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Database Management System Laboratory (10MCA38)

Ganesh Hegde [email protected] facebook.com/hegde.ganesh7

1

PROGRAM NO: 1

Consider the following relations for College Database:

Student (snum: integer, sname: string, major: string, level: string, age:

integer)

Class (name: string, meets at: string, room: string, fid: integer)

Enrolled (snum: integer, cname: string)

Faculty (fid: integer, fname: string, deptid: integer)

The meaning of these relations is straightforward; for example, Enrolled has

one record per student-class pair such that the student is enrolled in the

class. Level is a two character code with 4 different values (example:

Junior: JR etc)

Write the following queries in SQL. No duplicates should be printed in any of

the answers.

i. Find the names of all Juniors (level = JR) who are enrolled in a

class taught by Prof. Harshith

ii. Find the names of all classes that either meet in room R128 or

have five or more Students enrolled.

iii. Find the names of all students who are enrolled in two classes

that meet at the same time.

iv. Find the names of faculty members who teach in every room in

which some class is taught.

v. Find the names of faculty members for whom the combined

enrollment of the courses that they teach is less than five.

Table Creation:

SQL>CREATE TABLE GSTUDENT(snum number(5) primary key,sname varchar2(10),major

varchar2(10),

levels varchar2(2),age number(2));

SQL>CREATE TABLE FACULTY(fid number(5) primary key,fname varchar2(10),deptid

references dept(deptid));

SQL>CREATE TABLE CLASS(cname varchar2(10) primary key,meetsat varchar2(10),room

varchar2(5), fid references faculty(fid));

SQL>CREATE TABLE ENROLLED(snum references student(snum),cname references

class(cname));



2

Inserting Values:

INSERT INTO STUDENT VALUES(101,'adi','maths','jr','21');

SELECT * FROM STUDENT;

SNUM SNAME MAJOR LEVELS AGE

------- --------- --------- ------- -----

101 adi maths jr 21

102 arul math sr 22

103 bhavya se sr 27

104 divya or jr 20

105 kota se sr 27

INSERT INTO FACULTY VALUES(501,'Harshith','401');

SELECT * FROM FACULTY;

FID FNAME DEPTID

--------- ---------- ---------- ---

501 Harshith 401

502 sathya 402

503 kc 401

504 illangos 405

505 nirmala 403

506 asha 404

507 usha 402

INSERT INTO CLASS VALUES('3a','10.00','r128','501');

SELECT * FROM CLASS;

CNAME MEETSAT ROOM FID

--------- ---------- --- ------- --------- -----

3a 10.00 r128 501

3b 10.00 ro2 502

4a 11.00 r03 503

4b 11.00 r128 502

5a 12.00 r04 504



3

INSERT INTO ENROLLED VALUES(101,'3a');

SELECT * FROM ENROLLED;

SNUM CNAME

--------- ---------- ------

101 3a

101 3b

101 4a

101 4b

102 3a

102 3b

103 3a

104 4a

105 5a

101 5b

Queries:

I) Create the above tables by properly specifying the primary keys and the

foreign keys.

SQL> SELECT SNAME FROM STUDENT S,ENROLLEDE,CLASSC,FACULTY F WHERE

S.SNUM=E.SNUM AND E.CNAME=C.CNAME AND C.FID=F.FID AND

FNAME='HARSHITH' AND LEVELS='JR';

SNAME

--------- ---

ADI

II) Find the names of all classes that either meet in room R128 or have five or

more Students enrolled.

SQL> SELECT DISTINCT C.CNAME FROM CLASS C,ENROLLED E WHERE C.CNAME=E.CNAME

AND ROOM='R128' OR C.CNAME IN (SELECT CNAME FROM ENROLLED GROUP BY

CNAME HAVING COUNT(SNUM)>=5);

CNAME

--------- ---

3A

4B



4

III) Find the names of all students who are enrolled in two classes that meet at

the same time.

SQL>SELECT DISTINCT S.SNAME FROM STUDENT S WHERE S.SNUM IN

(SELECT E1.SNUM FROM ENROLLED E1,ENROLLED E2,CLASS C1,CLASS C2

WHERE E1.CNAME=C1.CNAME AND E2.CNAME=C2.CNAME AND E1.SNUM=E2.SNUM

AND C1.CNAME<>C2.CNAME AND C1.MEETSAT=C2.MEETSAT);

SNAME

--------- ----

ADI

ARUN

IV) Find the names of faculty members who teach in every room in which some

class is taught.

SQL>SELECT F.FNAME FROM FACULTY F WHERE NOT EXISTS

((SELECT C.ROOM FROM CLASS C) MINUS (SELECT C1.ROOM FROM CLASS C1 WHERE

C1.FID=F.FID));

FNAME

--------- ---

MANJU

KALA

V) Find the names of faculty members for whom the combined enrollment of the

courses that they teach is less than five.

SQL>SELECT DISTINCT F.FNAME FROM FACULTY F WHERE 5 > (SELECT COUNT (E.SNUM)

FROM CLASS C,ENROLLED E WHERE C.CNAME=E.CNAME AND F.FID=C.FID);

FNAME

--------- --

HARSHITH

KALA

SANTOSH



5

PROGRAM NO:2

The following relations keep track of airline flight information:

Flights (aid: integer, from: string, to: string, distance: integer, Departs:

time, arrives: time, price: real)

Aircraft (aid: integer, aname: string, cruisingrange: integer)

Certified (eid: integer, aid: integer)

Employees (eid: integer, ename: string, salary: integer)

(Note:that the Employees relation describes pilots and other kinds of

employees as well; Every pilot is certified for some aircraft, and only

pilots are certified to fly.)

Write each of the following queries in SQL.

i. Find the names of aircraft such that all pilots certified to operate

them have salaries more than Rs.80, 000.

ii. For each pilot who is certified for more than three aircrafts, find the

eidand the maximum cruisingrangeof the aircraft for which she or he

is certified.

iii. Find the names of pilots whose salary is less than the price of the

cheapest route from Bengaluru to Frankfurt.

iv. For all aircraft with cruisingrangeover 1000 Kms, .find the name of

the aircraft and the average salary of all pilots certified for this

aircraft.

v. Find the names of pilots certified for some Boeing aircraft.

vi. Find the aids of all aircraft that can be used on routes from Bengaluru

to New Delhi.

Table Creation:

SQL> CREATE TABLE AIRCRAFT(aid number(5) primary key,aname varchar2(10),cruisrange

number(10));

Table created.

SQL> CREATE TABLE EMPLOYEES(eid number(5) primary key,ename varchar2(10),salary

number(10));

Table created.

SQL> CREATE TABLE CERTIFIED(eid references employees(eid),aid references

aircraft(aid));

Table created.



6

SQL>CREATE TABLE FLIGHTS(aid references aircraft(aid),source varchar2(10),destn

varchar2(10),distance number(10),departs varchar2(5),arrives varchar2(5),price

number(10,2));

Table created.

Inserting Values:

INSERT INTO AIRCRAFT VALUES('1','Boeing',’500');

SELECT * FROM AIRCRAFT;

AID ANAME CRUISRANGE

--------- ---------- ----------

1 Boeing 500

2 Combat 20000

3 Invader 500

4 Mitchell 500

5 yale 30000

INSERT INTO EMPLOYEES VALUES('100','Arun','2000');

SELECT * FROM EMPLOYEES;

EID ENAME SALARY

--------- ---------- ---------

100 Arun 20000

200 Vijay 45000

300 Bruno 30000

400 Mithun 20000

500 Kunal 55000

600 Jack 85000

700 Jane 15000

INSERT INTO CERTIFIED VALUES('100','1');

SELECT * FROM CERTIFIED;

EID AID

--------- ---------

100 1

100 2

200 3

700 2

300 5

400 5

500 1

500 4

600 1

600 4

600 5

600 2



7

INSERT INTO FLIGHTS ALUES('1','Bangalore','Frankfurt','3000','10.00',’15.00’,’20000’);

SELECT * FROM FLIGHTS;

AID SOURCE DESTN DISTANCE DEPAR ARRIV PRICE

---- ------------ -------- ---------- ------- -------- ------

1 Bangalore Frankfurt 3000 10.00 15.00 20000

2 Delhi New York 5000 6.00 16.00 40000

3 Bangalore Newdelhi 5000 06.35 10.30 4000

4 Bangalor Chennai 1000 12.00 13.00 1200

5 Chennai new delhi 3000 13.00 15.30 4300

2 Bangalore Frankfurt 3000 12.30 18.00 18000

5 Bangalore Newdelhi 5000 13.00 15.30 3500

Queries:

I) Find the names of aircraft such that all pilots certified to operate them have

salaries more than Rs.80, 000.

SQL> SELECT ANAME FROM AIRCRAFT A,CERTIFIEDC,EMPLOYEES E WHERE A.AID=C.AID

AND C.EID=E.EID AND SALARY>80000;

ANAME

----------

BOEING

MITCHELL

COMBAT

II) For each pilot who is certified for more than three aircrafts, find the eidand the

maximum cruisingrangeof the aircraft for which she or he is certified.

SQL> SELECT EID,CRUISRANGE FROM AIRCRAFT A,CERTIFIED C WHERE A.AID=C.AID AND

EID IN (SELECT EID FROM CERTIFIED GROUP BY EID HAVING COUNT(AID)>3) AND

CRUISRANGE IN (SELECT MAX(CRUISRANGE) FROM AIRCRAFT A,CERTIFIED C WHERE

A.AID=C.AID GROUP BY EID HAVING COUNT(C.AID)>3);

EID CRUISRANGE

---- ----------

600 30000

III) Find the names of pilots whose salary is less than the price of the cheapest

route from Bengaluru to Frankfurt

SQL>SELECT DISTINCT ENAME FROM EMPLOYEES E,CERTIFIED C,FLIGHTS F WHERE

E.EID=C.EID AND C.AID=F.AID AND SALARY<(SELECT MIN(PRICE)

FROM FLIGHTS WHERE SOURCE='BANGALORE' AND DESTN='FRANKFORT');

ENAME

----------

JANE



8

IV) For all aircraft with cruisingrangeover 1000 Kms, .find the name of the aircraft

and the average salary of all pilots certified for this aircraft.

SQL>SELECT ANAME,AVG(SALARY) FROM AIRCRAFT A,EMPLOYEES E,CERTIFIED C WHERE

E.EID=C.EID AND C.AID=A.AID

AND CRUISRANGE>1000 GROUP BY ANAME;

ANAME AVG(SALARY)

---------- -----------

COMBAT 40000

YALE 45000

V) Find the names of pilots certified for some Boeing aircraft.

SQL> SELECT ENAME FROM AIRCRAFT A,EMPLOYEES E,CERTIFIED C WHERE E.EID=C.EID

AND C.AID=A.AID AND ANAME='BOIENG';

ENAME

----------

ARUN

KUNAL

JACK

VI) Find the aids of all aircraft that can be used on routes from Bengaluru to New

Delhi.

SQL> SELECT AID FROM FLIGHTS WHERE SOURCE=’BANGALORE’ AND

DESTN=’NEWDELHI’;

AID

---------

3

5



9

PROGRAM NO:3

Consider the following database of student enrollment in courses and books

adapted for each course.

STUDENT (regno:string, name:string, major:string, bdate:date)

COURSE (course#:int, cname:string, dept:string)

ENROLL (regno:string, course#:int, sem:int, marks:int)

BOOK-ADOPTION (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 keys.

(ii) Enter at least 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 two books.

(v) List any department that has all its adopted books published by a

specific publisher.

(vi) Generate suitable reports.

(vii) Create suitable front end for querying and displaying the results.

SOLUTION:

(i) Creating Tables:

CREATE TABLE STUDENT(regno varchar(10) primary key,name varchar(10),

major varchar(10),bdate date);

CREATE TABLE COURSE(cno number(4) primary key, cname varchar(20)

dept varchar(20));

CREATE TABLE ENROLL( regno varchar (10) references STUDENT,

cno number(5) references COURSE,

sem number(1) , marks number(8));

CREATE TABLE TEXT(book_isbn number(10) primary key,book_title varchar(20),

publisher varchar(20),author varchar(15));

CREATE TABLE BOOK_ADOPTION(cno number(5) references COURSE,

sem number(3) , book_isbn number(10) references TEXT);



10

(ii) Inserting Values:

INSERT INTO STUDENT VALUES(‘1MV10MCA09’,’JOHN’,’DATASTRUCTURES’,’20-JAN-88’);

SELECT * FROM STUDENT;

REGNO NAME MAJOR BDATE

--------------- -------------------- --------------------- ---------

1MV10MCA01 JOHN DATA STRUCTURES 20-JAN-88

1MV10MCA02 JOYCE NETWORKS 31-DEC-87

1MV10MCA03 ALEN OS 29-MAY-87

IMV10MCA04 SMITH SYSTEM SOFTWARE 25-NOV-87

IMV10MCA05 JAMES OS 2-DEC-86

INSERT INTO COURSE VALUES(1, 'MCA', 'CS');

SELECT * FROM COURSE;

CNO CNAME DEPT

-------- --------- ----------

1 MCA CS

2 MCA CS

3 MBA BUSINESS

4 CIVIL BE

5 MECH BE

INSERT INTO ENROLL VALUES(1MV10MCA01, 1,3, 90);

SELECT * FROM ENROLL;

REGNO CNO SEM MARKS

----------------- --------- --------- ------------

1MV10MCA01 1 3 90

1MV10MCA02 2 4 90

1MV10MCA03 3 3 90

IMV10MCA04 4 2 97

IMV10MCA05 5 2 93

INSERT INTO TEXT VALUES(111675,’DATA STRUCTURES’,’PEARSON’,’CHRIS);



11

SELECT * FROM TEXT;

BOOK_ISBN BOOK_TITLE PUBLISHER AUTHOR

--------------- ----------------------- ---------------- ------------------

111675 DATA STRUCTURES PEARSON CHRIS

111654 NETWORKS MC GRAW TANENBAUM

111425 OPERATING SYS PEARSON SOMER

111980 DATABASE PEARSON ELMASRI

111990 DS-2 SCHAND PADMA REDDY

INSERT INTO BOOK-ADOPTION VALUES(1,3,111675);

SELECT * FROM BOOK_ADOPTION;

CNO SEM BOOK_ISBN

-------- -------- ------------

1 3 111675

2 4 111654

3 3 111425

1 4 111980

4 2 111990

III) Demonstrate how you add a new text book to the database and make this

book to be adopted by some department

INSERT INTO TEXT VALUES (&book-isbn, '&book-title', '&publisher', '&author');

Enter value for book-isbn: 111449

Enter value for book-title: SYSTEM SOFTWARE

Enter value for publisher: MC GRAW

Enter value for author: LELAND

INSERT INTO BOOK_ADOPTION VALUES(&cno, &sem, &book-isbn);

Enter value for courseno: 5

Enter value for sem: 2

Enter value for book-isbn: 111449



12

SELECT * FROM TEXT;

BOOK_ISBN BOOK_TITLE PUBLISHER AUTHOR

--------------- ----------------------- ---------------- ------------------

111675 DATA STRUCTURES PEARSON CHRIS

111654 NETWORKS MC GRAW TANENBAUM

111425 OPERATING SYS PEARSON SOMER

111980 DATABASE PEARSON ELMASRI

111990 DS-2 SCHAND PADMA REDDY

111449 SYSTEM SOFTWARE MC GRAW LELAND

SELECT * FROM BOOK-ADOPTION;

CNO SEM BOOK_ISBN

------ ------ --------------

1 3 111675

2 4 111654

3 3 111425

1 4 111980

4 2 111990

5 2 111449

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 two books.

SELECT COURSE.CNO,TEXT.BOOK_ISBN,BOOK_TITLE FROM COURSE, TEXT,

BOOK_ADOPTION WHERE COURSE.CNO=BOOK_ADOPTION.CNO AND

TEXT.BOOK_ISBN=BOOK_ADOPTION.BOOK_ISBN AND COURSE.DEPT=’CS’

GROUP BY COURSE.CNO,TEXT.BOOK_ISBN,BOOK_TITLE HAVING

COUNT(DEPT)>=1;

CNO BOOK_ISBN BOOK_TITLE

----- -------------- ----------------------

1 111675 DATA STRUCTURES

1 111980 DATABASE

2 111654 NETWORKS

V) List any department that has all its adopted books published by a specific

publisher.

SELECT DEPT FROM COURSE C,BOOK_ADOPTION B,TEXT T WHERE C.CNO=B.CNO

AND T.BOOK_ISBN=B.BOOK_ISBN AND T.PUBLISHER='SUMUKHA' GROUP BY DEPT;

DEPT PUBLISHER

-------------- ----------------

CS PEARSON

BUSINESS PEARSON

CS PEARSON



13

(vi) Generate suitable reports

SQL> SET LINESIZE 70

SQL> TTITLE 'REPORT GENERATION|STUDENT DETAILS';

SQL> BTITLE 'END OF REPORT';

SQL> COLUMN REGNO HEADING 'REGISTER|NUMBER';

SQL> COLUMN NAME HEADING 'STUDENT|NAME';

SQL> COLUMN MAJOR HEADING 'MAJOR';

SQL> COLUMN BDATE HEADING 'BIRTH DATE';

SQL> SPOOL REPORT3;

SQL> SELECT * FROM STUDENT;

Thu Nov 5 page 1

REPORT GENERATION

STUDENT DETAILS

REGISTER STUDENT

NUMBER NAME MAJOR BIRTH DATE

----------------- ------------- ----------------------- ---------------

1MV10MCA01 JOHN DATA STRUCTURES 20-JAN-88

1MV10MCA02 JOYCE NETWORKS 31-DEC-87

1MV10MCA03 ALEN OS 29-MAY-87

1MV10MCA04 SMITH SYSTEM SOFTWARE 25-NOV-86

1MV10MCA05 JAMES OS 02-DEC-86

END OF REPORT

SQL> SPOOL OFF;

SQL> CLEAR COLUMN;

columns cleared

SQL> TTITLE OFF;

SQL> BTITLE OFF;



14

PROGRAM NO:4

The following tables are 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.


(iii) Give the details of authors who have two 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 publisherby10%.


(vii) Create suitable front end for querying and displaying the results.

SOLUTION:


CREATE TABLE AUTHOR(author_id number(5) primary key,

name varchar(20),city varchar(20),country varchar(20));

CREATE TABLE PUBLISHER(pub_id number(5) primary key,

name varchar(20),city varchar(20),country varchar(20));

CREATE TABLE CATEGORY(category_id number(5) primary key,

description varchar(20));

CREATE TABLE CATALOG(book_id number(5) primary key,

author_id number(5) references AUTHOR,

pub_id number(5) references PUBLISHER,

category_id number(5) references CATEGORY,

year number(4),

price number(10));

CREATE TABLE ORDER_DET(order_no number(5) primary key,

Book_id number(5) references CATALOG,quantity number(5));



15

(ii) Inserting Values:

INSERT INTO AUTHOR VALUES(1, ' BECK ', 'LONDON’,’UK’);

SELECT * FROM AUTHOR ;

AUTHOR_ID NAME CITY COUNTRY

------------- -------- ------------ -----------

1 BECK LONDON UK

2 JAVADEKAR BANGALORE INDIA

3 K V RAO BANGALORE INDIA

4 SOMERVILLE NEW YORK USA

5 TANENBAUM WASHINGTON USA

INSERT INTO PUBLISHER VALUES(11,’PHI’,’LONDON’,’UK’);

SELECT * FROM PUBLISHER ;

PUB_ID NAME CITY COUNTRY

---------- ----------- ------- -------------

11 PHI LONDON UK

12 MGHILL NEW DELHI INDIA

13 TMH BOMBAY INDIA

14 PEARSON WASINGTON USA

15 MBD NEW DELHI INDIA

INSERT INTO CATEGORY VALUES(‘1001’,’COMPUTER SCIENCE’);

SELECT * FROM CATEGORY ;

CATEGORY_ID DESCRIPTION

---------------- --------------------

1001 COMPUTER SCIENCE

1002 ALGORITHM DESIGN

1003 MECHANICS

1004 COMPUTER GRAPHICS

1005 DATA STRUCTURE



16

INSERT INTO CATALOG VALUES(21,1,11,1001,2002,300);

SELECT * FROM CATALOG ;

BOOK_ID AUTHOR_ID PUB_ID CATEGORY_ID YEAR PRICE

----------- -------------- --------- ---------------- ------- --------

21 1 11 1001 2002 300

22 1 11 1002 2003 250

23 2 11 1003 2000 400

24 3 12 1004 1999 500

25 4 14 1004 2004 540

26 5 15 1005 2001 399

27 4 11 1001 2004 300

28 1 11 1004 1998 850

INSERT INTO ORDER_DET VALUES(51,21,100);

SELECT * FROM ORDER-DET ;

ORDER_NO BOOK_ID QUANTITY

------------- ----------- -------------

51 21 100

52 21 150

53 21 200

54 22 39

55 23 35

56 23 49

57 24 35

58 25 100

III) Give the details of authors who have two 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.

SELECT * FROM AUTHOR WHERE AUTHOR_ID IN ( SELECT AUTHOR_ID FROM

CATALOG GROUP BY AUTHOR_ID HAVING COUNT(*)>2 AND AUTHOR_ID IN (

SELECT AUTHOR_ID FROM CATALOG

WHERE PRICE> (SELECT AVG(PRICE) FROM CATALOG) AND YEAR>2000));

AUTHOR_ID NAME CITY COUNTRY

------------- ------------ ---------- --------------

1 BECK LONDON UK




17

IV) Find the author of the book which has maximum sales.

SELECT NAME FROM AUTHOR WHERE AUTHOR_ID IN ( SELECT AUTHOR_ID FROM

CATALOG WHERE BOOK_ID =

(SELECT BOOK_ID FROM ORDER_DETAILS WHERE QUANTITY =

(SELECT MAX (QUANTITY) FROM ORDER_DETAILS)));

NAME

-------

BECK

V) Demonstrate how you increase the price of books published by a specific

publisher by 10%.

UPDATE CATALOG SET PRICE=PRICE+PRICE*0.1 WHERE PUB_ID IN ( SELECT

PUB_ID FROM PUBLISHER WHERE NAME='PHI');

BOOK_ID AUTHOR_ID PUB_ID CATEGORY_ID YEAR PRICE

----------- ------------- --------- ---------------- -------- ---------

21 1 11 1001 2002 330

22 1 11 1002 2003 275

23 2 11 1003 2000 440

27 4 11 1001 2004 330

28 1 11 1004 1998 935



18


SQL> SET LINESIZE 70

SQL> SET PAGESIZE 20

SQL> TTITLE 'REPORT GENERATION|AUTHOR DETAILS'

SQL> BTITLE CENTER 'END REPORT'

SQL> COLUMN AUTHOR_ID HEADING ‘AUTHOR|IDENTITY’;

SQL> COLUMN NAME HEADING ‘AUTHOR|NAME’;

SQL> COLUMN CITY HEADING ‘CITY’;

SQL> COLUMN COUNTRY HEADING ‘COUNTRY’;

SQL> SPOOL REPORT4;

SQL> SELECT * FROM AUTHOR;

Thu Nov 5 page 1

REPORT GENERATION

AUTHOR DETAILS

AUTHOR AUTHOR

IDENTITY NAME CITY COUNTRY

------------ ---------- ---------- ------------

1 BECK LONDON UK

2 JAVADEKAR BANGALORE INDIA

3 K V RAO BANGALORE INDIA


5 TANENBAUM WASHINGTON USA

END REPORT

SQL> SPOOL OFF;

SQL> CLEAR COLUMNS;

Columns cleared

SQL> TTITLE OFF;

SQL> BTITLE OFF;



19

PROGRAM NO: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)

DEPOSITOR (customer-name:string, accno:int)

CUSTOMER (customer-name:string, customer-steet:string, customer-

city:string)

LOAN (loan-number:int, branch-name:string, amount:real)

BORROWER (customer-name:string, loan-number:int)

(i) Create the above tables by properly specifying the primary keys an

the foreign keys.


(iii) Find all the customer who have at least two accounts at the main

branch.

(iv) Find all the customers who have an account tuples at every branch

located in a specific city.

(v) Demonstrate how you delete all account tuples at every branch

located in a specified city.


(vi) Create suitable front end for querying and displaying the results.

SOLUTION:


CREATE TABLE BRANCH(branch_name varchar(30) primary key,

branch_city varchar(30),asset number(20));

CREATE TABLE ACCOUNT(acc_no number(10) primary key,

branch_name varchar(30) references BRANCH,balance number(10));

CREATE TABLE CUSTOMER(cust_name varchar(30) primary key,

cust_street varchar(30),cust_city varchar(30));

CREATE TABLE DEPOSITOR(cust_name varchar(30) references CUSTOMER,

acc_no number(10) references ACCOUNT);

CREATE TABLE LOAN(loan_number number(5) primary key,

branch_name varchar(30) references BRANCH,amount number(10));

CREATE TABLE BORROWER(cust_name varchar(30) references CUSTOMER,

loan_number number (5) references LOAN unique);



20

(ii) Inserting values:

INSERT INTO BRANCH VALUES(' SBM BULL TEMPLE’,’BANGALORE’,500000);

SELECT * FROM BRANCH ;

BRANCH_NAME BRANCH_CITY ASSET

----------------- --------------------- ----------

SBM BULL TEMPLE BANGALORE 500000

SBM VIJAY NAGAR BANGALORE 200000

SBM JAY NAGAR BANGALORE 150000

SBM VIDYPEETHA BANGALORE 350000

SBI MAIN MYSORE 450000

INSERT INTO ACCOUNT VALUES(‘133550,’SBM BULL TEMPLE’,’45000);

SELECT * FROM ACCOUNT;

ACC_NO BRANCH_NAME BALANCE

--------- ----------------------- ----------

133550 SBM BULL TEMPLE 45000

133551 SBM VIJAY NAGAR 10000


2551 SBI MAIN 45000

2552 SBI MAIN 5000

2245 SBM VIDYPEETHA 5600

2246 SBM JAY NAGAR 20000




23451 SBI MAIN 4500




21

INSERT INTO CUSTOMER VALUES('PUSHPA', 'BULL TEMPLE ROAD', 'BANGALORE');

SELECT * FROM CUSTOMER;

CUST_NAME CUST_STREET CUST_CITY

-------------- ------------------------ --------------

PUSHPA BULL TEMPLE ROAD BANGALORE

RAMYA M G ROAD BANGALORE

SUDHA N R COLONY BANGALORE

SHYAMALA HANUMANTH NAGAR BANGALORE

ROOPA JALLAHALLI BANGALORE

UMA MAIN ROAD MYSORE

SANDYA COMMERCIAL STREET BANGALORE

INSERT INTO DEPOSITOR VALUES('PUAHPA',2551);

SELECT * FROM DEPOSITOR;

CUST_NAME ACC_NO

--------------- ----------

PUSHPA 2551

PUSHPA 2552

PUSHPA 23451

SANDYA 133550

SANDYA 2552

ROOPA 133551

SUDHA 25551

SUDHA 2245

RAMYA 2246

SHYAMALA 2257

SHYAMALA 22351

UMA 24466

UMA 23455

INSERT INTO LOAN VALUES(101,’ SBM BULL TEMPLE’, 30000’);

SELECT * FROM LOAN ;

LOAN_NUMBER BRANCH_NAME AMOUNT

---------------- --------------------- ----------



103 SBI MAIN 50000



INSERT INTO BORROWER VALUES(‘PUSHPA’,101);



22

SELECT * FROM BORROWER ;

CUST_NAME LOAN_NUMBER

-------------- -----------------

PUSHPA 101

SUDHA 102

ROOPA 103

SHYAMALA 104

UMA 105

III) Find all the customer who have at least two accounts at the main branch.

SELECT CUST_NAME FROM DEPOSITOR WHERE ACC_NO IN (SELECT ACC_NO

FROM ACCOUNT WHERE BRANCH_NAME = 'SBI MAIN') GROUP BY CUST_NAME

HAVING COUNT(*)>1;

CUST_NAME

------------------

PUSHPA

IV) Find all the customers who have an account tuples at every branch located ina

specific city.

SELECT DISTINCT CUST_NAME FROM DEPOSITOR WHERE ACC_NO IN (SELECT

ACC_NO FROM ACCOUNT WHERE BRANCH_NAME IN (SELECT BRANCH_NAME FROM

BRANCH WHERE BRANCH_CITY='MYSORE'));

CUST_NAME

--------------

SANDYA

PUSHPA

V) Demonstrate how you delete all account tuples at every branch located in a

specified city.

DELETE FROM ACCOUNT WHERE BRANCH_NAME IN (SELECT BRANCH_NAME

FROMBRANCH WHERE BRANCH_CITY='MYSORE');

SELECT * FROM ACCOUNT;

ACC_NO BRANCH_NAME BALANCE

--------- ----------------------- ----------













23

SQL> SET LINESIZE 70;

SQL> TTITLE ‘REPORT GENERATION|ACCOUNT DETAILS’;

SQL> BTITLE ‘END OF REPORT’;

SQL> COLUMN ACC_NO HEADING ‘ACCOUNT|NUMBER’;

SQL> COLUMN BRANCH_NAME HEADING ‘BRANCH|NAME’;

SQL> SPOOL REPORT5;

SQL> SELECT * FROM ACCOUNT;

Thu Nov 1 page 1

REPORT GENERATION

ACCOUNT DETAILS

ACCOUNT BRANCH

NUMBER NAME BALANCE

--------- ----------------------- ----------










END OF REPORT

SQL> SPOOL OFF;

SQL> CLEAR COLUMN;

Columns cleared

SQL> TTITLE OFF;

SQL> BTITLE OFF;

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