2006 HD-VW 2012 The Car is not registered!


BRAND ------------Fiat Volkswagen BMW Mercedes-Benz Mercedes-Benz Audi Audi Volkswagen Skoda BMW BMW Skoda Renault Mercedes-Benz Skoda Opel Audi Renault Volkswagen Audi


HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184

115

Duplicate Elimination

SELECT DISTINCT

©


116


A table with a primary key does not contain duplicates

SELECT * FROM Car; CARID ----… F04 F05 … F09 … F14 F15 …

©


PLATENUMBER ----------… HD-AL 1002 HD-MM 3206 … HD-UP 13 … HD-MM 1977 HD-MB 3030 …

BRAND -------… Mercedes Mercedes … Skoda … Mercedes Skoda …

COLOR ----… white black … red … white black …

HP --… 136 170 … 105 … 170 136 …

OWNER ----… H07 H03 … H02 … H03 H03 …

117


Duplicates can occur when a key column is not included in the projection list:

SELECT Brand FROM Car;

©



118


If duplicate elimination is not intended, you can specify the keyword ALL:

SELECT ALL Brand FROM Car;

©



119


The keyword

DISTINCT ensures that the result table contains no duplicates:

SELECT DISTINCT Brand FROM Car;

©


BRAND -------Fiat VW BMW Mercedes Audi Skoda Renault Opel

120


NULL values are treated in duplicate elimination as "normal" values The result table contains (at most) one row that consists entirely of NULL values:

SELECT DISTINCT Overtime FROM Official;

©


OVERTIME -------10 20 ? 18 22

121

Duplicate Elimination The "duplicate" term always refers to full result rows:

SELECT Brand, Color FROM Car;

Duplicates

©


BRAND -------… Mercedes Mercedes Skoda Skoda Mercedes Skoda …

COLOR ----… white black red red white black …

122


If a projection list contains multiple columns, DISTINCT always refers to the combination of all these columns:

SELECT DISTINCT Brand, Color FROM Car;

©


BRAND -------… Mercedes Mercedes Skoda Skoda …

COLOR ----… white black red black …

123


DISTINCT can also be used if the result table should contain all the columns If the (source) table has a (primary) key DISTINCT is not required: SELECT DISTINCT * FROM Car; CARID ----… F04 F05 … F09 … F14 F15 …

©


PLATENUMBER ----------… HD-AL 1002 HD-MM 3206 … HD-UP 13 … HD-MM 1977 HD-MB 3030 …


COLOR ----… white black … red … white black …

HP --… 136 170 … 105 … 170 136 …

OWNER ----… H07 H03 … H02 … H03 H03 …

124

Sorting

ORDER BY

©


125

Sorting

The result table can be sorted by a specific column:

SELECT Brand, Color FROM Car ORDER BY Brand;

©


BRAND -------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW

COLOR -----yellow blue orange green blue black green red white black white green red red red red black black black black 126

Sorting

A descending sorting is possible:

SELECT Brand, Color FROM Car ORDER BY Brand DESC;

©


BRAND -------VW VW VW Skoda Skoda Skoda Renault Renault Opel Mercedes Mercedes Mercedes Fiat BMW BMW BMW Audi Audi Audi Audi

COLOR -----black black black black red red red red green white black white red green black blue green orange blue yellow 127

Sorting

To sort ascending apply the optional keyword

SELECT Brand, Color FROM Car ORDER BY Brand ASC;

©


ASC:


COLOR -----yellow blue orange green blue black green red white black white green red red red red black black black black

128

Sorting

The sorting can be applied using a column that does not appear in the projection list:

SELECT Brand, Color FROM Car ORDER BY PlateNumber;

©


BRAND -------Opel Audi Renault Mercedes VW BMW Skoda Audi Mercedes BMW Mercedes Renault BMW Skoda Fiat Audi VW VW Skoda Audi

COLOR -----green green red white black blue black blue white green black red black red red yellow black black red orange 129

Sorting

You can sort using a combination of columns:

SELECT Brand, Color FROM Car ORDER BY Brand ASC, Color DESC;

©



COLOR -----yellow orange green blue green blue black red white white black green red red red red black black black black 130

Sorting

Instead of the column names in the ORDER BY clause, the column numbers (based on the projection list) can be used:

SELECT Brand, Color FROM Car ORDER BY 1 ASC, 2 DESC;

©




Sorting

If result columns are named explicitly, you can refer to the new name for sorting:

SELECT Brand AS Manufacturer, Color FROM Car ORDER BY Manufacturer ASC, Color DESC;

©


MANUFACTURER -----------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW


Sorting

If result columns are explicitly renamed, you can still reference the original name in the ORDER BY clause:

SELECT Brand AS Manufacturer, Color FROM Car ORDER BY Brand ASC, Color DESC;

©


MANUFACTURER -----------Audi Audi Audi Audi BMW BMW BMW Fiat Mercedes Mercedes Mercedes Opel Renault Renault Skoda Skoda Skoda VW VW VW


Sorting

You can sort based on calculated values.  Sorting criteria: How much is the power below 200 kW?

SELECT CarID, Brand, HP FROM Car ORDER BY 200 - HP / 1.36 ASC;

©



BRAND -------Audi BMW BMW Audi Audi Mercedes Mercedes VW BMW Mercedes Renault Skoda VW VW Opel Audi Skoda Skoda Renault Fiat

HP --260 184 184 184 184 170 170 160 140 136 136 136 125 120 120 116 105 105 90 75 134

Sorting

You can reference functions in the ORDER BY clause:

SELECT Name, Birthday FROM Owner ORDER BY YEAR(Birthday) DESC, Name ASC; NAME -----Mr X Mr Z Ms Y Ms U Ms W Mr V Ms T HDM AG IKEA SAP AG ©


BIRTHDAY ---------1986-08-30 1986-02-03 1986-02-10 1966-05-11 1957-06-01 1952-04-21 1934-06-20 ? ? ? 135

Top-n Clause

How many rows are in the result set?

©


136

Top-n Clause

You can determine how many rows should be included in the query result (maximum).  What are the ten most powerful vehicles (based on horse power)?

SELECT TOP 10 CarID, Brand, Color, HP FROM Car CARID BRAND ORDER BY 4 DESC; ----- -------F06 F20 F17 F11 F03 F14 F05 F08 F10 F15 ©


Audi Audi Audi BMW BMW Mercedes Mercedes VW BMW Skoda

COLOR -----yellow green orange green blue white black black black black

HP --260 184 184 184 184 170 170 160 140 136 137

Top-n Clause

You can also use the Top-n Clause if the result table should include all columns.  What are the 5 most powerful vehicles (based on horse power)?

SELECT TOP 5 * FROM Car ORDER BY HP DESC; CARID ----F06 F20 F17 F11 F03

©


PLATENUMBER ----------HD-VW 1999 ? HD-Y 333 HD-MM 208 HD-JA 1972

BRAND ----Audi Audi Audi BMW BMW

COLOR -----yellow green orange green blue

HP --260 184 184 184 184

OWNER ----H05 ? H09 H02 H03

138

Top-n Clause

You can combine the Top-n Clause with the keyword DISTINCT.  What are the 7 highest HP values?

SELECT TOP 7 DISTINCT HP FROM Car ORDER BY 1 DESC;

©


HP --260 184 170 160 140 136 125

139

Top-n Clause

No error is thrown if you request more rows than available. The result set will not be filled with additional, “artificial” rows.  570 different colors are requested:

SELECT TOP 570 DISTINCT Color FROM Car;

©


COLOR -----red black blue white yellow green orange

140

Top-n Clause

It is possible to request 0 result rows. In this case the result set does not contain any row.  Zero colors are requested:

SELECT TOP 0 Color FROM Car;

©


COLOR -----

141

Limit And Offset Clause

You can use the Limit clause as alternative to the Top-n clause. The Limit clause comes at the very end of the SELECT statement.  What are the 5 most powerful vehicles (based on horse power)?

SELECT * FROM Car ORDER BY HP DESC LIMIT 5; CARID ----F06 F20 F17 F11 F03

©


PLATENUMBER ----------HD-VW 1999 ? HD-Y 333 HD-MM 208 HD-JA 1972

BRAND ----Audi Audi Audi BMW BMW

COLOR -----yellow green orange green blue

HP --260 184 184 184 184

OWNER ----H05 ? H09 H02 H03

142

Limit And Offset Clause

The Limit clause can be combined with the Offset clause to skip records. This allows you to read result sets “page by page”.  What are the next 5 most powerful vehicles (based on horse power)?

SELECT * FROM Car ORDER BY HP DESC LIMIT 5 CARID OFFSET 5; ----F78 F77 F14 F05 F07

©


PLATENUMBER ----------HD-MT 2510 HD-MT 2509 HD-MM 1977 HD-MM 3206 HD-IK 1002

BRAND -------? ? Mercedes Mercedes VW

COLOR -----green red white black black

HP --260 184 184 184 184

OWNER ----? ? H03 H03 H07

143

WHERE Clause

Which rows are included in the result set?

©


144

WHERE Clause

The WHERE clause is used to filter rows. It is used to extract only those rows that fulfill a specified criterion.

SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'BMW'; PLATENUMBER ----------HD-JA 1972 HD-MT 507 HD-MM 208

©


BRAND ----BMW BMW BMW

COLOR ----blue black green

145

WHERE Clause

You can reference numeric columns in the WHERE clause.

SELECT HP, Brand, Color FROM Car WHERE HP >= 170;

©


HP --184 170 260 184 170 184 184

BRAND -------BMW Mercedes Audi BMW Mercedes Audi Audi

COLOR -----blue black yellow green white orange green

146

WHERE Clause

You can reference a column in the WHERE clause that is not included in the projection list.

SELECT Brand, Color FROM Car WHERE HP >= 170;

©



COLOR -----blue black yellow green white orange green

147

WHERE Clause

You can check for NULL in the WHERE clause:

SELECT CarID, Brand, Color FROM Car WHERE PlateNumber IS NULL; CARID ----F16 F20

©


BRAND ----Opel Audi

COLOR ----green green

148

WHERE Clause

You can check for IS in the WHERE clause:

NOT NULL

SELECT CarID, Brand, Color FROM Car WHERE PlateNumber IS NOT NULL;

©


CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F17 F18 F19

BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Audi Renault VW

COLOR -----red black blue white black yellow blue black red black green red red white black orange red black

149

WHERE Clause

You can check if values are included IN a value-list.

SELECT Brand, Color FROM Car WHERE Color IN ('red', 'blue', 'orange'); BRAND ------Fiat BMW Audi Skoda Skoda Renault Audi Renault ©


COLOR -----red blue blue red red red orange red 150

WHERE Clause

You can check if values are included in an interval.

SELECT PlateNumber, Brand, Color, HP FROM Car WHERE HP BETWEEN 140 AND 170; PLATENUMBER ----------HD-MM 3206 HD-IK 1002 HD-MT 507 HD-MM 1977

©


BRAND -------Mercedes VW BMW Mercedes

COLOR ----black black black white

HP --170 160 140 170

151

WHERE Clause

You can reference calculated values in the WHERE clause. We do not recommend this option for performance reasons.  Which cars have a power of at least 125 kW?

SELECT CarID, Brand, HP FROM Car WHERE HP / 1.36 >= 125;

©


CARID ----F03 F05 F06 F11 F14 F17 F20


HP --184 170 260 184 170 184 184

152

WHERE Clause

You can use functions in the WHERE clause.

SELECT * FROM Owner WHERE YEAR(Birthday) = 1986; OWNERID ------H08 H09 H10

©


NAME ---Mr X Ms Y Mr Z

BIRTHDAY ---------1986-08-30 1986-02-10 1986-02-03

CITY --------Walldorf Sinsheim Ladenburg

153

WHERE Clause

You can use search patterns in the WHERE clause.

SELECT PlateNumber, Brand, Color, HP FROM Car WHERE PlateNumber LIKE '%MM%'; PLATENUMBER ----------HD-MM 3206 HD-MM 208 HD-MM 1977

©


BRAND -------Mercedes BMW Mercedes

COLOR ----black green white

HP --170 184 170

154

LIKE Predicate The wildcard character % (percentage sign) represents any string containing no, one, or multiple characters. The wildcard character _ (underscore) represents any single character. WHERE MyColumn LIKE 'M%'

String starting with “M“

WHERE MyColumn LIKE 'M _'

String of two characters starting with “M“

WHERE MyColumn LIKE '%M'

String ending with “M“

WHERE MyColumn LIKE '%M%'

String containing “M“

WHERE MyColumn LIKE '_ _ _'

String with length 3

WHERE MyColumn LIKE '_ _ _ _T_M%'

“T“ on fifth and “M“ on seventh position

©


155

LIKE Predicate If you want to search for the percentage sign (%) or underscore (_) itself, you have to place an ESCAPE character in front. You can choose the ESCAPE character (with some restrictions). LIKE '$%%' ESCAPE '$'

String starting with a percentage sign

LIKE '$_ _' ESCAPE '$'

String of two characters starting with an underscore

LIKE '%$%' ESCAPE '$'

String ending with a percentage sign

LIKE '%$%%' ESCAPE '$'

String containing a percentage sign

LIKE '%$_$%%' ESCAPE '$'

String containing an underscore followed by a percentage sign

LIKE '_ _ _ _ $%_ $ _%' ESCAPE '$'

“%“ on fifth and “_“ on seventh position.

©


156

LIKE Predicate If you need to search for the escape character within a string, you need to put the escape character in front of itself to mask it. LIKE '$%%$$' ESCAPE '$'

String starting with a percentage sign and ending with $

LIKE '$_%$$%$_' ESCAPE '$'

String starting and ending with an underscore and containing the escape character $

LIKE '%$%$$%' ESCAPE '$'

String containing a percentage sign followed by $

LIKE '%$_%$$' ESCAPE '$'

String containing an underscore and ending with $

LIKE '_ _ _ _ $%_ $$%' ESCAPE '$'

“%“ on fifth and “$“ on seventh position

©


157

WHERE Clause

You can use compound WHERE clauses.

SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Skoda' AND Color = 'red'; PLATENUMBER ----------HD-UP 13 HD-XY 4711

©


BRAND ----Skoda Skoda

COLOR ----red red

158

WHERE Clause

You can reference the same column multiple times.

SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Skoda' OR Brand = 'BMW'; PLATENUMBER ----------HD-JA 1972 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-MB 3030

©


BRAND ----BMW Skoda BMW BMW Skoda Skoda

COLOR ----blue red black green red black

159

WHERE Clause

You can use brackets.

SELECT PlateNumber, Brand, Color FROM Car WHERE (Brand = 'Skoda' OR Brand = 'BMW') AND Color = 'black'; PLATENUMBER ----------HD-MT 507 HD-MB 3030

©


BRAND ----BMW Skoda

COLOR ----black black

160

WHERE Clause  The license plate started with "HD" for Heidelberg  I'm sure I saw an "M"  I clearly remember the digits "2" and "6"  The "2" was definitely left of the "6"  Between the "2" and the "6" was exactly one digit  It was neither Skoda, nor a VW  The car was blue or green CARID ----F07

PLATENUMBER ----------HD-ML 3206

BRAND ----Audi

COLOR ----blue

HP --116

SELECT * FROM Car WHERE PlateNumber LIKE 'HD-%M% %2_6%' AND Brand <> 'Skoda' AND Brand <> 'VW' AND (Color = 'blue' OR Color = 'green');

©


OWNER ----H03

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Operator Precedence

What is the precedence of the different operators?

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162

Operator Precedence

The operators have the precedence indicated by the table below: Precedence

Operator

Explanation

Highest

()

parentheses

‒

unary minus

*, /

multiplication, division

+, ‒

addition, subtraction

=, <, <=, >, >=, <>, IS NULL, LIKE, BETWEEN

Lowest

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comparison

NOT

logical negation

AND

conjunction

OR

disjunction

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Unit 2: Reading Data From A Table And View Summary

You should now be able to: • Write simple database queries using SQL‘s SELECT statement

• • • • • • •

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Project columns in and out of queries using the SELECT clause Avoid duplicates in SELECT statement result sets Include columns based on conditions Use built-in functions in column lists and WHERE clauses Limit results sets to the first N rows Ensure a specific order in SELECT statement result sets Restrict the result set using the WHERE clause


164

Unit 3 Aggregating Data

165

Unit 3: Aggregating Data Learning Objectives After completing this unit, you will be able to: • Determine aggregated values on table columns using a single SELECT statement • List the aggregate functions supported by HANA • Determine such aggregated values for groups of rows, using the GROUP BY clause • Filter such groups using the HAVING clause

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Aggregate Expressions

Calculations across multiple rows

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167


Count(*)



You can calculate the number of rows in the result set using COUNT .



Rows containing only NULL values are included.

 What is the quantity of cars with the brand “Audi“?

SELECT COUNT(*) FROM Car WHERE Brand = 'Audi';

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COUNT(*) -------4

168

Count(*)


You can explicitly name columns created by aggregate expressions.

 What is the quantity of cars with the brand “Audi“?

SELECT COUNT(*) AS "Quantity of Audi" FROM Car WHERE Brand = 'Audi'; Quantity of Audi ---------------4

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Count()

 

You can calculate the number of values within a single column. NULL values are not included.



You can only use a single column as parameter of COUNT.

 How many cars are registered to an owner?  This does not calculate the number of different owners!

SELECT COUNT(Owner) FROM Car;

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COUNT(OWNER) -----------18

170






Count(DISTINCT )

You can calculate the number of different NOT-NULL-values of a certain column. You can use only a single column as parameter for COUNT DISTINCT. NULL values are not included.

 How many different owners have a registered car?

SELECT COUNT(DISTINCT Owner) FROM Car; COUNT(DISTINCT OWNER) --------------------8

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SELECT * FROM T; SELECT S FROM T;

S ? ? ? X X

SELECT COUNT(*) FROM T;

COUNT(*) -------5

SELECT COUNT(S) FROM T;

COUNT(S) -------2

SELECT COUNT(DISTINCT *) FROM T; SELECT COUNT(DISTINCT S) FROM T;

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COUNT(DISTINCT S) ----------------1

172


MIN/MAX()

You can calculate the minimum or maximum value in a column.

 What is the horsepower range of the registered cars?

SELECT MIN(HP), MAX(HP) FROM Car;

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MIN(HP) ------75

MAX(HP) ------260

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You can combine aggregate expressions and “normal” functions.

In which year was the youngest owner born?

SELECT MAX(YEAR(Birthday)) AS Year FROM Owner;

YEAR ---1986

SELECT YEAR(MAX(Birthday)) AS Year FROM Owner;

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174


The sequence of nested functions can be relevant.

 What is the lowest or highest HP value?

SELECT ABS(MAX(0 - HP)) FROM Car;

SELECT MAX(ABS(0 - HP)) FROM Car;

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ABS(MAX(0-HP)) -------------75

MAX(ABS(0-HP)) -------------260

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The WHERE clause is included in the minimum and maximum calculation.

 What is the horsepower range of BMWs?

SELECT MIN(HP), MAX(HP) FROM Car WHERE Brand = 'BMW';

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MIN(HP) ------140

MAX(HP) ------184

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   

SUM()

You can sum up the values in a column. The WHERE clause is included in the summation. NULL values in the column are ignored. Individual NULL values contained in the column do not result in NULL for the sum. (as long as there is at least one numeric value)

 What is the sum of horsepower for all Audis?

SELECT SUM(HP) FROM Car WHERE Brand = 'Audi';

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SUM(HP) ------744

177


    

You can sum up the distinct values of a column. The WHERE clause is included in the summation. NULL values of the column are ignored. Individual NULL values contained in the column do not result in NULL for the sum. (as long as there is at least one numeric value) If a value occurs multiple times, it is still counted only once when summing

SELECT SUM(DISTINCT HP) FROM Car WHERE Brand = 'Audi';

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SUM(DISTINCT )


SUM(DISTINCT HP) ---------------560

178


   

AVG()

You can calculate the average of all values in a column. The WHERE clause is included in the calculation. NULL values in the column are completely ignored (in the numerator and denominator) Individual NULL values contained in the column do not result in NULL for the average.

 What is the average horsepower of Audi?

SELECT AVG(HP) FROM Car WHERE Brand = 'Audi';

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AVG(HP) ------186

179


    

AVG(DISTINCT )

You can calculate the average of all distinct values in a column. The WHERE clause is included in the calculation. NULL values in the column are completely ignored (in the numerator and denominator) Individual NULL values contained in the column do not result in NULL for the average. If a value occurs multiple times, it is still counted only once for the average calculation.

SELECT AVG(DISTINCT HP) FROM Car WHERE Brand = 'Audi'; AVG(DISTINCT HP) ----------------------------------186.6666666666666666666666666666667

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Which aggregate expressions are available in SAP HANA?

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181


SAP HANA provides the following aggregate expressions:

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Aggregate Name

Description

COUNT

Count

MIN

Minimum

MAX

Maximum

SUM

Sum

AVG

Average

STDDEV

Standard Deviation

VAR

Variance

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Group By

GROUP BY

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Group By

  

A table can be divided into (disjoint) groups of rows A group is represented in the query result by a single row Aggregate expressions will be evaluated separately for each group

 What is the number of cars per brand?

SELECT Brand, COUNT(*) FROM Car GROUP BY Brand;

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COUNT(*) -------1 3 3 3 4 3 2 1

184

Group By



What is the highest horse power value per brand?

SELECT Brand, MAX(HP) FROM Car GROUP BY Brand;

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MAX(HP) ------75 160 184 170 260 136 136 120

185

Group By

NULL values of the GROUP BY column are treated as “normal” values creating a single group.  How often a certain overtime value occurs?

SELECT Overtime, COUNT(*) AS Frequency FROM Official GROUP BY Overtime; OVERTIME FREQUENCY -------10 20 ? 18 22

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--------3 1 3 1 1

186

Group By

You can combine GROUP BY with ORDER BY for sorting.

SELECT FROM GROUP BY ORDER BY

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Brand, MAX(HP) Car Brand 2 DESC, 1 ASC;

BRAND -------Audi BMW Mercedes VW Renault Skoda Opel Fiat

MAX(HP) ------260 184 170 160 136 136 120 75

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Group By

You can use functions in the GROUP BY clause.



What is the number of owners per year of birth?

SELECT FROM GROUP BY ORDER BY

YEAR(Birthday), COUNT(*) Owner YEAR(Birthday) 2 DESC, 1 ASC; YEAR(BIRTHDAY) -------------? 1986 1934 1952 1957 1966

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COUNT(*) -------3 3 1 1 1 1

188

Group By

The WHERE clause is processed before the grouping.  What is the number of black cars per brand?  Only brands with black cars are included into the result set.

SELECT FROM WHERE GROUP BY

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Brand, COUNT(*) Car Color = 'black' Brand;


BRAND -------VW Mercedes BMW Skoda

COUNT(*) -------3 1 1 1

189

Group By

You can use a combination of columns in the GROUP BY clause.

SELECT Brand, Color, COUNT(*) FROM Car GROUP BY Brand, Color;

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BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi Skoda BMW BMW Renault Skoda Opel Audi Audi

COLOR -----red black blue white black yellow blue red black green red black green orange green

COUNT(*) -------1 3 1 2 1 1 1 2 1 1 2 1 1 1 1

190

Group By

You can explicitly rename result columns in combination with GROUP BY.

SELECT Brand AS Manufacturer, Color, COUNT(*) AS "# of cars" FROM Car GROUP BY Brand, Color; MANUFACTURER -----------Fiat VW BMW Mercedes Mercedes Audi Audi Skoda BMW BMW Renault …

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COLOR -----red black blue white black yellow blue red black green red …

# of cars --------1 3 1 2 1 1 1 2 1 1 2 …

191

HAVING Clause

What groups are included in the query result?

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192

HAVING

Using the HAVING clause you can specify which conditions a group must meet to be included in the result set.  Which combinations of brand and color occur at least twice?

SELECT FROM GROUP BY HAVING

Brand, Color, COUNT(*) Car Brand, Color COUNT(*) >= 2; BRAND -------VW Mercedes Skoda Renault

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COLOR ----black white red red

COUNT(*) -------3 2 2 2

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HAVING

The HAVING condition can reference columns not included in the projection list.  What is the number of brand-color combinations, where at least one car has less than 120 HP?  We want to analyze based on the quantity of all cars, not only the cars with less than 120 HP.

SELECT FROM GROUP BY HAVING

Brand, Color, COUNT(*) Car Brand, Color MIN(HP) < 120; BRAND ------Fiat Audi Skoda Renault

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COLOR ----red blue red red

COUNT(*) -------1 1 2 2

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SELECT statement

    

What is the number of cars per brand, which are black or red? Rename the Brand column to "Manufacturer" Display only those manufacturers with at least 2 cars Sort the result set first descending by number of cars Sort the result set second ascending by manufacturer.

SELECT FROM WHERE GROUP BY HAVING ORDER BY

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c.Brand AS "Manufacturer", COUNT(*) Car c Color IN ('black', 'red') Brand COUNT(*) >= 2 Manufacturer 2 DESC, Brand ASC;


-----------Skoda VW Renault

COUNT(*) -------3 3 2

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Unit 3: Aggregating Data Summary

You should now be able to: • Determine aggregated values on table columns using a single SELECT statement • List the aggregate functions supported by HANA • Determine such aggregated values for groups of rows, using the GROUP BY clause • Filter such groups using the HAVING clause

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196

Unit 4 Reading Data From Multiple Tables Part I

197

Unit 4: Reading Data From Multiple Tables Part I Learning Objectives After completing this unit, you will be able to: • Merge the result of several select statements using the UNION statement • • •

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Combine data from several tables when querying data using JOIN constructs List the various types of Joins Explain the differences between the various types of Joins, and apply the right type of Join depending on the problem


198

Access to multiple tables

How can I read data from multiple tables and views?

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199

Access to multiple tables

For read access to multiple database tables / views the following options are available:  Combination of results from several partial queries (UNION)  UNION ALL vs. UNION

 Combination of several tables (JOIN)  CROSS JOIN vs. INNER JOIN vs. OUTER JOIN  Implicit JOIN vs. Explicit JOIN

 Nested queries (Sub Query)  Uncorrelated Sub Query vs. Correlated Sub Query

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200

UNION [ALL]

Combination of results from several partial queries

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201

UNION [ALL]

SELECT Column, Column, Column FROM Table WHERE Condition UNION ALL SELECT Column, Column, Column FROM Table WHERE Condition;

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202

UNION [ALL]

You can combine the result tables of multiple queries using UNION [ALL].   

The individual result tables must have the same number of columns. The corresponding result columns must have compatible data types. The column names of the resulting output table are based on the first SELECT statement.

SELECT PNr, Name FROM Official WHERE Salary = 'A09' UNION ALL SELECT OwnerID, Name FROM Owner WHERE Birthday >= '1977-05-21';

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PNR --P01 P03 P06 H08 H09 H10

NAME ---Mr A Ms C Mr F Mr X Ms Y Mr Z

203

UNION [ALL]

 Almost the same statement as on the previous slide – but with a changed sequence of partial queries.  The column names of the resulting output table are based on the first SELECT statement.

SELECT OwnerID, Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';

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OWNERID ------H08 H09 H10 P01 P03 P06

NAME ---Mr X Ms Y Mr Z Mr A Ms C Mr F

204

UNION [ALL]

With UNION [ALL] you can also explicitly rename the result columns. SELECT OwnerID AS "Person ID", Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';

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Person ID --------H08 H09 H10 P01 P03 P06


205

UNION ALL

If the results of multiple partial queries overlap, the overall result includes duplicates with UNION ALL. SELECT PlateNumber, Brand, Color FROM Car WHERE Brand = 'Mercedes' UNION ALL SELECT PlateNumber, Brand, Color FROM Car WHERE Color = 'white';

Duplicates

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PLATENUMBER ----------HD-AL 1002 HD-MM 3206 HD-MM 1977 HD-AL 1002 HD-MM 1977

BRAND -------Mercedes Mercedes Mercedes Mercedes Mercedes

COLOR ----white black white white white

206

UNION

You use UNION instead of UNION ALL to eliminate duplicates.

SELECT FROM WHERE UNION SELECT FROM WHERE

PlateNumber, Brand, Color Car Brand = 'Mercedes' PlateNumber, Brand, Color Car Color = 'white'; PLATENUMBER ----------HD-AL 1002 HD-MM 3206 HD-MM 1977

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BRAND -------Mercedes Mercedes Mercedes

COLOR ----white black white 207

UNION [ALL]

You can use UNION [ALL] to combine result tables of multiple partial queries. SELECT FROM WHERE UNION SELECT FROM WHERE UNION SELECT FROM WHERE

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PlateNumber, Brand, Color Car Brand = 'BMW' PlateNumber, Brand, Color Car Color = 'yellow'

PlateNumber, Brand, Color Car Color = 'orange';


PLATENUMBER ----------HD-JA 1972 HD-MT 507 HD-MM 208 HD-VW 1999 HD-Y 333

BRAND ----BMW BMW BMW Audi Audi

COLOR -----blue black green yellow orange 208

Joining Tables

Joining tables

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209

Joining Tables

With JOIN semantics you can distinguish between:  CROSS JOIN  INNER JOIN  OUTER JOIN –

LEFT OUTER JOIN

–

RIGHT OUTER JOIN

–

FULL OUTER JOIN

With JOIN syntax you can distinguish between:  Implicit JOIN  Explicit JOIN

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210

Joining Tables

CROSS JOIN (Cartesian Product)

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211

CROSS JOIN

 

Each row of the left table is connected to each row of the right table. There is no CROSS JOIN join condition.

L

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… … … … … …

X 1 2 3 4 5


… … … … … …

… … … … … …

R

… … … … … …

Y 3 4 5 6 7

… … … … … …

… … … … … …

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CROSS JOIN

SELECT Column, Column, Column FROM Table, Table WHERE Condition; SELECT Column, Column, Column FROM Table CROSS JOIN Table WHERE Condition;

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213

Implicit CROSS JOIN

Each row of a table is connected to each row of the other table.  The result set contains 10 * 20 = 200 rows

SELECT * FROM Owner, Car; OWNERID ------H01 H02 H03 … H01 H02 H03 … H01 H02 H03 … H10

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NAME -----Ms T Ms U SAP AG … Ms T Ms U SAP AG … Ms T Ms U SAP AG … Mr Z

BIRTHDAY ---------1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1986-02-03

CITY ---------Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Ladenburg

CARID ----F01 F01 F01 … F02 F02 F02 … F03 F03 F03 … F20

PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-JA 1972 HD-JA 1972 HD-JA 1972 … ?

BRAND ----Fiat Fiat Fiat … VW VW VW … BMW BMW BMW … Audi

COLOR ----red red red … black black black … blue blue blue … green

HP --75 75 75 … 120 120 120 … 184 184 184 … 184

OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?

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Explicit CROSS JOIN

Each row of a table is connected to each row of the other table.  The result set contains 10 * 20 = 200 rows

SELECT * FROM Owner CROSS JOIN Car; OWNERID ------H01 H02 H03 … H01 H02 H03 … H01 H02 H03 … H10

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NAME -----Ms T Ms U SAP AG … Ms T Ms U SAP AG … Ms T Ms U SAP AG … Mr Z

BIRTHDAY ---------1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1934-06-20 1966-05-11 ? … 1986-02-03

CITY ---------Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Wiesloch Hockenheim Walldorf … Ladenburg





HP --75 75 75 … 120 120 120 … 184 184 184 … 184

OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?

215

CROSS JOIN

You can specify a WHERE clause for the implicit and explicit CROSS JOIN.

SELECT * FROM Owner, Car WHERE HP > 250; OWNERID ------H01 H02 H03 H04 H05 H06 H07 H08 H09 H10

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BIRTHDAY ---------1934-06-20 1966-05-11 ? ? 1952-04-21 1957-06-01 ? 1986-08-30 1986-02-10 1986-02-03

SELECT * FROM Owner CROSS JOIN Car WHERE HP > 250; CITY ---------Wiesloch Hockenheim Walldorf Heidelberg Leimen Wiesloch Walldorf Walldorf Sinsheim Ladenburg

CARID ----F06 F06 F06 F06 F06 F06 F06 F06 F06 F06

PLATENUMBER ----------HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999 HD-VW 1999

BRAND ----Audi Audi Audi Audi Audi Audi Audi Audi Audi Audi

COLOR -----yellow yellow yellow yellow yellow yellow yellow yellow yellow yellow

HP --260 260 260 260 260 260 260 260 260 260

OWNER ----H05 H05 H05 H05 H05 H05 H05 H05 H05 H05

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CROSS JOIN

You can join a table to itself. 

The result set contains 20 * 20 = 400 rows

SELECT * FROM Owner, Car WHERE HP > 250; CARID ----F01 F01 F01 … F02 F02 F02 … F03 F03 F03 … F20 ©





SELECT * FROM Owner CROSS JOIN Car WHERE HP > 250; HP --75 75 75 … 120 120 120 … 184 184 184 … 184

OWNER ----H06 H06 H06 … H03 H03 H03 … H03 H03 H03 … ?


PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 … HD-V 106 HD-VW 4711 HD-JA 1972 … HD-V 106 HD-VW 4711 HD-JA 1972 … ?

BRAND ----Fiat VW BMW … Fiat VW BMW … Fiat VW BMW … Audi

COLOR ----red black blue … red black blue … red black blue … green

HP --75 120 184 … 75 120 184 … 75 120 184 … 184

OWNER ----H06 H03 H03 … H06 H03 H03 … H06 H03 H03 … ? 217

CROSS JOIN

You can combine more than two tables.  The result set contains 10 * 20 * 3 = 600 rows

SELECT * FROM Owner CROSS JOIN Car CROSS JOIN Stolen;

SELECT * FROM Owner, Car, Stolen; OWNERID ------H01 H01 H01 … H01 H01 H01 … H02 H02 H02 … H10

©

NAME ---Ms T Ms T Ms T … Ms T Ms T Ms T … Ms U Ms U Ms U … Mr Z

BIRTHDAY ---------1934-06-20 1934-06-20 1934-06-20 … 1934-06-20 1934-06-20 1934-06-20 … 1966-05-11 1966-05-11 1966-05-11 … 1986-02-03


CITY ---------Wiesloch Wiesloch Wiesloch … Wiesloch Wiesloch Wiesloch … Hockenheim Hockenheim Hockenheim … Ladenburg


PLATENUMBER ----------HD-V 106 HD-V 106 HD-V 106 … HD-VW 4711 HD-VW 4711 HD-VW 4711 … HD-V 106 HD-V 106 HD-V 106 … ?

BRAND ----Fiat Fiat Fiat … VW VW VW … Fiat Fiat Fiat … Audi

COLOR ----red red red … black black black … red red red … green

HP --75 75 75 … 120 120 120 … 75 75 75 … 184

OWNER ----H06 H06 H06 … H03 H03 H03 … H06 H06 H06 … ?

PLATENUMBER ----------HD-VW 1999 HD-V 106 HD-Y 333 … HD-VW 1999 HD-V 106 HD-Y 333 … HD-VW 1999 HD-V 106 HD-Y 333 … HD-Y 333

REGISTERED_AT ------------2012-06-20 2012-06-01 2012-05-21 … 2012-06-20 2012-06-01 2012-05-21 … 2012-06-20 2012-06-01 2012-05-21 … 2012-05-21

218

Joining Tables

INNER JOIN

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219

INNER JOIN

One row of the left table and one row of the right table are always joined to a common result row - provided that the JOIN condition is fulfilled. JOIN Condition: L.X



L

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… … … … … …

X 1 2 3 4 5


… … … … … …

= R.Y

… … … … … …

R

… … … … … …

Y 3 4 5 6 7

… … … … … …

… … … … … …

220

INNER JOIN

SELECT Column, Column, Column FROM Table, Table WHERE JOIN_Condition AND Supplementary_Condition; SELECT Column, Column, Column FROM Table JOIN Table ON JOIN_Condition WHERE Supplementary_Condition;

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221

Implicit INNER JOIN One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled.  The JOIN condition is part of the WHERE clause  To whom which car is registered?

SELECT * FROM Owner, Car WHERE OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …

©

NAME -----Ms W SAP AG SAP AG IKEA SAP AG …

BIRTHDAY ---------1957-06-01 ? ? ? ? …


CITY -------Wiesloch Walldorf Walldorf Walldorf Walldorf …

CARID ----F01 F02 F03 F04 F05 …

PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 …

BRAND -------Fiat VW BMW Mercedes Mercedes …

COLOR ----red black blue white black …

HP --75 120 184 136 170 …

OWNER ----H06 H03 H03 H07 H03 …

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Explicit INNER JOIN 

One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled.  The JOIN condition is part of the JOIN operation (and not of the WHERE clause)  To whom is which car registered?

SELECT * FROM Owner INNER JOIN Car ON OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …

©


BIRTHDAY ---------1957-06-01 ? ? ? ? …



CARID ----F01 F02 F03 F04 F05 …




HP --75 120 184 136 170 …

OWNER ----H06 H03 H03 H07 H03 …

223

Explicit INNER JOIN  

You can simply write "JOIN“ instead of "INNER JOIN“. The INNER JOIN is the most important JOIN variant (and therefore the default)

 To whom is which car registered?

SELECT * FROM Owner JOIN Car ON OwnerID = Owner; OWNERID ------H06 H03 H03 H07 H03 …

©


BIRTHDAY ---------1957-06-01 ? ? ? ? …



CARID ----F01 F02 F03 F04 F05 …




HP --75 120 184 136 170 …

OWNER ----H06 H03 H03 H07 H03 …

224

Implicit INNER JOIN

You can specify certain columns in the projection list also for the implicit JOIN.  To whom is which car registered?

SELECT Name, Brand, Color FROM Owner, Car WHERE OwnerID = Owner;

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NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T



225

Explicit INNER JOIN

You can specify certain columns in the projection list also for the explicit JOIN.  To whom is which car registered?

SELECT Name, Brand, Color FROM Owner JOIN Car ON OwnerID = Owner;

©


NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T



226

Implicit INNER JOIN

You can also use table aliases or tuple variables in the projection list for the implicit JOIN.  

The JOIN condition can also refer to table aliases or tuple variables. If column names are not unique, you must qualify them NAME with the table aliases or tuple variables. -----Ms W  To whom is which car registered? SAP AG

SELECT o.Name, c.Brand, c.Color FROM Owner o, Car c WHERE o.OwnerID = c.Owner;

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SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T



227

Explicit INNER JOIN

You can use table aliases or tuple variables in the projection list for the explicit JOIN.  

The JOIN condition can also refer to table aliases or tuple variables. If column names are not unique, you must qualify them NAME with the table aliases or tuple variables. -----Ms W  To whom is which car registered? SAP AG

SELECT o.Name, c.Brand, c.Color FROM Owner o JOIN Car c ON o.OwnerID = c.Owner;

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SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG Ms Y SAP AG Ms T



228

Implicit INNER JOIN

You do not have to include a column from every involved table in the projection list. 

Besides the JOIN condition the WHERE clause can contain additional conditions.  To whom is a black car registered (at least one)?

SELECT DISTINCT o.Name FROM Owner o, Car c WHERE o.OwnerID = c.Owner AND c.Color = 'black';

©


NAME -----Ms T SAP AG HDM AG IKEA

229

Explicit INNER JOIN

You do not need to include a column from every involved table in the

projection list.  In addition to the JOIN condition you can specify a WHERE clause.  To whom is a black car registered (at least one)?

SELECT DISTINCT o.Name FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black';

©


NAME -----Ms T SAP AG HDM AG IKEA

230

Implicit INNER JOIN

You can explicitly rename result columns for the implicit JOIN.  Sorting the result table is also possible  To whom is a black car registered (at least one)?

SELECT DISTINCT o.Name AS "Owners' name" FROM Owner o, Car c WHERE o.OwnerID = c.Owner AND c.Color = 'black' ORDER BY o.Name; Owners'

name -----------HDM AG IKEA Ms T SAP AG

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231

Explicit INNER JOIN

You can explicitly rename result columns for the explicit JOIN.  Sorting the result table is also possible  To whom is a black car registered (at least one)?

SELECT DISTINCT o.Name AS "Owners' name" FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black' ORDER BY o.Name; Owners'

name -----------HDM AG IKEA Ms T SAP AG

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232

Implicit INNER JOIN

You can add a GROUP BY clause for the implicit Join.  How many black cars are registered to which owner?

SELECT FROM WHERE GROUP BY ORDER BY

o.Name, COUNT(*) AS "# black cars" Owner o, Car c o.OwnerID = c.Owner AND c.Color = 'black' o.OwnerID, o.Name 2 DESC, 1 ASC; NAME -----SAP AG HDM AG IKEA Ms T

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# black cars -----------3 1 1 1

233

Explicit INNER JOIN

You can add a GROUP BY clause for the explicit Join.  How many black cars are registered to which owner?

SELECT o.Name, COUNT(*) AS "# black cars" FROM Owner o JOIN Car c ON o.OwnerID = c.Owner WHERE c.Color = 'black' GROUP BY o.OwnerID, o.Name ORDER BY 2 DESC, 1 ASC; NAME # black -----SAP AG HDM AG IKEA Ms T

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cars -----------3 1 1 1

234

Implicit INNER JOIN

You can build the JOIN condition on multiple columns for the implicit JOIN.  To whom within the EU is a black car registered (at least one)?

SELECT DISTINCT o.Name AS "Owners' name" FROM Owner_EU o, Car_EU c WHERE o.Country = c.Country AND o.OwnerID = c.Owner AND c.Color = 'black' Owners' name ORDER BY o.Name; -----------HDM AG IKEA Ms O Ms T SAP AG Señora R

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235

Explicit INNER JOIN

You can build the JOIN condition on multiple columns for the explicit JOIN.  To whom within the EU is a black car registered (at least one)?

SELECT DISTINCT o.Name AS "Owners' name" FROM Owner_EU o JOIN Car_EU c ON o.Country = c.Country AND o.OwnerID = c.Owner WHERE c.Color = 'black' Owners' name ORDER BY o.Name; -----------HDM AG IKEA Ms O Ms T SAP AG Señora R

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Implicit INNER JOIN

You can join a table to itself.  Who is the manager of which employee?

SELECT e.Name AS Employee, m.Name AS Manager FROM Official e, Official m WHERE e.Manager = m.PNr; EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H

©


MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I

237

Explicit INNER JOIN

You can join a table to itself also for the explicit JOIN.  Who is the manager of which employee?

SELECT e.Name AS Employee, m.Name AS Manager FROM Official e JOIN Official m ON e.Manager = m.PNr; EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H

©


MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I

238

Implicit INNER JOIN

You can combine more than two tables.  To whom is which stolen car registered?

SELECT o.Name, c.Brand, c.Color, c.PlateNumber FROM Owner o, Car c, Stolen s WHERE o.OwnerID = c.Owner AND c.PlateNumber = s.PlateNumber; NAME ---Ms W Mr V Ms Y

©


BRAND ----Fiat Audi Audi

COLOR -----red yellow orange

PLATENUMBER ----------HD-V 106 HD-VW 1999 HD-Y 333

239

Explicit INNER JOIN

You can combine more than two tables also for the explicit JOIN.  To whom is which stolen car registered?

SELECT o.Name, c.Brand, c.Color, c.PlateNumber FROM Owner o JOIN Car c ON o.OwnerID = c.Owner JOIN Stolen s ON c.PlateNumber = s.PlateNumber;

NAME ---Ms W Mr V Ms Y

©





240

Explicit INNER JOIN

You can use different comparison operators other than EQUAL in the JOIN condition.  Which owner is older than which other owner?

SELECT o.Name AS "older", y.Name AS "younger" FROM Owner o JOIN Owner y ON o.Birthday < y.Birthday;

©


older ----Ms T Ms T Ms T Ms T Ms T Ms T Ms U Ms U Ms U Mr V Mr V Mr V Mr V Mr V Ms W Ms W Ms W Ms W Ms Y Mr Z Mr Z

younger ------Ms U Mr V Ms W Mr X Ms Y Mr Z Mr X Ms Y Mr Z Ms U Ms W Mr X Ms Y Mr Z Ms U Mr X Ms Y Mr Z Mr X Mr X Ms Y

241

Explicit INNER JOIN

You can use calculations in the JOIN condition.  Which car has at least three times the power than which other car?

SELECT m.*, l.* FROM Car m JOIN Car l ON m.HP >= l.HP * 3; CARID ----F06

©

PLATENUMBER ----------HD-VW 1999


BRAND ----Audi

COLOR -----yellow

HP --260

OWNER ----H05

CARID ----F01

PLATENUMBER ----------HD-V 106

BRAND ----Fiat

COLOR ----red

HP -75

OWNER ----H06

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Explicit INNER JOIN

You can use functions in the JOIN condition.  Which owner was born in the same year as which other owner?

SELECT o1.Name, o1.Birthday, o2.Name, o2.Birthday FROM Owner o1 JOIN Owner o2 ON YEAR(o1.Birthday) = YEAR(o2.Birthday) AND o1.OwnerID < o2.OwnerID;

NAME ---Mr X Mr X Ms Y

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BIRTHDAY ---------1986-08-30 1986-08-30 1986-02-10

NAME ---Ms Y Mr Z Mr Z

BIRTHDAY ---------1986-02-10 1986-02-03 1986-02-03

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Joining Tables

OUTER JOIN

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244

OUTER JOIN

The OUTER JOIN has three sub types:  LEFT OUTER JOIN  RIGHT OUTER JOIN  FULL OUTER JOIN

 For all three sub types of the OUTER JOIN SAP HANA only provides the explicit syntax variant.

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LEFT OUTER JOIN

 

One row of a table and one row of another table are always connected to a common result row - provided the JOIN condition is fulfilled. In addition, rows of the left table without matching row in the right table are copied to the query result. The missing values (from the right table) are filled with NULL values.

L

©

… … … … … …

X 1 2 3 4 5


… … … … … …

… … … … … …

NULL NULL NULL NULL

R

… … … … … …

Y 3 4 5 6 7

… … … … … …

… … … … … … 246

RIGHT OUTER JOIN

 

One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. In addition, rows of the right table without matching row in the left table are copied to the query result. The missing values (from the left table) are filled with NULL values.

L

… … … … … …

X 1 2 3 4 5

… … … … … …

… … … … … …

NULL NULL NULL NULL

©


R

… … … … … …

Y 3 4 5 6 7

… … … … … …

… … … … … … 247

FULL OUTER JOIN

 

One row of a table and one row of another table are always connected to a common result row - provided that the JOIN condition is fulfilled. In addition, rows of both tables without matching records are copied to the query result. The missing values (from the other table) are filled with NULL values.

L

… … … … … …

X 1 2 3 4 5

… … … … … …

… … … … … …

NULL NULL NULL NULL

©


NULL NULL NULL NULL

R

… … … … … …

Y 3 4 5 6 7

… … … … … …

… … … … … … 248

OUTER JOIN

SELECT Column, Column, Column FROM Table LEFT OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition; SELECT Column, Column, Column FROM Table RIGHT OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition; SELECT Column, Column, Column FROM Table FULL OUTER JOIN Table ON JOIN_Condition WHERE Additional_Condition;

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LEFT OUTER JOIN

 Which car is registered to which individual or company?  Individuals and companies that currently do not have a car registered should be included.  Cars without a registration should not be included in the result.

SELECT Name, CarID, Brand FROM Owner LEFT OUTER JOIN Car ON OwnerID = Owner;

©


NAME -----Ms T Ms U Ms U SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG HDM AG HDM AG Mr V Ms W IKEA IKEA IKEA Mr X Ms Y Mr Z

CARID ----F19 F09 F11 F02 F03 F18 F05 F15 F07 F14 F12 F10 F06 F01 F13 F08 F04 ? F17 ?

BRAND -------VW Skoda BMW VW BMW Renault Mercedes Skoda Audi Mercedes Skoda BMW Audi Fiat Renault VW Mercedes ? Audi ?

250

RIGHT OUTER JOIN

 Which car is registered to which individual or company?  Cars without a registration should be included.  Individuals and companies that currently do not have a car registered should not be included in the result.

SELECT Name, CarID, Brand FROM Owner RIGHT OUTER JOIN Car ON OwnerID = Owner;

©


NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ?



251

FULL OUTER JOIN

 Which car is registered to which individual or company?  Individuals and companies that currently do not have a car registered should be included.  Cars without a registration should be included in the result .

SELECT Name, CarID, Brand FROM Owner FULL OUTER JOIN Car ON OwnerID = Owner;

©


NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ? Mr X Mr Z

CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 ? ?

BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi ? ?

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LEFT OUTER JOIN

For the LEFT OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition.  Which car is registered to which individual?  Assume that companies don’t have a birthday.  Individuals that currently do not have a car registered should be included.  Cars without a registration should not be included in the result.

SELECT Name, CarID, Brand FROM Owner LEFT OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;

©


NAME -----Ms T Ms U Ms U SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG SAP AG HDM AG HDM AG Mr V Ms W IKEA IKEA IKEA Mr X Ms Y Mr Z

CARID ----F19 F09 F11 F02 F03 F18 F05 F15 F07 F14 F12 F10 F06 F01 F13 F08 F04 ? F17 ?

BRAND -------VW Skoda BMW VW BMW Renault Mercedes Skoda Audi Mercedes Skoda BMW Audi Fiat Renault VW Mercedes ? Audi ?

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RIGHT OUTER JOIN

For the RIGHT OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition.  Which car is registered to which individual?  Individuals who currently do not have a car registered should be excluded. (Realized by RIGHT OUTER JOIN)  Cars without a registration should not be included in the result. (Realized by the WHERE clause)

SELECT Name, CarID, Brand FROM Owner RIGHT OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;

©


NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ?



254

FULL OUTER JOIN

For the FULL OUTER JOIN you can also add a WHERE clause in addition to the JOIN condition.  Which car is registered to which individual?  Individuals that currently do not have a car registered should be included.  Cars without a registration should not be included in the result. (Realized by the WHERE clause)

SELECT Name, CarID, Brand FROM Owner FULL OUTER JOIN Car ON OwnerID = Owner WHERE Birthday IS NOT NULL;

©


NAME -----Ms W SAP AG SAP AG IKEA SAP AG Mr V SAP AG IKEA Ms U HDM AG Ms U HDM AG IKEA SAP AG SAP AG ? Ms Y SAP AG Ms T ? Mr X Mr Z

CARID ----F01 F02 F03 F04 F05 F06 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 ? ?

BRAND -------Fiat VW BMW Mercedes Mercedes Audi Audi VW Skoda BMW BMW Skoda Renault Mercedes Skoda Opel Audi Renault VW Audi ? ?

255

LEFT OUTER JOIN

You can LEFT OUTER JOIN a table with itself  Who is the manager of which employee?  Employees without a manager should be included.  Only Managers with employees should be included.

EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I

MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I ?

SELECT e.Name AS Employee, m.Name AS Manager FROM Official e LEFT OUTER JOIN Official m ON e.Manager = m.PNr;

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256

RIGHT OUTER JOIN

You can RIGHT OUTER JOIN a table with itself  Who is the manager of which employee?  The MANAGER column should also include officials that are not managing employees.  The EMPLOYEE column should only include officials with a manager assigned.

SELECT e.Name AS Employee, m.Name AS Manager FROM Official e RIGHT OUTER JOIN Official m ON e.Manager = m.PNr;

©


EMPLOYEE -------? ? ? Mr A Mr B Ms C ? ? ? Mr E Mr F Ms G Ms D Ms H

MANAGER ------Mr A Mr B Ms C Ms D Ms D Ms D Mr E Mr F Ms G Ms H Ms H Ms H Mr I Mr I

257

FULL OUTER JOIN

You can FULL OUTER JOIN a table with itself  Who is the manager of which employee?  The EMPLOYEE column should include officials without a manager.  The MANAGER column should also include officials that are not managing employees.

SELECT e.Name AS Employee, m.Name AS Manager FROM Official e FULL OUTER JOIN Official m ON e.Manager = m.PNr;

©


EMPLOYEE -------Mr A Mr B Ms C Ms D Mr E Mr F Ms G Ms H Mr I ? ? ? ? ? ?

MANAGER ------Ms D Ms D Ms D Mr I Ms H Ms H Ms H Mr I ? Mr A Mr B Ms C Mr E Mr F Ms G

258

LEFT/RIGHT/FULL OUTER JOIN You can also …  Project to certain columns  Reference table aliases or tuple variables in the projection list.  Use table aliases or tuple variables in the JOIN condition.  Explicitly rename result columns.  Sort the query result.  Add a GROUP BY clause.  Use aggregate expressions.  eliminate duplicates by using DISTINCT.  Specify a JOIN condition on multiple columns.  Use any comparison operator in the JOIN condition.  JOIN a table with itself.  JOIN more than two tables. ©


and:  You have to qualify column names with the table alias or with a tuple variable if the column names are not unique.  You do not need to include a column from every table involved in the JOIN into the projection list.

259

Unit 4: Reading Data From Multiple Tables Part I Summary

You should now be able to: •

Merge the result of several select statements using the UNION statement • Combine data from several tables when querying data using JOIN constructs • •

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List the various types of Joins Explain the differences between the various types of Joins, and apply the right type of Join depending on the problem


260

Unit 5 Reading Data From Multiple Tables Part II

261

Unit 5: Reading Data From Multiple Tables Part II Learning Objectives After completing this unit, you will be able to: • Use sub-queries to query data from multiple tables in a single select statement

• Explain the difference between uncorrelated and correlated subqueries

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262

Nested Queries

Nested Queries

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263

Nested Queries

 In addition to UNION and JOIN, Nested Queries also provide the option of reading from multiple tables and views.  Nested Queries contain a so called “Sub Query”.

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264

Sub Query

What is a sub query?

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265

Sub Query



A sub query is a query (SELECT-FROM-WHERE statement) that is used in another query (SELECT-FROM-WHERE statement). The "other Query" – containing the sub query - is called the "outer query". In this context the Sub Query is also referred to as "inner query".

 

Outer Query

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SELECT PlateNumber, Brand, Color Sub Query FROM Car WHERE Owner IN ( SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');

266

Sub Query

When is a sub query useful?

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267

Sub Query

A sub query is useful if ...  It makes the SELECT statement more "readable" - for example, because the use of nested queries is the "obvious way"  The SELECT statement has a better performance – however, with a "perfect" optimizer, this should not be the case.  The formulation of the SELECT statement is not possible (or only extremely cumbersome) without a sub query - which can be particularly the case for aggregate expressions.

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268

Sub Query

What is the difference between an uncorrelated and a correlated sub query?

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269

Sub Query

An uncorrelated sub query makes no reference to the outer query.

SELECT * FROM Car WHERE Owner IN (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch'); A correlated sub query refers to the outer query. SELECT * FROM Car c WHERE EXISTS (SELECT * FROM Owner o WHERE o.OwnerID = c.Owner AND o.City = 'Wiesloch');

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270

Sub Query

Uncorrelated sub queries

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271

Uncorrelated Sub Query

SELECT Column, Column, Column FROM Table WHERE Column IN (SELECT Column FROM Table WHERE Condition);

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272


You can use IN, if the outer value should be included in the result set of the sub query.  Which cars are registered to an owner from Wiesloch?

SELECT * FROM Car WHERE Owner IN (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');

CARID ----F01 F19

©


PLATENUMBER ----------HD-V 106 HD-VW 2012

BRAND ----Fiat VW

COLOR ----red black

HP --75 125

OWNER ----H06 H01

273


You can use =

ANY, if the outer value should match any result value of the sub query.

 Which cars are registered to an owner from Wiesloch?

SELECT * FROM Car WHERE Owner = ANY (SELECT OwnerID FROM Owner WHERE City = 'Wiesloch');

CARID ----F01 F19

©



BRAND ----Fiat VW

COLOR ----red black

HP --75 125

OWNER ----H06 H01

274


“= ANY” is equivalent to “IN”. So, where is the value of “= ANY” ?

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275


= ANY is equivalent to IN, but you can use ANY with other comparison operators: = ANY

The external value matches any value of the sub query.

< ANY

The external value is less than any value of the sub query.

<= ANY > ANY

©

The external value is less or equal than any value of the sub query. The external value is greater than any value of the sub query.

>= ANY

The external value is greater or equal than any value of the sub query.

<> ANY

The external value is different to any value of the sub query.


276


You can use > ANY, if the external value should be greater than any value of the sub query. 

Which officials have more than the minimum overtime hours?

SELECT Name, Overtime FROM Official WHERE Overtime > ANY (SELECT Overtime FROM Official); NAME ---Ms C Mr F Ms G

©


OVERTIME -------20 18 22

277


You can use > instead of > ANY, if the sub query results in only a single value. 

Which officials have more than the minimum overtime hours?

SELECT Name, Overtime FROM Official WHERE Overtime > (SELECT MIN(Overtime) FROM Official); NAME ---Ms C Mr F Ms G

©


OVERTIME -------20 18 22

278


You can use < ALL, if the outer value should be less than all result values of the sub query (other comparison operators are possible) = ALL

The external value matches each result value of the sub query. (This option is quite useful, as the sub query result can contain duplicates.)

< ALL

The external value is less than all values of the sub query.

<= ALL > ALL

©

The external value is less or equal than all values of the sub query. The external value is greater than all values of the sub query.

>= ALL

The external value is greater or equal than all values of the sub query.

<> ALL

The external value is different to all values of the sub query.


279


You can use <= ALL, if the outer value should be less or equal than all result values of the sub query (other comparison operators are possible)  Which officials have the fewest overtime hours?

SELECT Name, Overtime FROM Official WHERE Overtime <= ALL (SELECT Overtime FROM Official WHERE Overtime IS NOT NULL); NAME ---Mr A Mr B Mr E ©

OVERTIME -------10 10 10


280


You can use

<=

instead of

<= ALL, if the sub query results in only a single value.

 Which officials have the fewest overtime hours?  For the example below = would be possible instead of <= as well.

SELECT Name, Overtime FROM Official WHERE Overtime <= (SELECT MIN(Overtime) FROM Official); NAME ---Mr A Mr B Mr E

©

OVERTIME -------10 10 10


281


You can use functions in a sub query.

 Which owners were born in the same year as the youngest owner? SELECT * FROM Owner WHERE YEAR(Birthday) = (SELECT MAX(YEAR(Birthday)) FROM Owner);

OWNERID ------H08 H09 H10 ©


NAME ---Mr X Ms Y Mr Z

BIRTHDAY ---------1986-08-30 1986-02-10 1986-02-03

CITY --------Walldorf Sinsheim Ladenburg 282


 To whom is (at least) one blue car registered?

SELECT Name FROM Owner WHERE OwnerID IN (SELECT Owner FROM Car WHERE Color = 'blue');

NAME -----SAP AG

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283


You can use multiple columns in the sub query.  To whom within the EU is (at least) one blue car registered?

SELECT Name FROM Owner_EU WHERE (Country, OwnerID) IN (SELECT Country, Owner FROM Car_EU WHERE Color = 'blue');

NAME ------SAP AG Señor Q

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284


You can use the same table in both the FROM clause of the outer query and in the FROM clause of the sub query.  What is the name of Mr A’s manager?

SELECT Name FROM Official WHERE PNr IN (SELECT Manager FROM Official WHERE Name = 'Mr A');

NAME ---Ms D

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285


The WHERE clause of a sub query can contain another sub query.  Who are the owners of stolen cars?

SELECT Name FROM Owner WHERE OwnerID IN ( SELECT Owner FROM Car WHERE PlateNumber IN ( SELECT PlateNumber FROM Stolen ) ); NAME ---Mr V Ms W Ms Y ©


286


You can combine JOIN and Sub Query. To whom is the most powerful car per brand registered?

SELECT DISTINCT Brand, Name FROM Owner RIGHT OUTER JOIN Car ON OwnerID = Owner WHERE (Brand, HP) IN ( SELECT Brand, MAX(HP) FROM Car GROUP BY Brand ) ORDER BY Brand ASC, Name ASC;

©


BRAND -------Audi BMW BMW Fiat Mercedes Opel Renault Skoda VW

NAME -----Mr V Ms U SAP AG Ms W SAP AG ? IKEA SAP AG IKEA

287


 

A sub query is usually part of the WHERE clause. But you can also use a sub query in the FROM clause.

 Which blue cars have less than 120 HP?

SELECT * FROM (SELECT FROM WHERE WHERE Power <

PlateNumber, HP AS Power Car Color = 'blue') 120; PLATENUMBER ----------HD-ML 3206

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POWER ----116

288


 

A sub query is usually part of the WHERE clause. But you can also use a sub query in the SELECT clause.

 What is the HP deviation of each car when compared to the most powerful yellow car?

SELECT CarID, Brand, (SELECT MAX(HP) FROM Car WHERE Color = 'yellow') - HP AS Deviation FROM Car ORDER BY 3 DESC; CARID ----F01 F18 F09 … F06

©


BRAND ------Fiat Renault Skoda … Audi

DEVIATION --------185 170 155 … 0

289

Sub Query

Correlated sub queries

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290

Correlated Sub Query

SELECT Column, Column, Column FROM Table Tuple-Variable WHERE EXISTS (SELECT * FROM Table WHERE Condition);

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291


 

A correlated sub query refers to the outer query. Using EXISTS you can check that the query result of the sub query is not empty.

 Which vehicles are registered to an owner from Wiesloch? SELECT * FROM Car c WHERE EXISTS (SELECT * FROM Owner o WHERE o.City = 'Wiesloch' AND o.OwnerID = c.Owner);

CARID ----F01 F19

©



BRAND ----Fiat VW

COLOR ----red black

HP --75 125

OWNER ----H06 H01

292


Even with a correlated sub query, you can omit tuple variables if the column names are unique.  What vehicles are registered to an owner from Wiesloch? SELECT * FROM Car WHERE EXISTS (SELECT * FROM Owner WHERE City = 'Wiesloch' AND OwnerID = Owner);

CARID ----F01 F19

©



BRAND ----Fiat VW

COLOR ----red black

HP --75 125

OWNER ----H06 H01

293


Using NOT EXISTS you can check if the result of the sub query is empty.  Which cars are not reported as stolen?

SELECT CarID FROM Car c WHERE NOT EXISTS (SELECT * FROM Stolen s WHERE s.PlateNumber = c.PlateNumber);

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CARID ----F02 F03 F04 F05 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F18 F19 F20

294


You can use the same table in both the FROM clause of the outer query and the FROM clause of the correlated sub query.  Which vehicle has the most HP?

SELECT * FROM Car c1 WHERE NOT EXISTS (SELECT * FROM Car c2 WHERE c2.HP > c1.HP);

CARID ----F06

©


PLATENUMBER ----------HD-VW 1999

BRAND ----Audi

COLOR -----yellow

HP --260

OWNER ----H05 295


Even when using a correlated sub query, you can for example use ">= ALL"  How much horsepower has the most powerful car of each brand?

SELECT DISTINCT Brand, HP FROM Car c1 WHERE c1.HP >= ALL (SELECT c2.HP FROM Car c2 WHERE c2.Brand = c1.Brand);

©


BRAND -------Fiat BMW Mercedes Audi VW Renault Skoda Opel

HP --75 184 170 260 160 136 136 120

296


The WHERE clause of a correlated sub query can contain an additional correlated sub query.  To whom (at least) one stolen car is registered? SELECT o.Name FROM Owner o WHERE EXISTS ( SELECT FROM WHERE AND

NAME ---Mr V Ms W Ms Y

* Car c c.Owner = o.OwnerID EXISTS ( SELECT * FROM Stolen s WHERE s.PlateNumber = c.PlateNumber )

);

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297


You can also use a correlated sub query in the SELECT clause.  How much horsepower each car is lacking compared to the most powerful car of the same brand?

SELECT CarID, Brand, (SELECT MAX(HP) FROM Car c2 WHERE c2.Brand = c1.Brand) - HP AS Difference FROM Car c1; CARID BRAND DIFFERENCE ----F01 F02 F03 F04 F05 F06 F07 … F20 ©


-------Fiat VW BMW Mercedes Mercedes Audi Audi … Audi

---------0 40 0 34 0 0 144 … 76 298

Sub Query

Using Nested Queries you also can …        

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Restrict the projection list to certain columns Explicitly rename the result columns Sort the query result Use grouping Include aggregate expressions Eliminate duplicates using DISTINCT Involve the same table multiple times Involve more than two tables


299

Unit 5: Reading Data From Multiple Tables Part II Summary


Merge the result of several select statements using the UNION statement • Combine data from several tables when querying data using JOIN constructs • •

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List the various types of Joins Explain the differences between the various types of Joins, and apply the right type of Join depending on the problem


300

Unit 6 Understanding NULL values

301

Unit 6: Understanding NULL Values Learning Objectives After completing this unit, you will be able to: • Explain how to interpret NULL values in databases • Understand why the presences of NULL values can lead to unexpected query results • Analyze if two seemingly equivalent SQL statements return different results because of NULL values

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302

NULL Values

What is a NULL value?

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303

NULL Values

A NULL value signals that the corresponding value is unknown or does not exist.  The NULL value is a special value reserved by the DBMS. – For numeric data types the NULL value is not equal to 0. – For string based data types the NULL value is not equal to an empty string or to the space character.

A NULL value is used, … – If the corresponding value exists in principle, but is (temporarily or permanently) unknown (such as a birthday of a person) – If the corresponding value does not exist (for example the owner of an unregistered car).

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304

NULL Values

How are NULL values created?

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305

NULL Values

NULL values are “created” when reading rows (SELECT statement)  If the source table already contains NULL values  When using outer joins NULL values are created in a table when inserting rows (INSERT statement) if  No value for a column is provided, NULL values are allowed and no default value is defined  A NULL value is explicitly specified for a column and NULL values are allowed NULL values are created in a table when changing rows (UPDATE statement) if  A NULL value is explicitly specified for a column and NULL values are allowed NULL values are created in a table when when adding a column (ALTER TABLE statement), if  The table has at least one row, NULL values are allowed for the added column and no default value is defined for the added column.

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306

NULL Values

NULL values enforce a trivalent logic

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NULL Values

 By allowing NULL values a third truth value is necessary.  In addition to "true" and “false" there is also the truth value "unknown“ in SQL .  Any comparison with a NULL value results in the truth value "unknown“. – This applies to all comparison operators: =, >, >=, <, <=, <>

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A

A=7

A <> 7

A >= 6

5

false

true

false

7

true

false

true

NULL

unknown

unknown

unknown


308

Trivalent Logic: NOT

©


X

NOT X

true

false

false

true

unknown

unknown

309

Trivalent Logic: AND

©

X

Y

X AND Y

true

true

true

true

false

false

true

unknown

unknown

false

true

false

false

false

false

false

unknown

false

unknown

true

unknown

unknown

false

false

unknown

unknown

unknown


310

Trivalent Logic: OR

©

X

Y

X OR Y

true

true

true

true

false

true

true

unknown

true

false

true

true

false

false

false

false

unknown

unknown

unknown

true

true

unknown

false

unknown

unknown

unknown

unknown


311

Trivalent Logic

The trivalent logic has only one more truth value, so where is the problem?

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312

Trivalent Logic 39 = 19,683 binary logical operators are conceivable

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X

Y

X op Y

true

true

true / false / unknown

true

false


true

unknown


false

true


false

false


false

unknown


unknown

true


unknown

false


unknown

unknown



313

NULLs & Trivalent Logic

Many "obviously true" statements are no longer necessarily true!!!

 A=A  (A < 5) OR ( A >= 5)  (A = ‘Walldorf’) OR (A <> ‘Walldorf’)

 A+1>A  A * A >= 0  0*A=0  2*A=A+A  MAXIMUM >= ALL ( set of all values)  MINIMUM <= ALL ( set of all values)  A OR (NOT A)

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314

NULL Values

 Do the following two SQL queries have the same result? SELECT Name, Overtime FROM Official WHERE Overtime <= 10 OR Overtime > 10; SELECT Name, Overtime FROM Official;

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315

NULL Values

 The two SQL queries have different results!

 Only officials where Overtime IS NOT NULL applies: SELECT Name, Overtime FROM Official WHERE Overtime <= 10 OR Overtime > 10;  All officials (including those with: Overtime IS NULL ) SELECT Name, Overtime FROM Official;

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NAME ---Mr A Mr B Ms C Mr E Mr F Ms G

OVERTIME -------10 10 20 10 18 22


OVERTIME -------10 10 20 ? 10 18 22 ? ?

316

NULL Values

 Do the following two SQL queries have the same result? SELECT * FROM Official WHERE PNr <> ALL (SELECT PNr FROM Official WHERE Overtime = Overtime); SELECT * FROM Official WHERE Overtime IS NULL;

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317

NULL Values

 The two SQL queries have the same result! SELECT * FROM Official WHERE PNr <> ALL (SELECT PNr FROM Official WHERE Overtime = Overtime);

SELECT * FROM Official WHERE Overtime IS NULL;

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PNR --P04 P08 P09

NAME ---Ms D Ms H Mr I

OVERTIME -------? ? ?

SALARY -----A12 A13 A14

MANAGER ------P09 P09 ?

318

NULL Values

 The sub query returns all personnel numbers of the officials, where the number of overtime hours is known. SELECT * FROM Official WHERE PNr <> ALL (SELECT PNr FROM Official WHERE Overtime = Overtime);

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PNR --P01 P02 P03 P05 P06 P07

319

NULL Values

 The outer SELECT query returns all officials, whose PersNumber differs from all PersNumbers of the sub query.  So the outer SELECT query returns all officials, whose PersNumber differs from the PersNumbers of those officials whose overtime is known.  So the outer SELECT query returns all officials whose overtime is unknown.

SELECT * FROM Official WHERE PNr <> ALL (SELECT PNr FROM Official WHERE Overtime = Overtime); PNR --P04 P08 P09

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NAME ---Ms D Ms H Mr I

OVERTIME -------? ? ?

SALARY -----A12 A13 A14

MANAGER ------P09 P09 ? 320

NULL Values

 Do the following two SQL queries have the same result? SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official); SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

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321

NULL Values

 The two SQL queries have different results! SELECT Name, Overtime The FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official); SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

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query result is empty.

NAME ---Ms G

OVERTIME -------22

322

NULL Values

 The sub query also returns NULL values.

SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official);

OVERTIME -------10 10 20 ? 10 18 22 ? ?

 Hence there is no value which is greater or equal to all sub query values! SELECT Name, Overtime The FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official);

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query result is empty.

323

NULL Values

 Do the following two SQL queries have the same result?

SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official WHERE Overtime IS NOT NULL); SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

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324

NULL Values

 For the existing data, both queries return the same result.

SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official WHERE Overtime IS NOT

NAME ---Ms G

SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

NAME ---Ms G

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OVERTIME -------22

NULL); OVERTIME -------22

325

NULL Values

 But: If the overtime values were unknown for all officials , the two SQL queries would return different results!


SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official WHERE Overtime IS NOT NULL); SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

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OVERTIME -------? ? ? ? ? ? ? ? ?

The query result is empty.

326

NULL Values

 The result of the sub query is empty.  Hence the WHERE clause of the outer query is fulfilled for all rows. SELECT Name, Overtime FROM Official WHERE Overtime >= ALL (SELECT Overtime FROM Official WHERE Overtime IS NOT NULL);


OVERTIME -------? ? ? ? ? ? ? ? ?

 The sub query returns a NULL value.  So the WHERE clause of the outer query is not fulfilled for any row. SELECT Name, Overtime FROM Official WHERE Overtime = (SELECT MAX(Overtime) FROM Official);

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The query result is empty.

327

Unit 6: Understanding NULL Values Summary


Explain how to interpret NULL values in databases • Understand why the presences of NULL values can lead to unexpected query results • Analyze if two seemingly equivalent SQL statements return different results because of NULL values

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328

Unit 7 Changing Data Stored In Tables

329

Unit 7: Changing Data Stored In Tables Learning Objectives After completing this unit, you will be able to: • Add rows to database tables using SQL • Change existing rows of a database table

• Remove existing rows from a database table

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330

Changing Database Access

Which SQL statements can be used to insert, change or delete data?

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331


INSERT statement  New rows are inserted into a table

UPDATE statement  Existing table rows are changed

DELETE statement  Existing table rows are deleted

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332


INSERT statement

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333

INSERT

INSERT INTO Table VALUES (Value, Value, Value);

INSERT INTO Table(Column, Column) VALUES (Value, Value);

INSERT INTO Table SELECT … FROM … WHERE …;

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334

INSERT

Using INSERT you can specify a value for each column. 

If the sequence of the values corresponds to the sequence of columns, you do not need to list the column names.  The order of the columns above mentioned is set implicitly when you create the table.  Insert a new row into the Official table:

INSERT INTO Official VALUES ('P10', 'Ms J', 20, 'A10',

PNR --… P10 ©

NAME ---… Ms J

OVERTIME -------… 20


SALARY -----… A10

'P04');

MANAGER ------… P04 335

INSERT

Using INSERT you can specify a value for each column. 

If the sequence of the values does not match the sequence of the corresponding columns, you must list the column names in the correct sequence of the values.

 Insert a new row into the Official table:

INSERT INTO Official( Salary, Overtime, Name, PNr, Manager) VALUES ('A12', 50, 'Mr K', 'P11', 'P09');

PNR --… P11 ©

NAME ---… Mr K

OVERTIME -------… 50


SALARY -----… A12

MANAGER ------… P09 336

INSERT

You do not have to specify a value for each column of the INSERT.  

You must list the column names in the correct sequence for each column, where you specify a value. Missing values will be replaced by NULL or the default value of the column

 Insert a new row into the Official table:

INSERT INTO Official(PNr, Name) VALUES ('P12', 'Ms L');

PNR --… P12 ©

NAME ---… Ms L

OVERTIME -------… ?


SALARY -----… A06

MANAGER ------… ? 337

INSERT

You can explicitly specify NULL values (for certain columns) for the INSERT.  Insert a new row into the Official table:

INSERT INTO Official VALUES ('P13', 'Mr M', NULL, NULL, NULL);

PNR --… P13 ©

NAME ---… Mr M

OVERTIME -------… ?


SALARY -----… ?

MANAGER ------… ? 338

INSERT

You can INSERT data read from another table.  INSERT all owner IDs and names from city Wiesloch into the Official table:

INSERT INTO Official(PNr, Name) SELECT OwnerID, Name FROM Owner WHERE City = 'Wiesloch'; PNR --… H01 H06

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NAME ---… Ms T Ms W

OVERTIME -------… ? ?


SALARY -----… A06 A06

MANAGER ------… ? ? 339


The UPDATE statement

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340

UPDATE

UPDATE Table SET Column = Value, Column = Value, Column = Value WHERE Condition;

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341

UPDATE

You can use the UPDATE statement to change all rows in a table.  The overtime value of all officials should be increased by 10:

UPDATE Official SET Overtime = Overtime + 10;

PNR --P01 P02 P03 P04 P05 … ©


NAME ---Mr A Mr B Ms C Ms D Mr E …

OVERTIME -------20 20 30 ? 20 …

SALARY -----A09 A10 A09 A12 A08 …

MANAGER ------P04 P04 P04 P09 P08 … 342

UPDATE

You can use the UPDATE statement to change only certain rows in a table by adding a WHERE clause.  The overtime value of officials in salary group “A09” should be doubled:

UPDATE Official SET Overtime = Overtime * 2 WHERE Salary = 'A09'; PNR --P01 P02 P03 P04 … ©



OVERTIME -------20 10 40 ? …

SALARY -----A09 A10 A09 A12 …

MANAGER ------P04 P04 P04 P09 … 343

UPDATE

You can use the UPDATE statement to change values of multiple columns.  The overtime value of officials in salary group “A09” should be tripled. In addition their salary group should be changed to “A10”.

UPDATE Official SET Overtime = Overtime * 3, Salary = 'A10' WHERE Salary = 'A09'; PNR --P01 P02 P03 … ©


NAME ---Mr A Mr B Ms C …

OVERTIME -------30 10 60 …

SALARY -----A10 A10 A10 …

MANAGER ------P04 P04 P04 … 344

UPDATE

You can use an uncorrelated sub query on the WHERE clause of the UPDATE statement.  Officials who are contacts for the car owners are moved to salary group “A15”.

UPDATE Official SET Salary = 'A15' WHERE PNr IN (SELECT PersNumber FROM Contact); PNR --… P07 P08 P09 ©


NAME ---… Ms G Ms H Mr I

OVERTIME -------… 22 ? ?

SALARY -----… A11 A15 A15

MANAGER ------… P08 P09 ? 345

UPDATE

You can use a correlated sub query on the WHERE clause of the UPDATE statement.  Officials who are contacts for the car owners are moved to salary group “A15”.

UPDATE Official o SET Salary = 'A15' WHERE EXISTS (SELECT * FROM Contact c WHERE c.PersNumber = o.PNr); PNR --… P07 P08 P09 ©


NAME ---… Ms G Ms H Mr I

OVERTIME -------… 22 ? ?

SALARY -----… A11 A15 A15

MANAGER ------… P08 P09 ? 346

UPDATE

  

Usually a sub query is part of the WHERE clause. You can also use a (correlated or uncorrelated) sub query in the SET clause. For the following SQL statement we assume, that the table Car has an additional column Ownername.

UPDATE Car c SET Ownername = (SELECT o.Name FROM Owner o WHERE o.OwnerID = c.Owner); CARID ----F01 F02 F03 …

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PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 …

BRAND ----Fiat VW BMW …

COLOR ----red black blue …

HP --75 120 184 …

OWNER ----H06 H03 H03 …

OWNERNAME --------Ms W SAP AG SAP AG …

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The DELETE statement

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348

DELETE

DELETE FROM Table WHERE Condition;

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349

DELETE

You can DELETE all rows in a table:  Delete all entries from table Car:

DELETE FROM Car;

CARID

PLATENUMBER

BRAND

COLOR

HP

OWNER

-----

-----------

-----

-----

--

-----

©


350

DELETE

You can DELETE selected rows from a table:  Delete all cars with horsepower less than 180 :

DELETE FROM Car WHERE HP < 180; CARID ----F03 F06 F11 F17 F20 ©

PLATENUMBER ----------HD-JA 1972 HD-VW 1999 HD-MM 208 HD-Y 333 ?


BRAND ----BMW Audi BMW Audi Audi

COLOR -----blue yellow green orange green

HP --184 260 184 184 84

OWNER ----H03 H05 H02 H09 ? 351

DELETE

You can use an uncorrelated sub query in the DELETE statement.  Delete all cars that are reported as stolen.

DELETE FROM Car WHERE PlateNumber IN (SELECT PlateNumber FROM Stolen);

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CARID ----F02 F03 F04 F05 F07 F08 F09 F10 F11 F12 F13 F14 F15 F16 F18 F19 F20

PLATENUMBER ----------HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 ? HD-MQ 2006 HD-VW 2012 ?

352

DELETE

CARID ----F02  Delete all cars that are reported as stolen. F03 F04 F05 DELETE F07 FROM Car c F08 WHERE EXISTS (SELECT * F09 FROM Stolen s F10 WHERE s.PlateNumber = c.PlateNumber); F11 F12 F13 F14 F15 F16 F18 F19 F20

You can use a correlated sub query in the DELETE statement.

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PLATENUMBER ----------HD-VW 4711 HD-JA 1972 HD-AL 1002 HD-MM 3206 HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 HD-MM 1977 HD-MB 3030 ? HD-MQ 2006 HD-VW 2012 ?

353

Unit 7: Changing Data Stored In Tables Summary

You should now be able to: • Add rows to database tables using SQL • Change existing rows of a database table • Remove existing rows from a database table

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354

Unit 8 Defining How Data Is Stored

355

Unit 8: Defining How Data Is Stored Learning Objectives After completing this unit, you will be able to: • List the most important data types SAP HANA supports • Create new database tables in HANA

• Change such tables, e.g. by adding, removing or renaming columns

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356

Data Definition

Which data types are available in SAP HANA?

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357

Data Types

SAP HANA provides the following Data Types …  Numeric types – TINYINT, SMALLINT, INTEGER, BIGINT, SMALLDECIMAL, DECIMAL, REAL, DOUBLE

 Character string types – VARCHAR, NVARCHAR, ALPHANUM, SHORTTEXT

 Date time types – DATE, TIME, SECONDDATE, TIMESTAMP

 Binary types – VARBINARY

 Large object types – BLOB, CLOB, NCLOB, TEXT

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358

Data Types

SAP HANA data types for integer values:

TINYINT

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0 to 255

SMALLINT

-32,768 to 32,767

INTEGER

-2,147,483,648 to 2,147,483,647

BIGINT

-9,223,372,036,854,775,808 to 9,223,372,036,854,775,807


359

Data Types

SAP HANA data types for fixed-point values:

DECIMAL(p,s) The DECIMAL(p, s) data type specifies a fixed-point decimal number with precision p and scale s. The precision is the total number of significant digits. The scale is the number of digits from the decimal point to the least significant digit. Precision: p (1 ≤ p ≤ 34) Scale: s decimals (-6,111 ≤ s ≤ 6,176)  0.0000259  259 2590000  1.0000259  123.4567

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→ → → → →

DECIMAL(3,7) DECIMAL(3,0) DECIMAL(3,-4) DECIMAL(8,7) DECIMAL(7,4)

360

Data Types

SAP HANA data types for floating-point values:

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SMALLDECIMAL

Floating-point decimal number Precision: variable (1 to 16) Scale: variable (-369 to 368)

DECIMAL

Floating-point decimal number Precision: variable (1 to 34) Scale: variable (-6,111 to 6,176 )

REAL

Floating-point binary number (32 Bit)

DOUBLE

Floating-point binary number (64 Bit)


361

Data Types

When using numeric data types please note that ...



DECIMAL(3,7) is suitable for storing fixed-point decimals numbers with a precision of 3 and a scale of 7, while DECIMAL is used to store floating point decimal numbers with variable precision and variable scale.



Certain decimal values (for example, 0.2 and 0.4) using binary-based floating-point numbers (REAL or DOUBLE) cannot be represented in an exact way.



Therefore columns of type REAL or DOUBLE should not be used in WHERE or JOIN conditions.

• • •

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0.4 is stored in a REAL column as 0.4000000059604645 10*(0.04/0.2) results in REAL value 1.9999998807907104 (instead of 2) 10*(0.04/0.2) results in DOUBLE value 1.9999999999999998 (instead of 2)


362

Data Types

SAP HANA data types for character strings:

VARCHAR(n)

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ASCII character string with maximum length n (n ≤ 5,000)

NVARCHAR(n)

Unicode character string with maximum length n (n ≤ 5,000)

ALPHANUM(n)

Alphanumeric character string with maximum length n (n ≤ 127)

SHORTTEXT(n)

Unicode character string with maximum length n (n ≤ 5,000) Special data type supporting text- and string-search features (e. g. Fuzzy Search) based on NVARCHAR(n)


363

Data Types

SAP HANA data types for date and time:

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DATE

Consists of year, month, day '2012-05-21'

TIME

Consists of hour, minute, second '18:00:57'

SECONDDATE

Combination of data and time '2012-05-21 18:00:57'

TIMESTAMP

Precision: ten millionth of a second '2012-05-21 18:00:57.1234567'


364

Data Types

SAP HANA data types for binary data:

VARBINARY(n)

Binary data, maximum length n Bytes (n ≤ 5,000)

CREATE COLUMN TABLE MyTable ("VARBINARY-Column" VARBINARY(8)); INSERT INTO MyTable VALUES (TO_BINARY('Walldorf')); INSERT INTO MyTable VALUES (TO_BINARY('SAP')); INSERT INTO MyTable VALUES (x'00075341500700FF'); SELECT * FROM MyTable;

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VARBINARY-Column ---------------57616C6C646F7266 534150 00075341500700FF 365

Data Types

SAP HANA data types for large objects:

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CLOB

Long ASCII character string (maximum 2 GB)

NCLOB

Long Unicode character string (maximum 2 GB)

BLOB

Large binary data (maximum 2 GB)

TEXT

Long Unicode character string (maximum 2 GB) Special data type supporting text- and string-search features (e. g. Fuzzy Search) based on NCLOB


366

Data Types

 The term LOB data type (LOB = large object) is used as a generic term for data types such as CLOB (character large object) or BLOB (binary large object).  When using LOB data types it is important to note that ...

• • • • • • • •

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LOB columns can not be part of the primary key LOB columns can not be used in the ORDER BY clause LOB columns can not be used in the GROUP BY clause LOB columns may not be part of the JOIN condition (explicit JOIN) LOB columns can not be used as an argument for an aggregate function LOB columns can not be used in the SELECT DISTINCT clause LOB columns can not occur in an UNION statement LOB columns can not be part of a database index


367

Data Definition

Which SQL statements can be used to create, change, rename and delete tables?

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368

Data Definition

CREATE COLUMN TABLE  A new (empty) table is created.

ALTER TABLE  The definition of an existing table is changed .

RENAME TABLE  An existing table is renamed.

DROP TABLE  An existing table is deleted.

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369

Data Definition

CREATE COLUMN TABLE

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370

CREATE COLUMN TABLE

CREATE COLUMN TABLE Table

(Column Data_Type, Column Data_Type, Column Data_Type NOT NULL,

Column Data_Type NOT NULL, Column Data_Type DEFAULT Default_Value, Column Data_Type NOT NULL DEFAULT Default_Value,

PRIMARY KEY(Column, Column));

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371

CREATE COLUMN TABLE

 

It is not mandatory to define a primary key. If you do not define a key, the table con contain duplicates.

CREATE COLUMN TABLE Official (PNr VARCHAR(3), Name VARCHAR(20), Overtime INTEGER, Salary VARCHAR(3), Manager VARCHAR(3));

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372

CREATE COLUMN TABLE



Table names and column names usually do not distinguish between uppercase and lowercase letters.  If you want to distinguish between uppercase and lowercase letters, you must use double quotes.  We recommend not to use the option of lowercase letters.

CREATE COLUMN ("PNr" "Name" "Overtime" "Salary" "Manager"

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TABLE "Official" VARCHAR(3), VARCHAR(20), INTEGER, VARCHAR(3), VARCHAR(3));

373

CREATE COLUMN TABLE

 Normally, table and column names may not contain spaces.  If table or column names should contain spaces, you must use double quotes.  We recommend not to use the option of spaces in table or column names.

CREATE COLUMN TABLE "Our Officials" ("Official PersNumber" VARCHAR(3), "Official Name" VARCHAR(20), "Overtime Hours" INTEGER, "Salary Group" VARCHAR(3), "Manager PersNumber" VARCHAR(3));

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374

CREATE COLUMN TABLE



If double quotes are used, table and column names may be different, but only by the use of upper and lower case.  We do not recommend using this option.

CREATE COLUMN TABLE "Don’t do it" ("sap" INTEGER, "saP" INTEGER, "sAp" INTEGER, "sAP" INTEGER, "Sap" INTEGER, "SaP" INTEGER, "SAp" INTEGER, "SAP" INTEGER);

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375

CREATE COLUMN TABLE

 If double quotes are used, table and column names can solely consist of spaces.  We do not recommend using this option.

CREATE COLUMN TABLE " " (" " INTEGER, " " INTEGER, " " INTEGER, " " INTEGER); SELECT " ", " FROM " " WHERE " " = "

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", "

"

";

376

CREATE COLUMN TABLE

Condition

"1" = "2"

The value of the column named "1" must match the value of the column named "2" (based on the same row).

"2" = '3'

The value of the column named "2" must match the character string '3' (based on a certain row).

"3" = 4

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Explanation


The numeric value of column named "3" must be equal to 4 (based on a certain row).

377

CREATE COLUMN TABLE

  

You can define a primary key. If a primary key is defined, the table does not contain duplicates. You can use the primary key definition as part of the column definition, if the primary key consists of only one column.

CREATE COLUMN TABLE Official (PNr VARCHAR(3) PRIMARY KEY, Name VARCHAR(20), Overtime INTEGER, Salary VARCHAR(3), Manager VARCHAR(3));

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378

CREATE COLUMN TABLE

 

You can use a separate PRIMARY KEY clause. You can define a maximum of one primary key per table.

CREATE COLUMN TABLE Official (PNr VARCHAR(3), Name VARCHAR(20), Overtime INTEGER, Salary VARCHAR(3), Manager VARCHAR(3), PRIMARY KEY(PNr));

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379

CREATE COLUMN TABLE

 

You must use a separate PRIMARY KEY clause to define a multi-column primary key. For all columns of the PRIMARY KEY NULL values are (implicitly) prohibited.

CREATE COLUMN TABLE Owner_EU (Country VARCHAR(3), OwnerID VARCHAR(3), Name VARCHAR(20), Birthday DATE, City VARCHAR(20), PRIMARY KEY(Country, OwnerID));

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380

CREATE COLUMN TABLE



When defining a column you can specify that no NULL values are allowed by adding the words NOT NULL



For primary key columns, NULL values are implicitly prohibited. For these columns, adding NOT NULL is permitted but not required.

CREATE COLUMN TABLE Official (PNr VARCHAR(3), Name VARCHAR(20) NOT NULL, Overtime INTEGER, Salary VARCHAR(3) NOT NULL, Manager VARCHAR(3), PRIMARY KEY(PNr));

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381

CREATE COLUMN TABLE



You can define a default value for a column by adding DEFAULT .



The default value is used during INSERT, if no value is specified for the corresponding column. The default value will not be used during INSERT, if a NULL value is explicitly specified.



CREATE COLUMN TABLE Official (PNr VARCHAR(3), Name VARCHAR(20), Overtime INTEGER DEFAULT 0, Salary VARCHAR(3) DEFAULT 'A06', Manager VARCHAR(3), PRIMARY KEY(PNr));

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382

CREATE COLUMN TABLE





NULL values are not excluded when setting a default value. Hence the words NOT NULL can also be useful for columns that do have a default value set. The order of the key words "DEFAULT" and "NOT NULL" is irrelevant.

CREATE COLUMN TABLE Official (PNr VARCHAR(3), Name VARCHAR(20), Overtime INTEGER NOT NULL DEFAULT 0, Salary VARCHAR(3) DEFAULT 'A06' NOT NULL, Manager VARCHAR(3), PRIMARY KEY(PNr));

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383

CREATE COLUMN TABLE



When defining a column, you can specify that this column must not contain double values by adding UNIQUE .



In contrast to the PRIMARY

KEY, UNIQUE allows NULL values.

CREATE COLUMN TABLE Car (CarID VARCHAR(3) PRIMARY KEY, PlateNumber VARCHAR(10) UNIQUE, Brand VARCHAR(20), Color VARCHAR(10), HP INTEGER, Owner VARCHAR(3));

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384

CREATE COLUMN TABLE



NULL values are not excluded when setting UNIQUE . Hence the words



NOT NULL can also be useful for columns defined as UNIQUE .

The order of the key word “UNIQUE" and "NOT NULL" is irrelevant

CREATE COLUMN TABLE Car (CarID VARCHAR(3) PRIMARY KEY, PlateNumber VARCHAR(10) UNIQUE NOT NULL, Brand VARCHAR(20), Color VARCHAR(10), HP INTEGER, Owner VARCHAR(3));

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385

CREATE COLUMN TABLE

 

It is possible to prohibit duplicate values for more than one column. Therefore you can define several keys for one and the same table.

 Using

UNIQUE

two single-column keys are defined:

CREATE COLUMN TABLE Car (CarID VARCHAR(3) PRIMARY KEY, PlateNumber VARCHAR(10), Brand VARCHAR(20) UNIQUE, Color VARCHAR(10) UNIQUE, HP INTEGER, Owner VARCHAR(3));

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386

CREATE COLUMN TABLE

You can use a separate  Using

UNIQUE

UNIQUE

clause for the key definition.

two single-column keys are defined:

CREATE COLUMN TABLE Car (CarID VARCHAR(3) PRIMARY KEY, PlateNumber VARCHAR(10), Brand VARCHAR(20), Color VARCHAR(10), HP INTEGER, Owner VARCHAR(3), UNIQUE(Brand), UNIQUE(Color));

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387

CREATE COLUMN TABLE

You must use a separate

UNIQUE

clause to define a multi-column key.

 Using UNIQUE a two-column key is defined:

CREATE COLUMN TABLE Car (CarID VARCHAR(3) PRIMARY KEY, PlateNumber VARCHAR(10), Brand VARCHAR(20), Color VARCHAR(10), HP INTEGER, Owner VARCHAR(3), UNIQUE(Brand, Color));

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388

Data Definition

ALTER TABLE

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389

ALTER TABLE

ALTER TABLE Table ADD (Column Data_Type, Column Data_Type NOT NULL DEFAULT Default_Value);

ALTER TABLE Table DROP (Column, Column, Column);

ALTER TABLE Table ALTER (Column Data_Type DEFAULT Default_Value, Column Data_Type NULL);

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390

ALTER TABLE



You can ADD columns to an existing table.



You can use NOT NULL only if the table is empty or if you define a DEFAULT value in addition. Newly added columns are filled with NULL values, or – if available – with the corresponding default value.



ALTER TABLE Official ADD (RemainderDays INTEGER, AnnualLeave INTEGER NOT NULL DEFAULT 30);

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391

ALTER TABLE

DROP



You can

 

You cannot drop columns that are part of the primary key. Also for tables without a primary key at least one column must be left.

columns from an existing table.

ALTER TABLE Official DROP (Salary, Manager, Name);

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392

ALTER TABLE

   

You can change the definition of an existing column. For columns without a default value, you can specify one. For columns with a default value, you can change the existing value. The new (or first time) defined default value only affects newly inserted rows.



For columns that do not allow NULL values so far, NULL values can be allowed.

ALTER TABLE Official ALTER (Overtime Salary Name

©


INTEGER DEFAULT 7, VARCHAR(3) DEFAULT 'A01', VARCHAR(20) NULL);

393

ALTER TABLE

  

For columns with a default value you can delete the existing default value. The deletion of the default value only affects newly inserted rows. Already existing values of a table are not affected by the deletion of the default value.

ALTER TABLE Official ALTER (Salary VARCHAR(3) DEFAULT NULL);

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394

ALTER TABLE

 

You can change the data type of existing columns. Type compatibility must be ensured. For string-based data types the new data type must have at least the same maximum length (in comparison to the old data type).

 VARCHAR(20) → VARCHAR(25) is allowed, but not VARCHAR(25) → VARCHAR(20) ALTER TABLE Official ALTER (Name Salary

©


VARCHAR(25), VARCHAR(5));

395

ALTER TABLE

  

You can delete the definition of the primary key. Only the primary key property is deleted. The corresponding columns of the (previous) primary key are not deleted.

ALTER TABLE Official DROP PRIMARY KEY;

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396

ALTER TABLE

  

You can subsequently define a primary key for a table without one. The corresponding columns must not contain NULL values. The key property must not be violated by the existing data.

ALTER TABLE Official ADD PRIMARY KEY(PNr);

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397

Data Definition

RENAME TABLE

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398

RENAME TABLE

RENAME TABLE old_name TO new_name;

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399

RENAME TABLE

RENAME



You can

 

The table you want to rename can be empty or can contain data. The new table name must be different to the old table name.

an existing table.

RENAME TABLE Official TO Employee;

©


400

Data Definition

DROP TABLE

©


401

DROP TABLE

DROP TABLE Table;

©


402

DROP TABLE

DROP



You can

 

The table you want to drop can be empty or contain data. Both, the table content and the table itself will be deleted.

an existing table.

DROP TABLE Official;

©


403

Data Definition

Renaming a table column

©


404

Data Definition

RENAME COLUMN  An existing table column is renamed.  This statement is used to "permanently" rename a column of an existing table (by which the table definition is changed).  Hence, we are not simply renaming a column of the query-result (for which the keyword "AS" can be used in the SELECT clause).

©


405

Data Definition

RENAME COLUMN

©


406

RENAME COLUMN

RENAME COLUMN Table.Column TO new_Column_Name;

©


407

RENAME COLUMN

  

You can (subsequently) rename a table column. The respective table can be empty or can contain data. The new column name must be different to the old column name.

 Column “PNr“ of Table “Official“ is renamed into “PersNumber“:

RENAME COLUMN Official.PNr TO PersNumber;

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408

Unit 8: Defining Data And Access Control Summary


List the most important data types SAP HANA supports • Create new database tables in HANA • Change such tables, e.g. by adding, removing or renaming columns

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409

Unit 9 Using Views for Data Access

410

Unit 9: Using Views For Data Access Learning Objectives

After completing this unit, you will be able to: • Describe the use cases for and advantages of using database views

• Define database views using SQL • Query data through views • Delete database views using SQL

©


411

Data Definition

Why do we need views? What are the advantages of the view concept?

©


412

Advantages of Views

 Decoupling applications from lower levels (relative to the 3 level schema architecture) – Changes to the conceptual schema (such as adding new columns of the table) have no effect on application

 Tailored views, individually customized for the users and their tasks – The user (or application programmer) does not know the entire database structure in detail - but the excerpt that is relevant to them.

 Simplification of queries – Complex queries referencing views are easier to formulate, provided that the view provides an appropriate pre-selection of data.

 Possibility of access restriction – Views can be used to realize value-dependent access privileges (for example, managers can access only the data of their employees).

©


413

Data Definition

Leveraging Views

©


414

VIEWs

 Example of a view containing only red cars:

CREATE VIEW SELECT FROM WHERE

redCars AS CarID, Color, PlateNumber, HP AS Power Car Color = 'red';

CARID ----F01 F09 F12 F13 F18

©


COLOR ----red red red red red

PLATENUMBER ----------HD-V 106 HD-UP 13 HD-XY 4711 HD-IK 1001 HD-MQ 2006

POWER ----75 105 105 136 90

415

VIEWs

You can select data from a view in the same way you select from a “normal” table.  Which red cars have more than 100 HP?

SELECT FROM WHERE ORDER BY

CarID, PlateNumber, Power redCars Power > 100 Power DESC, CarID ASC;

CARID ----F13 F09 F12

©


PLATENUMBER ----------HD-IK 1001 HD-UP 13 HD-XY 4711

POWER ----136 105 105

416

VIEWs

You can INSERT new rows into a view. 

The new rows are not inserted into the view, but into the underlying base table!

INSERT INTO redCars VALUES ('F77', 'red', 'HD-MT 2509', 170);

SELECT * FROM Car; CARID ----F01 F02 F03 … F18 F19 F20 F77

©


PLATENUMBER ----------HD-V 106 HD-VW 4711 HD-JA 1972 … HD-MQ 2006 HD-VW 2012 ? HD-MT 2509

BRAND ------Fiat VW BMW … Renault VW Audi ?

COLOR ----red black blue … red black green red

HP --75 120 184 … 90 125 184 170

OWNER ----H06 H03 H03 … H03 H01 ? ? 417

VIEWs You can UPDATE rows in a view. 

The rows are not changed in the view, but in the underlying base table!

UPDATE redCars SET Power = 120 WHERE Power < 120; SELECT * FROM Car;

©


CARID ----F01 … F07 F08 F09 F10 F11 F12 F13 … F18 …

PLATENUMBER ----------HD-V 106 … HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 … HD-MQ 2006 …

BRAND ------Fiat … Audi VW Skoda BMW BMW Skoda Renault … Renault …

COLOR ----red … blue black red black green red red … red …

HP --120 … 116 160 120 140 184 120 136 … 120 …

OWNER ----H06 … H03 H07 H02 H04 H02 H04 H07 … H03 … 418

VIEWs You can DELETE rows from a view. 

The rows are not deleted from the view, but from the underlying base table!

DELETE FROM redCars WHERE Power < 125; SELECT * FROM Car;

©


CARID ----F01 … F07 F08 F09 F10 F11 F12 F13 … F18 …

PLATENUMBER ----------HD-V 106 … HD-ML 3206 HD-IK 1002 HD-UP 13 HD-MT 507 HD-MM 208 HD-XY 4711 HD-IK 1001 … HD-MQ 2006 …

BRAND ------Fiat … Audi VW Skoda BMW BMW Skoda Renault … Renault …

COLOR ----red … blue black red black green red red … red …

HP --120 … 116 160 120 140 184 120 136 … 120 …

OWNER ----H06 … H03 H07 H02 H04 H02 H04 H07 … H03 … 419

Data Definition

Which SQL statements can I use to create or delete views?

©


420

Data Definition

CREATE VIEW  Creates a new view.

DROP VIEW  Deletes an existing view.

©


421

Data Definition

CREATE VIEW

©


422

CREATE VIEW


©


View AS Column, Column, Column Table Condition;

423

CREATE VIEW



You can restrict a view to specific columns of the underlying base table.

CREATE VIEW Autos AS SELECT Brand, Color, HP FROM Car;

©




HP --75 120 184 136 170 260 116 160 105 140 184 105 136 170 136 120 184 90 125 184

424

CREATE VIEW



You can restrict a view to specific rows of the underlying base table.


Audis AS Brand, Color, HP Car Brand = 'Audi';

BRAND ----Audi Audi Audi Audi

©


COLOR -----yellow blue orange green

HP --260 116 184 184

425

CREATE VIEW



You can restrict a view to a certain amount of rows:


"Top3 Audis" AS TOP 3 Brand, Color, HP Car Brand = 'Audi';

BRAND ----Audi Audi Audi

©


COLOR -----yellow blue orange

HP --260 116 184

426

CREATE VIEW



The columns of a view can have different names than the columns of the underlying base table.


Audis (Manufacturer, Color, Power) AS Brand, Color, HP Car Brand = 'Audi';

MANUFACTURER -----------Audi Audi Audi Audi

©



POWER ----260 116 184 184

427

CREATE VIEW



You can (and should) rename the columns in the SELECT clause


Audis AS Brand AS Manufacturer, Color, HP AS Power Car Brand = 'Audi';


©



POWER ----260 116 184 184

428

CREATE VIEW

 

If you rename columns in both the CREATE VIEW clause and in the SELECT clause, the name of the CREATE VIEW clause "wins“. We recommend renaming columns only in the SELECT clause.


Audis (Manufacturer, Color, Power) AS Brand AS Description, Color, HP AS HorsePower Car Brand = 'Audi';


©



POWER ----260 116 184 184

429

CREATE VIEW



A view can contain duplicates.

CREATE VIEW Colors AS SELECT Color FROM Car;

©



430

CREATE VIEW

 

You can use DISTINCT in the view definition. Thus the view does not contain duplicates.

CREATE VIEW Colors AS SELECT DISTINCT Color FROM Car;

©


COLOR -----red black blue white yellow green orange

431

CREATE VIEW

   

You can use ORDER BY in the view definition. From a theoretical perspective the use of ORDER BY is highly questionable, because the rows of a view are unsorted! There is no guarantee that the specified ORDER BY sorting is taken into account when reading the view. We recommend avoiding ORDER BY in the view definition.

CREATE VIEW SELECT FROM ORDER BY

©


Colors AS DISTINCT Color Car Color;

COLOR -----blue yellow green orange red black white 432

CREATE VIEW



You can use calculated columns in a view.


Audis AS CarID, Color, ROUND(HP / 1.35962162, 0) AS "kW" Car Brand = 'Audi';

CARID ----F06 F07 F17 F20

©



kW --191 85 135 135

433

CREATE VIEW



You can use functions in the view definition.

CREATE VIEW OwnerOverview AS SELECT Name, YEAR(Birthday) AS YearOfBirth FROM Owner; NAME -----Ms T Ms U SAP AG HDM AG Mr V Ms W IKEA Mr X Ms Y Mr Z ©


YEAROFBIRTH ----------1934 1966 ? ? 1952 1957 ? 1986 1986 1986 434

CREATE VIEW



You can use GROUP BY and aggregate expressions in the view definition.

CREATE VIEW SELECT FROM GROUP BY HAVING

Cars AS Brand, COUNT(*) AS Quantity Car Brand COUNT(*) > 2; BRAND -------VW BMW Mercedes Audi Skoda

©


QUANTITY -------3 3 3 4 3

435

CREATE VIEW



You can use UNION [ALL] in the view definition.

CREATE VIEW Persons AS SELECT OwnerID AS "Pers ID", Name FROM Owner WHERE Birthday >= '1977-05-21' UNION ALL SELECT PNr, Name FROM Official WHERE Salary = 'A09';

©


Pers ID ------H08 H09 H10 P01 P03 P06


436

CREATE VIEW



You can use (any) JOIN in the view definition.


StolenCars AS c.PlateNumber, c.Brand, c.Color Car c, Stolen s c.PlateNumber = s.PlateNumber;


©




437

CREATE VIEW



You can use an uncorrelated sub query in the view definition.


StolenCars AS PlateNumber, Brand, Color Car PlateNumber IN (SELECT PlateNumber FROM Stolen);


©




438

CREATE VIEW



You can use a correlated sub query in the view definition.


StolenCars AS c.PlateNumber, Car c EXISTS (SELECT FROM WHERE

c.Brand, c.Color * Stolen s s.PlateNumber = c.PlateNumber);


©




439

CREATE VIEW



You can define a view based on another view.


Volkswagen AS CarID, PlateNumber, Color, HP Car Brand = 'VW';


blackVolkswagen AS CarID, PlateNumber, HP Volkswagen Color = 'black';

©


CARID ----F02 F08 F19

PLATENUMBER ----------HD-VW 4711 HD-IK 1002 HD-VW 2012

HP --120 160 125

440

Data Definition

DROP VIEW

©


441

DROP VIEW

DROP VIEW View;

©


442

DROP VIEW

  

You can drop an existing view. Only the view definition is deleted. The underlying data of the view is not deleted.

DROP VIEW redCars;

©


443

Views and Changeability

Is it always possible to change the data of the underlying table by referencing a view?

©


444

Views and Changeability

Referencing a view, the rows of the underlying base table cannot be changed if (at minimum) any of the following statements is true:  The view definition contains a calculation in the projection list (such as HP / 1.36).  The view definition uses a function in the projection list (such as YEAR).  The view definition uses DISTINCT to eliminate duplicates.  The view definition contains the TOP n clause to restrict the number of rows to n.  The view definition uses an aggregate function in the projection list (such as MAX).  The view definition is based on a GROUP BY clause.

 The view definition uses UNION.  The view definition contains an (implicit or explicit) JOIN.  The view definition uses a sub query in the projection list.

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445

Unit 9: Using Views For Data Access Summary

You should now be able to: • Describe the use cases for and advantages of using database views • Define database views using SQL • Query data through views

• Delete database views using SQL

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446

Unit 10 Defining Data Access

447

Unit 10: Defining Data Access Learning Objectives

After completing this unit, you will be able to: • Access tables in other schemas (assuming suitable permissions) • Control how other users can access data in your own schema

• Explain when database indexes make sense even in HANA • Create and delete database indexes using SQL

© 2013

SAP AG. All rights reserved.

448

Data Definition

Why are there no naming collisions when multiple users create the same table or view?

© 2013


449

Schemata

Please note: The following information applies to direct (SQL-based) work on SAP HANA - but not to database access using ABAP!  User 1 creates table “Car”.  User 2 also creates table “Car”.  Why is there no naming collision?  The name of the database object (table, view etc.) implicitly contains a schema name as prefix: .  “USER1.Car” ≠ “USER2.Car”

© 2013


450

Schemata

 When creating a user an identically named schema is created implicitly.  By default a user works in his own schema (and namespace).  You can explicitly specify the schema name if necessary.  Which brand has (at least) one black car? SELECT FROM WHERE ORDER BY

© 2013

DISTINCT User1.Car.Brand User1.Car User1.Car.Color = 'black' User1.Car.Brand;


BRAND -------BMW Mercedes Skoda VW 451

Schemata

 Queries across multiple schemata are possible.  To whom is (at least) one black car registered? SELECT DISTINCT Schema1.Owner.Name FROM Schema1.Owner, Schema2.Car WHERE Schema1.Owner.OwnerID = Schema2.Car.Owner AND Schema2.Car.Color = 'black';

NAME -----SAP AG IKEA HDM AG Ms T © 2013


452

Access Control

How can I define who has access to which data?

© 2013


453

Access Control

To specify who can access which data, you can use the following two main (SQL based) options. They can be used in combination:  Create views that represent a portion of the data: – You provide a view that contains only the owners of CityA to a user for whom the owners outside of CityA are not relevant.

 Grant specific access permissions to selected users: – You grant read only permissions to a user, who should be able to read the owner data but not to change it.

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454

Access Control

Which SQL statements can I use for access control?

© 2013


455

Access Control

GRANT  You can selectively grant access permissions to a user (or role).

REVOKE  You can selectively revoke access permissions from a user (or role).

© 2013


456

Access Control

The SQL statement GRANT

© 2013


457

GRANT

GRANT Privilege, Privilege ON Database_Object TO Database_User WITH GRANT OPTION;

© 2013


458

GRANT



You can define which user is allowed to SELECT from which table.



By adding the WITH GRANT OPTION the recipient of the privilege (grantee) is allowed to grant this privilege to other users (without losing it).

 User1 is allowed to read (SELECT) from table Official of schema Schema2.  User1 is allowed to grant this SELECT privilege to other users.

GRANT ON TO WITH

© 2013

SELECT Schema2.Official User1 GRANT OPTION;


459

GRANT



You can exactly specify which user is allowed to INSERT rows into which table.



The INSERT privilege does not necessarily require the SELECT privilege.



The INSERT privilege does not automatically include the SELECT privilege.

 User1 is allowed to INSERT rows into table Owner of schema Schema2. As User 1 has no SELECT privilege, he is not allowed to read the rows he has inserted.

GRANT INSERT ON Schema2.Owner TO User1;

© 2013


460

GRANT



You can exactly specify which user is allowed to UPDATE rows of which table.



The UPDATE privilege does not necessarily require the SELECT privilege.



The UPDATE privilege does not automatically include the SELECT privilege.

 User1 is allowed to UPDATE table Car_EU of schema Schema2.  Because User1 has no SELECT privileges for the table Car_EU the WHERE clause of his UPDATE statement cannot contain a reference to the columns of this table.

GRANT UPDATE ON Schema2.Car_EU TO User1;

© 2013


461

GRANT



You can exactly specify which user is allowed to DELETE rows from which table.



The DELETE privilege does not necessarily require the SELECT privilege.



The DELETE privilege does not automatically include the SELECT privilege.

 User1 is allowed to DELETE rows from table Stolen of schema Schema2.  Because User1 has no SELECT privileges for the table Stolen the WHERE clause of his DELETE statement cannot contain a reference to the columns of this table.

GRANT DELETE ON Schema2.Stolen TO User1;

© 2013


462

GRANT

 You can assign multiple privileges for the same database object by using a single GRANT statement.  User1 is allowed to SELECT, INSERT, UPDATE, DELETE on table Contact of schema Schema2.  User1 is allowed to grant these privileges to other users (in whole or in part). GRANT ON TO WITH

© 2013

SELECT, INSERT, UPDATE, DELETE Schema2.Contact User1 GRANT OPTION;


463

GRANT

You can assign privileges for all tables and views of a schema by using a single GRANT statement.  User1 is allowed to SELECT from all tables and views of schema Schema2.  User1 is allowed to grant this privilege to other users (for all or selected tables or views). 

GRANT ON TO WITH

© 2013

SELECT SCHEMA Schema2 User1 GRANT OPTION;


464

GRANT

Depending on the granularity (or the type of database object) you can assign different access privileges to users.

Granularity

Access Privileges

(single) Schema

SELECT, INSERT, UPDATE, DELETE, DROP, ALTER, CREATE ANY

(single) Table

SELECT, INSERT, UPDATE, DELETE, DROP, ALTER

(single) View

SELECT, INSERT, UPDATE, DELETE, DROP

© 2013


465

Access Control

The SQL statement REVOKE

© 2013


466

REVOKE

REVOKE Privilege, Privilege, Privilege ON Database_Object FROM Database_User;

© 2013


467

REVOKE



Using the REVOKE statement you can selectively revoke access privileges from a user.

 User1 will no longer be allowed to read from table Official of schema Schema2 (assuming that this privilege is not granted by other privileges)

REVOKE SELECT ON Schema2.Official FROM User1;

© 2013


468

REVOKE



You can revoke multiple privileges for the same database object by using a single REVOKE statement

 User1 will no longer be allowed to SELECT, INSERT, UPDATE, DELETE from table Contact of schema Schema2 (assuming that these privileges are not granted by other privileges)

REVOKE SELECT, INSERT, UPDATE, DELETE ON Schema2.Contact FROM User1;

© 2013


469

REVOKE



Access privileges on different hierarchical levels of database objects are independent.

 Despite the REVOKE statement User1 still has read access to the table Official of schema Schema2 , because he still has a (not withdrawn) read permission on the entire schema Schema2 .

GRANT SELECT ON SCHEMA Schema2 TO User1; REVOKE SELECT ON Schema2.Official FROM User1;

© 2013


470

REVOKE



The independence of access privileges at various hierarchical levels is also valid in the contrary case.



Despite the REVOKE statement User1 still has read access to the table Official of schema Schema2 , because he still has a (not withdrawn) read permission on the table.

GRANT SELECT ON Schema2.Official TO User1; REVOKE SELECT ON SCHEMA Schema2 FROM User1;

© 2013


471

Access Paths

What is a database index?

© 2013


472

Access Paths

 An index (plural: indexes or indices) is an access path.  An index may speed up search and sorting. – Index access instead of "full table scan" or "full column scan" – Index usage with HANA COLUMN TABLES only in exceptional cases

 An index usually slows down inserts, update and deletion.  An index has no influence on the query result. – Effect "only" on performance

 Indexes are not covered by the SQL standard. – CREATE / DROP INDEX are strictly speaking no SQL statements.

© 2013


473

HANA: Why Indexes on Column Store?

– Sorted dictionary of values – (Bit-)Vector of value IDs

Are indexes necessary to speed up data access?

© 2013


Column „Name“ conceptual

Column “Name” – internal representation Value ID vector

Dictionary

Miller

4

0

Baker

Jones

1

1

Jones

Millman

5

2

John

Zsuwalski

772

3

Johnson

Baker

0

4

Miller

Miller

4

5

Millman

John

2

Miller

4

Johnson

3

Jones

1

…

…

sorted

 HANA stores data in columns, not rows  Columns are stored as

… 772

Zsuwalski …

474

HANA: Table Scan With Value ID

 Value ID lookup is fast

Dictionary

–

WHERE Name = 'Zsuwalski'

 Row ID lookup can still involve “full column scan” – No issue for most value ID vectors, even less if sorted – But may be slow for very large value ID vectors

© 2013


binary search on dictionary

– binary search on sorted dictionary

Value ID vector

ID

Value

Row ID

Value ID

0

Baker

0

4

1

Jones

1

1

2

John

2

5

3

Johnson

3

772

4

Miller

4

0

5

Millman

772

full column scan with value ID

…

…

254621

772

Zsuwalski

…

…

…

475

HANA: Inverted Index on Column Store

 An inverted index can be reasonable in exceptional cases if column scan performance is not sufficient

WHERE Name = 'Zsuwalski'

© 2013


binary search on dictionary

Dictionary

Inverted Index

Value ID vector

ID

Value

Value ID

Rows

Row ID

Value ID

0

Baker

0

56756

0

4

1

Jones

1

345, 76876

…

…

3

772

… 772

Zsuwalski …

…

…

772

3, 254621

…

…

… 254621

772

…

…

476

Access Paths

Which SQL statements can I use to create a database index?

© 2013


477

Access Paths

CREATE INDEX  A new database index is created.

DROP INDEX  An existing database index is deleted.

© 2013


478

Access Paths

CREATE INDEX

© 2013


479

Access Paths

CREATE INDEX Access_Path ON Table (Column, Column);

CREATE UNIQUE INDEX Access_Path ON Table (Column, Column);

© 2013


480

CREATE INDEX

 

To speed up the read access you can create an index on a table column. The respective table can be empty or contain data.

 An index on column PlateNumber of table Car will be created:

CREATE INDEX PlateNumberIndex ON Car (PlateNumber);

© 2013


481

CREATE INDEX



You can create a UNIQUE INDEX to ensure that the corresponding column cannot contain duplicate values.



Unlike PRIMARY KEY, UNIQUE INDEX does not prohibit NULL values.



A UNIQUE INDEX can only be created when the column contains no duplicate values.

 In table Car no duplicated plate numbers are allowed:

CREATE UNIQUE INDEX PlateNumberIndex ON Car (PlateNumber);

© 2013


482

CREATE INDEX

You can create multiple indexes on the same table.  A single-column index is created on column Brand of table Car.  A single-column index is created on column Color of the same table.

CREATE INDEX BrandsIndex ON Car (Brand); CREATE INDEX ColorsIndex ON Car (Color);

© 2013


483

CREATE INDEX

You can create a multi-column index.  Here a two-column index is created:

CREATE INDEX BrandsColorsCombinationsIndex ON Car (Brand, Color);

© 2013


484

CREATE INDEX



By applying a multi-column UNIQUE INDEX, the relevant combination of columns can be defined as key.



In contrast to the PRIMARY KEY, UNIQUE INDEX does not prohibit NULL values in the key columns.



You can only create a multi-column UNIQUE INDEX, if the relevant column combination does not contain duplicates. Here, the value uniqueness of Country and PlateNumber combinations is enforced:



CREATE UNIQUE INDEX CountryPlateNumberCombiIndex ON Car_EU (Country, PlateNumber);

© 2013


485

Access Paths

DROP INDEX

© 2013


486

Access Paths

DROP INDEX Access_Path;

© 2013


487

DROP INDEX

  

You can drop an existing index. Only the access path (index) is deleted. The underlying data of the index is not deleted.

 DROP INDEX does not syntactically distinguish between NON-UNIQUE and UNIQUE indexes.

DROP INDEX PlateNumberIndex;

© 2013


488

Unit 10: Defining Data Access Summary


Access tables in other schemas (assuming suitable permissions) • Control how other users can access data in your own schema • Explain when database indexes make sense even in HANA • Create and delete database indexes using SQL

© 2013


489

Unit 11 Database Transactions

490

Unit 11: Database Transactions Learning Objectives

After completing this unit, you will be able to: • Explain what a database transaction is and why it is needed • Explain the acronym ACID

• Finish database transactions in HANA using SQL statements • Describe issues that arise if transactions are not mutually isolated • Understand how classical database systems synchronize concurrent transactions • Understand how HANA synchronizes concurrent transactions

© 2013


491

Database Transactions

•

Transaction (TA) = Sequence of associated database operations (SQL statements) for which the ACID requirements must be met.

•

All database operations always take place within transactions (in the extreme case, each operation has its own transaction).

• • • •

A C I D

© 2013

= = = =

Atomicity Consistency Isolation Durability


492


• Atomicity (A): A transaction is either executed completely or not at all.  For Example: Transfer from Account1 to Account2

• Consistency (C): A transaction will bring the database from one consistent state to an(other) consistent state.  During a transaction inconsistent states are possible, for example account transfers or marriage registration.

• Isolation (I): The database changes performed within a transaction shall only be visible to the outside after the completion of the transaction Important only for multi-user operations. Problematic if changes of canceled transactions can be seen.

• Durability (D): If a transaction is successfully completed (COMMIT), all changes from the transaction must permanently stay even in case of failures, or can be restored automatically. For example: Booking a deposit

© 2013


493

Database Transactions •

The ACID requirements must be ensured by the DBMS (not trivial).

•

The DBMS must provide mechanisms/automation for:  Logging & Recovery for A + D  Synchronization for I (such as locks)  Integrity control for C

•

ACID requirements „cost“:  With respect to the DBMS implementation effort  With regard to performance at runtime

© 2013


494


Which SQL statements can I use to start a transaction?

© 2013


495


SAP HANA does not provide a SQL statement to explicitly start a transaction.

A transaction is implicitly started, if the previous transaction was canceled or successfully completed (or there is no previous transaction) AND one of the following SQL statements is executed SELECT INSERT UPDATE DELETE The transaction begins directly before the SQL statement, by which it is implicitly started.

© 2013


496


Which SQL statements can I use to finish a transaction?

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497


COMMIT

 The currently running transaction should be completed successfully.  The durability requirement must be ensured.

ROLLBACK

 The currently running transaction should be canceled.  The atomicity requirement must be ensured.

© 2013


498


DELETE FROM Car; COMMIT; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) COMMIT; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) ROLLBACK; INSERT INTO Car(CarID) INSERT INTO Car(CarID) INSERT INTO Car(CarID) COMMIT; SELECT CarID FROM Car;

© 2013


VALUES ('F01'); VALUES ('F02'); VALUES ('F03'); VALUES ('F04'); VALUES ('F05'); VALUES ('F06'); VALUES ('F07'); VALUES ('F08'); VALUES ('F09');

CARID ----F01 F02 F03 F07 F08 F09

499


DELETE FROM Car; COMMIT; INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES CREATE COLUMN TABLE T(S INT); INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES ROLLBACK; INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES INSERT INTO Car(CarID) VALUES COMMIT; SELECT CarID FROM Car;

© 2013


Certain SQL statements cause an implicit COMMIT ('F01'); ('F02'); ('F03'); ('F04'); ('F05'); ('F06'); ('F07'); ('F08'); ('F09');

CARID ----F01 F02 F03 F07 F08 F09

500

Multi-user Mode

Multi-user operations are absolutely necessary - but not without problems

© 2013


501

Multi-user Mode

Multi-user mode is absolutely necessary! •

If the transactions took place strictly in sequence one after another, this would be a huge waste of resources, since waiting times could not be used. • • •

Waiting for user input Waiting for disk access (for classical DBMS products) Waiting times caused by the application program

Multiple transaction must run time-shared (“interlocked”) •

When this happens in an uncontrolled manner (for example without synchronization), many problems can occur.

•

The problems can be divided into 5 different error classes (see following slides)

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Uncontrolled Multi-user Mode 1/5 Lost-update Problem  The changes of a transaction are overwritten by another transaction and therefore lost.

Point in time

Transaction 1

t1

Read x

t2 t3

Read x x := x + 5,000; x := x – 100;

t4 t5 t6

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Transaction 2

Write x Write x

503

Uncontrolled Multi-user Mode 2/5 Inconsistent Analyses  A transaction sees an apparent inconsistency, since it refers to two different consistent database states.

Point in time

Transaction 1

t1

Read marital status of Mandy (Result: “unmarried”)

t2

Change marital status of Mandy to “married”

t3

Change marital status of Thomas to “married”

t4

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Transaction 2


Read marital status of Thomas (Result: “married”)

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Uncontrolled Multi-user Mode 3/5 Phantom Problem (“malicious”, as it cannot be prevented by row-locking)  A transaction sees an apparent inconsistency, since another transaction inserted a new row in the meantime. Point in time t1

Transaction 2

Read all SAP employees

t2

Insert Mr Z as a new SAP employee

t3

Register Mr Z as course participant

t4

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Transaction 1


Read all course participants (Mr Z is a participant of an internal course, though he apparently seems to be no SAP employee!)

505

Uncontrolled Multi-user Mode 4/5 Non-repeatable read  During the same transaction different results are obtained for repeated reading of the same facts. Point in time

Transaction 1

t1

Read salary of Mr A

t2 t3

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Transaction 2

Raise salary of Mr A Read salary of Mr A

506

Uncontrolled Multi-user Mode 5/5 Dependency on uncommitted changes (Dirty reads)  A transaction sees and uses a state that officially never existed.

Point in time

Transaction 1

t1

Raise salary to € 10,000,000.-

t2

Read salary

t3

Donate € 5,000,000.-

t4

COMMIT;

t5

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Transaction 2


ROLLBACK; (transaction canceled)

507

Synchronisation

 To avoid the problems shown before, multi-user mode must be controlled or synrchonized – Synchronization = time based coordination of processes

 A common option for synchronization is the use of locking – Locks are the most important (but not the only) way – “Multi-version Concurrency Control" can also be used – Alternatively, there is the “Optimistic Synchronization Method“

 The goal of synchronization is the logical single-user mode with physical multi-user mode.

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Logical Single-user Mode

What does logical single-user mode with physical multi-user mode mean?

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Logical Single-user Mode

The user should have the impression of being the only active user of the database. This goal cannot be reached to 100%.  If access to a (currently locked) object takes longer, the users will suspect that they are not alone.  If the transaction is aborted by the DBMS to resolve a deadlock, the users will suspect that they are not alone.

Externally there should be the impression that the transactions are strictly executed in series (that means one after another - Serializability criterion)  In reality: time-shared processing of several transactions – But the impression of serial processing should be created

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Serializability

What is the Serializability Criterion?

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Serializability

 Serializability = There is a serial execution order of the transactions that has the same result – Result in terms of database state or database content

 Serializability is therefore a correctness criterion for the "correct" execution of the synchronization (that means for the "right" control of the multi-user mode)  Ensuring Serializability is the responsibility of the DBMS (and not of the user or of the application program)  What Serializability means, you can best understand with reference to a concrete example … (see following slides).

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Transaction 1

UPDATE Official SET Overtime = Overtime + 2 WHERE PNr = 'P07'; UPDATE Official SET Overtime = Overtime + 2 WHERE PNr = 'P07'; COMMIT;

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Transaction 2

--Rounding to full ten UPDATE Official SET Overtime = ROUND(Overtime, -1) WHERE PNr = 'P07';

SELECT * FROM Contact; COMMIT;

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Transaction 3

UPDATE Official SET Overtime = Overtime * 4 WHERE PNr = 'P07'; UPDATE Official SET Overtime = Overtime / 2 WHERE PNr = 'P07'; COMMIT;

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Effect of the 3 Transactions

TA1 increases the overtime value by 4

1a) o = o + 2; 1b) o = o + 2; TA2 rounds the overtime value to full ten

2a) o = round(o); 2b) read access; TA3 doubles the overtime value

3a) o = o * 4; 3b) o = o / 2;

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Serial Execution

  

TA1 increases the overtime value by 4 TA2 rounds the overtime value to full ten TA3 doubles the overtime value

TA1 → TA2 → TA3

22 → 26 → 30 → 60

TA1 → TA3 → TA2

22 → 26 → 52 → 50

TA2 → TA1 → TA3

22 → 20 → 24 → 48

TA2 → TA3 → TA1

22 → 20 → 40 → 44

TA3 → TA1 → TA2

22 → 44 → 48 → 50

TA3 → TA2 → TA1

22 → 44 → 40 → 44



Initial value = 22

Correct result (according to Serializability): 44, 48, 50 und 60

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Incorrect Interleaving

Initial Value

22

1.

(1a)

o = o + 2;

24

2.

(2a)

o = round(o);

20

3.

(2b)

read access;

20

4.

(3a)

o = o * 4;

80

5.

(1b)

o = o + 2;

82

6.

(3b)

o = o / 2;

41



The interleaving shown above is incorrect, because the result (41) does not correspond to any value that can occur in serial execution (44, 48, 50 and 60)

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Correct Interleaving

Initial Value

22

1.

(3a)

o = o * 4;

88

2.

(2a)

o = round(o);

90 

3.

(1a)

o = o + 2;

92

4.

(3b)

o = o / 2;

46

5.

(2b)

read access;

46

6.

(1b)

o = o + 2;

48

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The interleaving shown here is correct, because the result (48) corresponds to one of the values that can occur by serial execution (44, 48, 50 and 60).

519

Multi-user Mode

What are Isolation Levels?

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Isolation Levels

Isolation Levels define … •

Whether rows that are read, inserted, updated, or deleted by a transaction are available to other transactions running in parallel.

•

Whether read, insert, update, or delete activities of another transaction running in parallel have an impact on the current transaction.

•

Which problems of uncontrolled multi-user mode may occur and which problems must be prevented (by the DBMS)

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Isolation Levels

 4 isolation levels are defined by the SQL Standard in such a way that each level defines which problems may occur and which problems must be prevented by the DBMS.

Isolation Level

Uncommitted Dependency (Dirty Reads)

Non-repeatable Reads

Phantom Problem

0

READ UNCOMMITTED

possible

possible

possible

1

READ COMMITTED

impossible

possible

possible

2

REPEATABLE READ

impossible

impossible

possible

3

SERIALIZABLE

impossible

impossible

impossible

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Multi-user Mode

Which SQL statement can I use to define the isolation level?

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Multi-user Mode

SET TRANSACTION ISOLATION LEVEL    

A certain isolation level is set. You must set the isolation level before the transaction starts. The isolation level is only valid for a single (i.e. next to run) transaction. If different isolation levels are set before the transaction starts, only the last setting is decisive.  The isolation level “SERIALIZABLE” guarantees serializability.

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Multi-user Mode

SET TRANSACTION ISOLATION LEVEL

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Isolation Levels

SET TRANSACTION ISOLATION LEVEL IsolationLevel;

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Isolation Levels

 According to the SQL Standard isolation level SERIALIZABLE is the default.  In SAP HANA the isolation level READ COMITTED is the default.  It is not possible to set isolation level READ UNCOMMITED in SAP HANA.  You can set the remaining three isolation levels as follows: SET TRANSACTION ISOLATION LEVEL READ COMMITTED; SET TRANSACTION ISOLATION LEVEL REPEATABLE READ; SET TRANSACTION ISOLATION LEVEL SERIALIZABLE;

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Synchronization

How do classical DBMS products synchronize the multi-user mode?

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Classical Synchronization

 Classical DBMS products usually use locking for multi-user mode synchronization.  Database objects (schemata, tables, rows) are locked prior to any access (and therefore cannot or only in a limited way be used by other transactions).  Locks are set and released by the DBMS.  User or applications do not need to care of the locks (automation of the DBMS).

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Synchronization

How does SAP HANA synchronize its multi-user mode?

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Synchronization in SAP HANA

The consistency of write access is ensured by X-locks  The affected rows are locked “as usual”. The consistency of read access is ensured by “Multi Version Concurrency Control“ (MVCC)  No S-locks are set when reading rows .  By default, all SELECT statements of the same transaction see the same (possibly obsolete) consistent version of the database state (≈ isolation level REPEATABLE READ).  This version contains all changes that were committed before the start of the transaction (together with all changes done by the transaction itself). This is called “transaction-level snapshot isolation”.  Write access for other transactions running in parallel creates a new version of the database state (which is not visible to the transaction mentioned above.)  Isolation level READ COMMITED corresponds to statement-level snapshot isolation

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Multi-Version Concurreny Control – An Example

Timeline of 5 transactions T1-T5 accessing same record „item D“

T2 T4

T5 starts after T3 is committed  sees V2 When T4 ends (last transaction that still sees V1), V1 becomes obsolete

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V2 open V1 open

committed

read

read

insert

update

T5

item D versions



T3

T1 writes V1 of item D T3 starts and writes V2. V1 still remains visible for transactions that started before T3 ends.  With „transaction-level“ snapshot isolation, T2 and T4 keep seeing V1 of item D (≈ repeatable read)



commit

Tx T1

read

 

commit

committed out-dated

532

Write Locks & Snapshot Isolation – Additional Remarks

 Snapshot Isolation can lead to updates a transaction that started earlier doesn’t see  Transaction-level snapshot isolation increases probability of conflicts on application level

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Transaction T1

write

write

read

re-read

Transaction T2

data record versions

 Primary key & uniqueness constraints are checked considering all existing versions  Deadlock situations can not be prevented

commit

V3 V2 V1

Committed Committed

533

Unit 11: Database Transactions Summary


Explain what a database transaction is and why it is needed • Explain the acronym ACID • Finish database transactions in HANA using SQL statements • Describe issues that arise if transactions are not mutually isolated •

Understand how classical database systems synchronize concurrent transactions • Understand how HANA synchronizes concurrent transactions

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Summary

535

Summary

Motivation And Basic Concepts Reading Data From A Table Or View Aggregating Data

Reading Data From Multiple Tables Changing Data Stored in Tables Defining How Data Is Stored Using Views For Data Access Defining Data Access

Understanding NULL Values Database Transactions © 2013


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Open Questions

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Done!

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Meta Data

How can I access meta data?

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Accessing Meta Data

 

The DBMS provides metadata views (catalog views). This is a requirement of the SQL standard. You can use SQL to read the meta data.



Which tables belong to schema Schema1?

SELECT Table_Name FROM Sys.Tables WHERE Schema_Name = 'SCHEMA1';

© 2013


TABLE_NAME ---------CONTACT OFFICIAL OWNER STOLEN OWNER_EU CAR CAR_EU

540

Accessing Meta Data

 Which columns are in table Car of schema Schema1?

SELECT FROM WHERE ORDER BY

Column_Name, Position, Data_Type_Name Sys.Table_Columns Schema_Name = 'SCHEMA1' AND Table_Name ='CAR' Position;

COLUMN_NAME ----------CARID PLATENUMBER BRAND COLOR HP OWNER

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POSITION -------1 2 3 4 5 6

DATA_TYPE_NAME -------------VARCHAR VARCHAR VARCHAR VARCHAR INTEGER VARCHAR

541

Accessing Meta Data

 How many columns do the tables of schema Schema1 have?

SELECT FROM WHERE GROUP BY ORDER BY

Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name 2 TABLE_NAME ---------CONTACT STOLEN OWNER OWNER OWNER_EU CAR CAR_EU

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NUMCOLUMNS ---------2 2 4 5 5 6 7 542

Accessing Meta Data

 Which tables of schema Schema1 have the fewest columns?

SELECT FROM WHERE GROUP BY HAVING

TABLE_NAME ---------CONTACT STOLEN © 2013

Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name COUNT(*) = (SELECT TOP 1 COUNT(*) FROM Sys.Table_Columns WHERE Schema_Name = 'SCHEMA1' GROUP BY Table_Name ORDER BY 1); NUMCOLUMNS ---------2 2


543

Accessing Meta Data

 Which tables of schema Schema1 have the most columns?

SELECT FROM WHERE GROUP BY HAVING

TABLE_NAME ---------CAR_EU

© 2013

Table_Name, COUNT(*) AS NumColumns Sys.Table_Columns Schema_Name = 'SCHEMA1' Table_Name COUNT(*) = (SELECT TOP 1 COUNT(*) FROM Sys.Table_Columns WHERE Schema_Name = 'SCHEMA1' GROUP BY Table_Name ORDER BY 1 DESC); NUMCOLUMNS ---------7


544

Meta Data

Which meta data views are available?

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Accessing Meta Data

 There are several meta data views …

Meta Data View

Information contained

Sys.Tables

Which tables are available?

Sys.Table_Columns

Which columns are in the tables?

Sys.Views

Which views are available?

Sys.View_Columns

Which columns are in the views?

Sys.Indexes

Which indexes are available?

Sys.Index_Columns

Which columns are in the indexes?

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HA150.pdf

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