5th Sem Dbms Lab Programs

Anjuman Engineering College, Bhatkal 1 Database Applications Laboratory

1. Consider the Insurance database given below. The primary keys are underlined and the data types are specified: PERSON (driver-id:string,name:string,address:string) CAR (Regno:string,model:string,year:int) ACCIDENT (report-number:int,date:date,location:string) OWNS (driver-id:string,regno:string) PARTICIPATED (driver-id:string,regno:string,report-number:int,damage-amount:int) i) create the above tables by properly specifying the primary keys and the foreign keys ii) Enter atleast five tuples for each relation iii) Demonstrate how you a. Update the damage amount for the car with a specific regno in accident with report number 12 to 25000 b. Add a new accident to the database iv) Find the total number of people who owned cars that were involved in accidents in 2006. v) Find the number of accidents in which cars belonging to a specific model were involved. vi) Generation of suitable reports vii) Create suitable front end for querying and display the results create table person ( driverid varchar(5) primary key, name varchar(30), address varchar(50) ); create table car ( regno varchar(5) primary key, model varchar(25), year int ); create table accident ( reportno int primary key, date1 date, location varchar(30) ); create table owns ( driverid references person, regno references car );


create table participated ( driverid references person, regno references car, reportno references accident, damages int ); insert into person values (1,'abc','hkbkce,nagawara'); insert into person values (2,'bcd','atria,yelahanka'); insert into person values (3,'ade','ramahiah,near iisc'); insert into person values (4,'cfg','mvjce,whitefield'); insert into person values (5,'rte','pesit,somewhere'); insert into car values ('a1','merc',2007); insert into car values ('a2','porsche',2005); insert into car values ('b2','rolls royce',2006); insert into car values ('b4','ferrari',1999); insert into car values ('c1','bentley',2000); insert into car values ('d2','maybach',2007); insert into accident values (1,'1-jan-2006','domlur'); insert into accident values (2,'28-mar-2006','m.g.road'); insert into accident values (3,'2-dec-2006','r.t.nagar'); insert into accident values (4,'5-jan-2007','yelahanka'); insert into accident values (5,'26-jan-2007','koramangla'); insert into accident values (6,'4-feb-2007','majestic'); insert into owns values (1,'a1'); insert into owns values (2,'a2'); insert into owns values (2,'c1'); insert into owns values (3,'b2'); insert into owns values (4,'b4'); insert into owns values (5,'d2'); insert into participated values (1,'a1',1,500); insert into participated values (2,'a2',1,2000); insert into participated values (3,'b2',1,1000); insert into participated values (1,'a1',2,1500); insert into participated values (4,'c1',2,800); insert into participated values (5,'d2',3,750); insert into participated values (2,'c1',3,600); insert into participated values (1,'a1',3,200); insert into participated values (5,'d2',4,1000); insert into participated values (2,'a2',5,1200); insert into participated values (3,'b2',6,10000); insert into participated values (2,'a1',6,5000);


select * from person; DRIVERID NAME ---------------- ---------1 abc 2 bcd 3 ade 4 cfg 5 rte

ADDRESS ---------------------hkbkce,nagawara atria,yelahanka ramahiah,near iisc mvjce,whitefield pesit,somewhere

select * from car; REGNO MODEL YEAR ----------- ---------------- -----------a1 merc 2007 a2 porsche 2005 b2 rolls royce 2006 b4 ferrari 1999 c1 bentley 2000 d2 maybach 2007 select * from accident; REPORTNO DATE1 ----------------- -------------1 01-JAN-06 2 28-MAR-06 3 02-DEC-06 4 05-JAN-07 5 26-JAN-07 6 04-FEB-07

LOCATION ----------------------domlur m.g.road r.t.nagar yelahanka koramangla majestic

select * from owns; DRIVERID REGNO -------------- ----------1 a1 2 a2 2 c1 3 b2 4 b4 5 d2 select * from participated; DRIVERID REGNO REPORTNO DAMAGES ---------------- ----------- ----------------- ---------------1 a1 1 500 2 a2 1 2000 3 b2 1 1000 1 a1 2 1500 4 c1 2 800 5 d2 3 750 2 c1 3 600 1 a1 3 200


5 2 3 2

d2 a2 b2 a1

4 5 6 6

1000 1200 10000 5000

iii.a Update the damage amount for the car with a specific regno in accident with report number 12 to 25000 update participated set damages=25000 where reportno=4 and regno='d2'; 1 row updated select * from participated; DRIVERID REGNO REPORTNO DAMAGES ---------------- ----------- ----------------- ---------------1 a1 1 500 2 a2 1 2000 3 b2 1 1000 1 a1 2 1500 4 c1 2 800 5 d2 3 750 2 c1 3 600 1 a1 3 200 5 d2 4 25000 2 a2 5 1200 3 b2 6 10000 2 a1 6 5000 iii.b Add a new accident to the database insert into accident values (7,'16-mar-2007','nagawara'); 1 row updated insert into participated values (2,'a1',7,15000); 1 row updated select * from accident; REPORTNO DATE1 ----------------- -------------1 01-JAN-06 2 28-MAR-06 3 02-DEC-06 4 05-JAN-07 5 26-JAN-07 6 04-FEB-07 7 16-MAR-07

LOCATION ----------------------domlur m.g.road r.t.nagar yelahanka koramangla majestic nagawara


select * from participated; DRIVERID REGNO REPORTNO DAMAGES ---------------- ----------- ----------------- ---------------1 a1 1 500 2 a2 1 2000 3 b2 1 1000 1 a1 2 1500 4 c1 2 800 5 d2 3 750 2 c1 3 600 1 a1 3 200 5 d2 4 25000 2 a2 5 1200 3 b2 6 10000 2 a1 6 5000 2 a1 7 15000 iv. Find the total number of people who owned cars that were involved in accidents in 2006 select count(distinct p.driverid) from accident a,owns o,participated p where date1 like '%06' and a.reportno=p.reportno and p.regno=o.regno and o.driverid=p.driverid; COUNT(DISTINCTP.DRIVERID) --------------------------------------------4 v. Find the number of accidents in which cars belonging to a specific model were involved. select count(*) from car c,participated p where model='porsche' and c.regno=p.regno; COUNT(*) -------------2


2. Consider the following relations for an order processing database application in a company. CUSTOMER (Cust #: int, Cname: string, City: string) ORDER (Order #: int, Odate: date, Cust #: int, Ord-Amt: int) ORDER-ITEM (Order #: int, Item #: int, qty: int) ITEM (Item #: int, Unit Price: int) SHIPMENT (Order #: int, Warehouse #: int, Ship-Date: date) WAREHOUSE (Warehouse #: int, City: string) i) iii) iv) v) vi) vii)

Create the above tables by properly specifying the primary keys and the foreign keys. ii) Enter at least five tuples for each relation. Produce a listing: CUSTNAME, NO_OF_ORDERS, AVG_ORDER_AMT, where the middle column is the total number of orders by the customer and the last column is the average order amount for that customer. List the Order# for the orders that were shipped from all the warehouses that the company has in a specific city. Demonstrate how you delete Item# 10 from the ITEM table and make that field null in the ORDER-ITEM table. Generation of suitable reports. Create a suitable front end for querying and displaying the results.

create table customer ( cno int primary key, cname varchar(10), city varchar(15) ); create table order1 ( ono int primary key, odate date, cno references customer, oamt int ); create table item ( ino int primary key, uprice int ); create table oitem ( ono references order1, ino references item on delete set null, qty int );


create table warehouse ( wno int primary key, city varchar(15) ); create table shipment ( ono references order1, wno references warehouse, sdate date ); insert into customer values(1,'abc','bangalore'); insert into customer values(2,'cde','mysore'); insert into customer values(3,'def','chennai'); insert into customer values(4,'efg','mumbai'); insert into customer values(5,'adf','kolkata'); insert into order1 values(1,'1-jan-2006',1,20000); insert into order1 values(2,'26-mar-2006',2,10000); insert into order1 values(3,'12-jun-2006',1,5000); insert into order1 values(4,'15-sep-2006',3,9000); insert into order1 values(5,'5-jan-2007',4,2500); insert into order1 values(6,'10-jan-2007',4,2400); insert into order1 values(7,'3-mar-2007',5,3500); insert into item values(1,500); insert into item values(2,300); insert into item values(3,2500); insert into item values(4,800); insert into item values(5,700); insert into oitem values(1,1,40); insert into oitem values(2,1,20); insert into oitem values(3,3,2); insert into oitem values(5,3,1); insert into oitem values(4,2,30); insert into oitem values(6,4,3); insert into oitem values(7,5,5); insert into warehouse values(100,'bangalore'); insert into warehouse values(101,'chennai'); insert into warehouse values(102,'mumbai'); insert into warehouse values(103,'kolkata'); insert into warehouse values(104,'mysore'); insert into shipment values(1,100,'3-jan-2006'); insert into shipment values(2,100,'28-mar-2006'); insert into shipment values(3,101,'13-jun-2006');


insert into shipment values(4,102,'18-sep-2006'); insert into shipment values(5,103,'11-jan-2007'); insert into shipment values(6,104,'13-jan-2007'); insert into shipment values(7,103,'3-mar-2007'); select * from customer; CNO CNAME CITY ---------- ---------- --------------1 abc bangalore 2 cde mysore 3 def chennai 4 efg mumbai 5 adf kolkata select * from order1; ONO ODATE CNO OAMT ---------- ---------------- ---------- ---------1 01-JAN-06 1 20000 2 26-MAR-06 2 10000 3 12-JUN-06 1 5000 4 15-SEP-06 3 9000 5 05-JAN-07 4 2500 6 10-JAN-07 4 2400 7 03-MAR-07 5 3500 select * from item; INO UPRICE ---------- ---------1 500 2 300 3 2500 4 800 5 700 select * from oitem; ONO INO QTY ---------- ---------- ---------1 1 40 2 1 20 3 3 2 5 3 1 4 2 30 6 4 3 7 5 5


select * from warehouse; WNO CITY ---------- --------------100 bangalore 101 chennai 102 mumbai 103 kolkata 104 mysore select * from shipment; ONO WNO SDATE ---------- -------- -----------1 100 03-JAN-06 2 100 28-MAR-06 3 101 13-JUN-06 4 102 18-SEP-06 5 103 11-JAN-07 6 104 13-JAN-07 7 103 03-MAR-07 iii. Produce a listing: CUSTNAME, NO_OF_ORDERS, AVG_ORDER_AMT, where the middle column is the total number of orders by the customer and the last column is the average order amount for that customer. select c.cname,count(q.qty),avg(oamt) from customer c, order1 o, oitem q where q.ono=o.ono and c.cno=o.cno group by cname; CNAME COUNT(Q.QTY) ------------ ---------------------abc 2 adf 1 cde 1 def 1 efg 2

AVG(OAMT) -------------------12500 3500 10000 9000 2450

iv. List the Order# for the orders that were shipped from all the warehouses that the company has in a specific city. select w.city, s.wno, s.ono from warehouse w, shipment s where w.wno=s.wno and w.city='mumbai'; CITY WNO ONO --------------- ---------- ---------mumbai 102 4


v. Demonstrate how you delete Item# 10 from the ITEM table and make that field null in the ORDER-ITEM table delete from item where ino=5; 1 row deleted select * from item; INO UPRICE ---------- ---------1 500 2 300 3 2500 4 800


3. Consider the following database of student enrollement in courses and books adopted for each course . STUDENT (regno :string , name : string , major : string , bdate : int) COURSE (course# : int , cname : string , dept : string) ENROLL ( regno : string , course#: int , sem : int , marks : int ) BOOK_ADAPTION ( course#: int , sem : int , book_isbn :int) TEXT( book_isbn : int , book-title : string , publisher : string , author : string). i) Create the above tables by properly specifying the primary keys and the foreign key . ii) Enter atleast five tuples for each relation . iii)Demonstrate how you add a new text book to the database and make this book be adopted by some department. iv) Produce a list of text books( include course # ,book_isbn,book-title) in the alphabetical order for courses offered by the cs department that use more than 2 books. v) List any department that has all its adopted books published by specific publisher. vi) Generation of suitable reports. vii) Create suitable front end for querying and display the results create table student ( regno varchar(5) primary key, name varchar(10), major varchar(10), bdate date ); create table course ( cno int primary key, cname varchar(10), dept varchar(10) ); create table enroll ( regno references student, cno references course, sem int primary key, marks int ); create table text ( isbn int primary key, title varchar(15), publisher varchar(10), author varchar(10) );


create table bad ( cno references course, sem references enroll, isbn references text ); insert into student values('cs42','mikhil','cse','17-dec-1986'); insert into student values('cs48','mujeeb','cse','02-sep-1986'); insert into student values('ec26','pradeep','cse','16-aug-1987'); insert into student values('ee37','majid','ise','28-may-1986'); insert into student values('is48','wajid','ise','28-may-1986'); insert into course values(1,'.net','computer'); insert into course values(2,'j2ee','computer'); insert into course values(3,'mis','infosci'); insert into course values(4,'fs','infosci'); insert into course values(5,'oracle','computer'); insert into enroll values('cs42',1,6,98); insert into enroll values('cs48',2,3,97); insert into enroll values('ec26',5,5,50); insert into enroll values('is48',3,7,90); insert into enroll values('ee37',4,4,80); insert into text values(101,'let us c','lpe','fahad'); insert into text values(102,'c++','abc','mujeeb'); insert into text values(103,'oracle','def','othman'); insert into text values(104,'.net','lpe','naushad'); insert into text values(105,'j2ee','pearson','mikhil'); insert into bad values(1,6,101); insert into bad values(1,6,103); insert into bad values(1,6,102); insert into bad values(4,4,104); insert into bad values(5,7,105); select * from student; REGNO NAME MAJOR BDATE ---------- ---------- ------------- ------------cs42 mikhil cse 17-DEC-86 cs48 mujeeb cse 02-SEP-86 ec26 pradeep cse 16-AUG-87 ee37 majid ise 28-MAY-86 is48 wajid ise 28-MAY-86


select * from course; CNO CNAME DEPT ---------- -------------- -----------1 .net computer 2 j2ee computer 3 mis infosci 4 fs infosci 5 oracle computer select * from enroll; REGNO CNO SEM MARKS ------------ ---------- ---------- -----------cs42 1 6 98 cs48 2 3 97 ec26 5 5 50 is48 3 7 90 ee37 4 4 80 select * from text; ISBN TITLE PUBLISHER AUTHOR ---------- --------------- ------------------ --------------101 let us c lpe fahad 102 c++ abc mujeeb 103 oracle def othman 104 .net lpe naushad 105 j2ee pearson mikhil select * from bad; CNO SEM ISBN ---------- ---------- ---------1 6 101 1 6 103 1 6 102 4 4 104 5 7 105 iii.Demonstrate how you add a new text book to the database and make this book be adopted by some department. insert into text values(106,'JAVA','pearson','Avril'); insert into bad values(2,7,106); select * from text; ISBN TITLE PUBLISHER AUTHOR ---------- --------------- ------------------ --------------101 let us c lpe fahad 102 c++ abc mujeeb 103 oracle def othman 104 .net lpe naushad 105 j2ee pearson mikhil 106 JAVA pearson Avril


select * from bad; CNO SEM ISBN ---------- ---------- ---------1 6 101 1 6 103 1 6 102 4 4 104 5 7 105 2 7 106 iv. Produce a list of text books( include course # ,book_isbn,book-title) in the alphabetical order for courses offered by the cs department that use more than 2 books. SELECT s.cno,t.title,b.isbn FROM course s,bad b,text t where s.cno=b.cno AND b.isbn=t.isbn AND s.dept='computer' AND b.cno in (select b.cno from bad b,course s where b.cno=s.cno group by b.cno having count(*)>2) order by t.title; CNO TITLE ISBN ---------- --------------- ---------1 c++ 102 1 let us c 101 1 oracle 103 v. List any department that has all its adopted books published by specific publisher. select s.dept, t.publisher from course s, bad b, text t where s.cno=b.cno and t.isbn=b.isbn and publisher='lpe'; DEPT ------------computer infosci

PUBLISHER -------------------lpe lpe


4. Consider the following relations for the details maintained by a book dealer. AUTHOR (Author-id: int, Name: string, City: string, Country: string) PUBLISHER (Publisher-id: int, Name: string, City: string, Country: string) CATALOG (Book-id: int, title: string, author-id: int, Publisher-id: int, Category-id: int, Year: int, Price: int) CATEGORY (Category-id: int, Description: string) ORDER-DETAILS (Order-no : int, Book-id: int, Quantity: int) i. Create the above tables by properly specifying the primary keys and the foreign keys. ii. Enter at least five tuples for each relation. iii. Give the details of the authors who have 2 or more books in the catalog and the price of the books is greater than the average price of the books in the catalog and the year of publication is after 2000. iv. Find the author of the book which has maximum sales. v. Demonstrate how you increase the price of books published by a specific publisher by 10%. vi. Generation of suitable reports. vii. Create a suitable front end for querying and displaying the results. create table author ( aid int primary key, name varchar(15), city varchar(15), country varchar(15) ); create table publisher ( pid int primary key, name varchar(15) , city varchar(15) , country varchar(15) ); create table category ( cid int primary key, description varchar(15) ); create table catalog ( bid int primary key, title varchar(15), aid references author, pid references publisher, cid references category, year int, price int );


create table odetails ( ono int, bid references catalog, qty int ); insert into author values (1,'mik','bang','ind'); insert into author values (2,'muj','bang','ind'); insert into author values (3,'prad','tri','aus'); insert into author values (4,'maj','anan','ame'); insert into author values (5,'waj','anan','euro'); insert into publisher values (101,'pearson','bang','ind'); insert into publisher values (102,'tata','mumbai','aus'); insert into publisher values (103,'sapna','che','euro'); insert into publisher values (104,'abc','tri','ame'); insert into publisher values (105,'xyz','anan','ind'); insert into category values (1001,'computer'); insert into category values (1002,'electronics'); insert into category values (1003,'maths'); insert into category values (1004,'science'); insert into category values (1005,'electrical'); insert into catalog values (111,'lib1',1,101,1001,2002,500); insert into catalog values (112,'lib2',2,102,1002,2000,800); insert into catalog values (113,'lib3',3,103,1003,2003,200); insert into catalog values (114,'lib4',4,104,1001,2006,350); insert into catalog values (115,'lib5',5,105,1004,2007,100); insert into catalog values (116,'lib6',2,103,1005,2007,600); insert into catalog values (117,'lib7',2,105,1002,2007,450); insert into catalog values (118,'lib8',1,101,1001,2002,500); insert into odetails values (1,111,2); insert into odetails values (2,112,3); insert into odetails values (3,111,5); insert into odetails values (4,113,1); insert into odetails values (5,114,2); insert into odetails values (6,115,1); insert into odetails values (1,114,2); insert into odetails values (2,113,2); select * from author; AID NAME CITY COUNTRY ---------- --------------- --------------- --------------1 mik bang ind 2 muj bang ind 3 prad tri aus 4 maj anan ame 5 waj anan euro


select * from publisher; PID NAME CITY COUNTRY ---------- --------------- --------------- --------------101 pearson bang ind 102 tata mumbai aus 103 sapna che euro 104 abc tri ame 105 xyz anan ind select * from category; CID DESCRIPTION ---------- --------------------1001 computer 1002 electronics 1003 maths 1004 science 1005 electrical select * from catalog; BID TITLE AID PID CID YEAR PRICE ------- ------------ --------- -------- -------- ---------- ---------111 lib1 1 101 1001 2002 500 112 lib2 2 102 1002 2000 800 113 lib3 3 103 1003 2003 200 114 lib4 4 104 1001 2006 350 115 lib5 5 105 1004 2007 100 116 lib6 2 103 1005 2007 600 117 lib7 2 105 1002 2007 450 118 lib8 1 101 1001 2002 500 select * from odetails; ONO BID QTY ---------- ---------- ---------1 111 2 2 112 3 3 111 5 4 113 1 5 114 2 6 115 1 1 114 2 2 113 2 iii. Give the detail of the author who have 2 or more books in the catalog and the price of the book is greater than the average price of the book in the catalog and the year of publication after 2000 select name,city,country from author where aid in (select aid from catalog where year>2000 and price>(select avg(price) from catalog)


group by aid having count(*)>1); NAME CITY COUNTRY ----------- --------------- --------------mik bang ind muj bang ind iv. Find the author of the book which has maximum sales select a.aid, name from author a, catalog c where a.aid=c.aid and c.bid=(select bid from odetails group by bid having sum(qty)=(select max(sum(qty)) from odetails group by bid)); AID NAME ---------- --------------1 mik v. Demonstrate how to increase the price of books published by a specific publisher by 10% update catalog set price=1.1*price where pid in (select pid from publisher where name='abc'); select * from catalog; BID TITLE AID PID CID YEAR PRICE ------- ------------ --------- -------- -------- ---------- ---------111 lib1 1 101 1001 2002 500 112 lib2 2 102 1002 2000 800 113 lib3 3 103 1003 2003 200 114 lib4 4 104 1001 2006 385 115 lib5 5 105 1004 2007 100 116 lib6 2 103 1005 2007 600 117 lib7 2 105 1002 2007 450 118 lib8 1 101 1001 2002 500


5. Consider the following database for a banking enterprise BRANCH (branch_name: string, branch_city: string, assets: real) ACCOUNT (accno: int, branch_name: string, balance: real) CUSTOMER (customer_name: string, customer_street: string, city:string) DEPOSITOR (customer_name: string, accno: int) LOAN (loan_number: int, branch_name: string, amount: real) BORROWER (customer_name: string, loan_number: int) i) ii) iii) iv) v)

Create the above tables by properly specifying the primary keys and the foreign keys. Enter atleast five tuples for each relation. Find all the customers who atleast two accounts at the MAIN branch. Find all the customers who have an account at all branches located in a specific city. Demonstrate how you delete all account tuples at every branch located in a specific city. vi) Generation of suitable reports. vii) Create suitable front end for querying and displaying the results. create table branch ( bname varchar(10) primary key, city varchar(10), assets real ); create table account ( accno int primary key, bname varchar(10), balance real, foreign key(bname) references branch(bname) ); create table cust ( cname varchar(10) primary key, cstreet varchar(10) not null, city varchar(10) not null ); create table depositor ( cname varchar(10), accno int, primary key(cname,accno), foreign key(accno) references account(accno) on delete cascade, foreign key(cname) references cust(cname) on delete cascade );


create table loan ( lno int primary key, bname varchar(10) not null, amt real, foreign key(bname) references branch(bname) on delete cascade ); create table borrower ( cname varchar(10), lno int, primary key(cname,lno), foreign key(cname) references cust(cname) on delete cascade, foreign key(lno) references loan(lno) on delete cascade ); insert into branch values('abc','bang',1200000); insert into branch values('def','che',2000000); insert into branch values('abn','mum',330000); insert into branch values('xyz','hyd',555555); insert into branch values('mno','bang',9999999); insert into insert into insert into insert into insert into insert into

account values(1,'abc',25000); account values(2,'def',12000); account values(3,'def',1000); account values(4,'abn',10000); account values(5,'mno',600000); account values(6,'xyz',50000);

insert into cust values('mik','ab','bang'); insert into cust values('muj','cd','bang'); insert into cust values('maj','ef','che'); insert into cust values('waj','xy','del'); insert into cust values('prad','lm','mum'); insert into cust values('now','op','hyd'); insert into depositor values('mik',2); insert into depositor values('muj',1); insert into depositor values('muj',5); insert into depositor values('prad',4); insert into depositor values('maj',3); insert into depositor values('waj',6); insert into depositor values('mik',3); insert into loan values(1,'abc',5000); insert into loan values(2,'def',1500); insert into loan values(3,'abn',10000); insert into loan values(4,'xyz',3500); insert into loan values(5,'mno',20000);


insert into borrower values('mik',2); insert into borrower values('muj',1); insert into borrower values('prad',3); insert into borrower values('maj',4); insert into borrower values('waj',5); select *from branch; BNAME CITY ------------------abc bang def che abn mum xyz hyd mno bang

ASSETS ---------1200000 2000000 330000 555555 9999999

select *from account; ACCNO BNAME BALANCE ---------------------------1 abc 25000 2 def 12000 3 def 1000 4 abn 10000 5 mno 600000 6 xyz 50000 select *from cust; CNAME CSTREET --------------------mik ab muj cd maj ef waj xy prad lm now op select *from depositor; CNAME ACCNO ---------- -----------mik 2 muj 1 muj 5 prad 4 maj 3 waj 6 mik 3

CITY ---------bang bang che del mum hyd


select *from loan; LNO BNAME AMT ---------- ------------ ---------1 abc 5000 2 def 1500 3 abn 10000 4 xyz 3500 5 mno 20000 select *from borrower; CNAME LNO ------------- ---------mik 2 muj 1 prad 3 maj 4 waj 5 iii. Find all the customers who atleast two accounts at the MAIN branch. select cname from account a,depositor d where a.accno=d.accno and bname='def' group by cname having count(*)>1; CNAME ---------mik iv. Find all the customers who have an account at all branches located in a specific city select cname from cust c where not exists (select bname from branch where city='bang' minus select bname from depositor d,account a where d.accno=a.accno and d.cname=c.cname) and exists (select bname from branch where city='bang'); CNAME -----------muj v. Demonstrate how you delete all account tuples at every branch located in a specific city. delete from account where bname in (select bname from branch where city='che'); 2 rows deleted. select *from account; ACCNO BNAME BALANCE ----------- ------------ --------------1 abc 25000 4 abn 10000 5 mno 600000 6 xyz 50000 **************

5th Sem Dbms Lab Programs

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