10 csl57 dbms lab

Database Management Systems Lab Manual

Session: Aug – Nov 2012

V Semester

Laboratory Manual
on

DATABASE MANAGEMENT SYSTEMS LABORATORY
[10CSL 57]

By

Mr. R. RAJKUMAR
Assistant Professor
Department of Information Science & Engineering

For V Semester CSE / ISE

RNS Institute of Technology
Rajarajeshwarinagar Post, Channasandra,
Bengaluru – 560098

Department of Information Science

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DATABASE APPLICATIONS LABORATORY
Subject Code: 10CSL57
Hours/Week : 03
Total Hours : 42

I.A. Marks : 25
Exam Hours: 03
Exam Marks: 50

1. Consider the following relations:
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.
2. The following relations keep track of airline flight information:
Flights (no: 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 eid and the maximum cruisingrange of
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 cruisingrange over 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.
3. Consider the following database of student enrollment in courses & books adopted 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.


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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.
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)
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%.
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-street: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 and the foreign keys
ii. Enter at least five tuples for each relation
iii. Find all the customers who have at least two accounts at the Main branch.
iv. Find all the customers who have an account at all the branches located in a specific city.
v. Demonstrate how you delete all account tuples at every branch located in a specific city.
Instructions:
1. The exercises are to be solved in an RDBMS environment like Oracle or DB2.
2. Suitable tuples have to be entered so that queries are executed correctly.
3. Front end may be created using either VB or VAJ or any other similar tool.
4. The student need not create the front end in the examination. The results of the queries may be displayed
directly.
5. Relevant queries other than the ones listed along with the exercises may also be asked in the examination.
6. Questions must be asked based on lots.


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1. Consider the following relations:
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 (Junior: JR,
Senior:SR, Super Senior: SS, Fresher: FR)
CREATE TABLE STUDENT
(SNUM
INTEGER
NOT NULL,
SNAME
VARCHAR (15),
MAJOR
VARCHAR (10),
SLEVEL
CHAR (2)
NOT NULL,
AGE
NUMBER (3),
PRIMARY KEY (SNUM));
CREATE TABLE FACULTY
(FID
INTEGER
NOT NULL,
FNAME
VARCHAR (12) NOT NULL,
DEPTID
NUMBER (3),
PRIMARY KEY (FID));
CREATE TABLE CLASS
(NAME
MEETS_AT VARCHAR (10),
ROOM
VARCHAR (5),
FID
INTEGER,
PRIMARY KEY (NAME),
FOREIGN KEY (FID) REFERENCES FACULTY (FID));
CREATE TABLE ENROLLED
(SNUM
INTEGER
NOT NULL,
CNAME
PRIMARY KEY (SNUM, CNAME),
FOREIGN KEY (SNUM) REFERENCES STUDENT (SNUM),
FOREIGN KEY (CNAME) REFERENCES CLASS (NAME));
INSERT INTO STUDENT VALUES (&SNUM, ‘&SNAME’, ‘&MAJOR’, ‘&SLEVEL’, &AGE);
INSERT INTO FACULTY VALUES (&FID, ‘&FNAME’, &DEPTID);
INSERT INTO CLASS VALUES (‘&CNAME’, ‘&MEETS_AT’, ‘&ROOM’, &FID);
INSERT INTO ENROLLED VALUES (&SNUM, ‘&CNAME’);
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
SELECT DISTINCT S.SNAME
FROM STUDENT S, CLASS C, ENROLLED E, FACULTY F
WHERE S.SNUM = E.SNUM AND E.CNAME = C.NAME AND C.FID = F.FID AND
F.FNAME = ‘Prof. Harshith’ AND S.LEVEL = ‘JR’;


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ii) Find the names of all classes that either meet in room R128 or have five or more Students enrolled.
SELECT C.NAME
FROM CLASS C
WHERE C.ROOM = ‘R128’
OR C.NAME IN (SELECT E.CNAME
FROM ENROLLED E
GROUP BY E.CNAME
HAVING COUNT (*) >= 5);
iii) Find the names of all students who are enrolled in two classes that meet at the same time.
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.SNUM = E2.SNUM AND E1.CNAME <> E2.CNAME
AND E1.CNAME = C1.NAME
AND E2.CNAME = C2.NAME AND C1.MEETS_AT = C2.MEETS_AT);
iv) Find the names of faculty members who teach in every room in which some class is taught.
SELECT DISTINCT F.FNAME
FROM FACULTY F
WHERE NOT EXISTS (( SELECT *
FROM CLASS C
EXCEPT
(SELECT C1.ROOM
FROM CLASS C1
WHERE C1.FID = F.FID));
v) Find the names of faculty members for whom the combined enrollment of the courses that they teach is less
than five.
SELECT
FROM
WHERE

DISTINCT F.FNAME
FACULTY F
5 > (SELECT COUNT (E.SNUM)
FROM
CLASS C, ENROLLED E
WHERE C.CNAME = E.CNAME
AND
C.FID = F.FID);

2. The following relations keep track of airline flight information:
Flights (flno: 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.


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CREATE TABLE FLIGHTS
(FLNO
INTEGER
FFROM
VARCHAR(15)
TTO
VARCHAR(15)
DISTANCE
INTEGER,
DEPARTS
TIMESTAMP,
ARRIVES
TIMESTAMP,
PRICE
NUMBER(10,2));


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PRIMARY KEY,
NOT NULL,
NOT NULL,

CREATE TABLE AIRCRAFT
(AID
INTEGER
PRIMARY KEY,
ANAME
VARCHAR(10),
CRUISINGRANGE
INTEGER);
CREATE TABLE EMPLOYEES
(EID
INTEGER
PRIMARY KEY,
ENAME
VARCHAR(15),
SALARY
NUMBER(10,2));
CREATE TABLE CERTIFIED
(EID INTEGER NOT NULL,
AID
INTEGER NOT NULL,
PRIMARY KEY (EID, AID),
FOREIGN KEY (EID) REFERENCES EMPLOYEES (EID),
FOREIGN KEY (AID) REFERENCES AIRCRAFT (AID));
INSERT INTO FLIGHTS VALUES (&FLNO, ‘&FFROM’, ‘&TTO’, &DISTANCE, ‘&DEPARTS’,
‘&ARRIVES’, &PRICE);
INSERT INTO AIRCRAFT VALUES (&AID, ‘&ANAME’, &CRUISRANGE);
INSERT INTO EMPLOYEES VALUES (&EID, ‘&ENAME’, &SALARY);
INSERT INTO CERTIFIED VALUES (&EID, &AID);
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.
SELECT
FROM
WHERE

DISTINCT A.ANAME
AIRCRAFT A
A.AID IN
(SELECT
FROM
WHERE
NOT EXISTS

C.AID
CERTIFIED C, EMPLOYEES E
C.EID = E.EID AND
(SELECT
*
FROM
EMPLOYEES E1
WHERE
E1.EID = E.EID AND E1.SALARY < 80000));

ii) For each pilot who is certified for more than three aircrafts, find the eid and the maximum cruisingrange of
the aircraft for which she or he is certified.
SELECT
FROM
WHERE
GROUP BY
HAVING

C.EID, MAX (A.CRUISINGRANGE)
CERTIFIED C, AIRCRAFT A
C.AID = A.AID
C.EID
COUNT (*) > 3;


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iii) Find the names of pilots whose salary is less than the price of the cheapest route from Bengaluru to
Frankfurt.
SELECT
FROM
WHERE

DISTINCT E.ANAME
EMPLOYEE E
E.SALARY < ( SELECT MIN (F.PRICE)
FROM FLIGHTS F
WHERE F.FFROM = ‘Bengaluru’ AND F.TTO = ‘Frankfurt’);

iv) For all aircraft with cruisingrange over 1000 Kms,. Find the name of the aircraft and the average salary of all
pilots certified for this aircraft.
Observe that aid is the key for Aircraft, but the question asks for aircraft names; we deal with this complication
by using an intermediate relation Temp;
SELECT
FROM

TEMP.NAME, TEMP.AVGSALARY
(SELECT
A.AID, A.ANAME AS NAME,
AVG (E.SALARY) AS AVGSALARY
FROM
AIRCRAFT A, CERTIFIED C, EMPLOYEES E
WHERE
A.AID = C.AID AND
C.EID = E.EID AND A.CRUISINGRANGE > 1000
GROUP BY
A.AID, A.ANAME) AS TEMP;

v) Find the names of pilots certified for some Boeing aircraft.
SELECT
FROM
WHERE

DISTINCT E.ENAME
EMPLOYEES E, CERTIFIED C, AIRCRAFT A
E.EID = C.EID AND
C.AID = A.AID AND
A.ANAME = ‘Boeing’;

vi) Find the aids of all aircraft that can be used on routes from Bengaluru to New Delhi.
SELECT A.AID
FROM AIRCRAFT A
WHERE A.CRUISINGRANGE

>

(SELECT MIN (F.DISTANCE)
FROM FLIGHTS F
WHERE F.FFROM = ‘Bengaluru’ AND F.TTO = ‘New Delhi’);

3. Consider the following database of student enrollment in courses & books adopted 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)
CREATE TABLE SSTUDENT
(RREGNO
VARCHAR(30)
NAME
VARCHAR(30)


PRIMARY KEY,
NOT NULL,

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MAJOR
BDATE

VARCHAR(30)
DATE


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NOT NULL,
NOT NULL);

CREATE TABLE CCOURSE
(COURSE
INTEGER
PRIMARY KEY,
CCNAME
VARCHAR(30) NOT NULL,
DEPT
VARCHAR(30) NOT NULL);
CREATE TABLE EENROLL
(RREGNO
COURSE
INTEGER
NOT NULL,
SEM
INTEGER
NOT NULL,
MARKS
INTEGER
NOT NULL,
PRIMARY KEY (RREGNO, COURSE, SEM),
FOREIGN KEY (RREGNO) REFERENCES SSTUDENT(RREGNO),
FOREIGN KEY (COURSE) REFERENCES CCOURSE(COURSE));
CREATE TABLE TTEXT
(BOOKISBN INTEGER
BOOKTITLE VARCHAR(30)
PUBLISHER VARCHAR(30)
AUTHOR
VARCHAR(30)

PRIMARY KEY,
NOT NULL,
NOT NULL,
NOT NULL);

CREATE TABLE BBOOKADOPTION
(COURSE
INTEGER
NOT NULL,
SEM
INTEGER
NOT NULL,
BOOKISBN INTEGER
NOT NULL,
PRIMARY KEY (COURSE, SEM, BOOKISBN),
FOREIGN KEY (COURSE) REFERENCES CCOURSE (COURSE),
FOREIGN KEY (BOOKISBN) REFERENCES TTEXT (BOOKISBN));
INSERT INTO SSTUDENT VALUES('1RN10IS012','ANN','DATBASE','15-JAN-84');
INSERT INTO SSTUDENT VALUES('1 RN10CS012','MARY','DMS','25-FEB-84');
INSERT INTO SSTUDENT VALUES('1 RN10TC012','TOM','SSDT','11-DEC-84');
INSERT INTO SSTUDENT VALUES('1 RN10EE012','EVE','POWER GENERATION','1-APR-84');
INSERT INTO SSTUDENT VALUES('1 RN10EC012','GEORGE','POWER ELECTRONICS','5-NOV-84');
INSERT INTO CCOURSE VALUES(1,'DATABASE','CS');
INSERT INTO CCOURSE VALUES(2,'DMS','CS');
INSERT INTO CCOURSE VALUES(3,'SSDT','TC');
INSERT INTO CCOURSE VALUES(4,'POWER GENERATION','EE');
INSERT INTO CCOURSE VALUES(5,'POWER ELECTRONICS','EC');
INSERT INTO CCOURSE VALUES(6,'DATASTRUCTURE','CS');
INSERT INTO TTEXT VALUES(1,'DATABASE A SYSTEMATIC APPROACH','JOHN WILEY','R ASHOK KUMAR');

INSERT INTO TTEXT VALUES(2,'DMS FOR DUMMIES','JOHN WILEY','MADHUPRIYA');
INSERT INTO TTEXT VALUES(3,'SSDT NO ONE CAN TEACH BETTER','PEARSON','GAURA');
INSERT INTO TTEXT VALUES(4,'POWER GENERATION BIBLE','TMH','MEENA');
INSERT INTO TTEXT VALUES(5,'POWER OF POWER ELECTRONICS','O REILLY','GG THE GREAT');
INSERT INTO TTEXT VALUES(6,'POWER OF DATASTRUCTURES','JOHN WILEY','DENNISRITCHIE');


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INSERT INTO TTEXT VALUES(7,'ELEMENTARY DATASTUCTURES1','JOHN WILEY','HERBERT SHIELD');
INSERT INTO TTEXT VALUES(8,'ELEMENTARY DATASTUCTURES2','JOHN WILEY','HERBERT SHIELD');

INSERT INTO TTEXT VALUES(9,'DATABASE','JOHN WILEY','MAYOR');
INSERT INTO EENROLL VALUES ('1RN10IS012', 1, 5, 98);
INSERT INTO EENROLL VALUES ('1RN10CS012', 2, 3, 88);
INSERT INTO EENROLL VALUES ('1RN10TC012', 3, 5, 88);
INSERT INTO EENROLL VALUES ('1RN10EE012', 4, 5, 88);
INSERT INTO EENROLL VALUES ('1RN10EC012', 5, 5, 88);
INSERT INTO BBOOKADOPTION VALUES (1, 5, 1);
iii. Demonstrate how you add a new text book to the database and make this book be adopted by some
department.
INSERT INTO TTEXT VALUES (10, 'DATABASE FUNDAS', 'PEARSON', 'SCHIELD');
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
FROM
WHERE

C.COURSE, T.BOOKISBN, T.BOOKTITLE
CCOURSE C, BBOOKADOPTION BA, TTEXT T
C.COURSE=BA.COURSE AND
BA.BOOKISBN=T.BOOKISBN
AND C.DEPT='CS' AND
EXISTS
(SELECT
COUNT (COURSE)
FROM
BBOOKADOPTION
WHERE
COURSE=C.COURSE
GROUP BY COURSE
HAVING
COUNT (COURSE)>=2)
ORDER BY T.BOOKTITLE;

v. List any department that has all its adopted books published by a specific publisher.
SELECT
FROM
WHERE

C.DEPT, T.BOOKTITLE, T.PUBLISHER
CCOURSE C, TTEXT T, BBOOKADOPTION BA
C.COURSE=BA.COURSE AND T.BOOKISBN=BA.BOOKISBN
AND
T.PUBLISHER ='JOHN WILEY'AND
T.PUBLISHER= ALL (SELECT
T1.PUBLISHER
FROM
CCOURSE C1, BBOOKADOPTION BA1, TTEXT T1
WHERE
BA1.BOOKISBN=T1.BOOKISBN AND


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BA1.COURSE=C1.COURSE AND
C.DEPT=C1.DEPT);
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)
CREATE TABLE AAUTHOR
( AUTHORID
INTEGER
PRIMARY KEY,
NAME
CITY
COUNTRY
CREATE TABLE PPUBLISHER
(PUBLISHERID
INTEGER
PRIMARY KEY,
NAME
CITY
COUNTRY
CREATE TABLE BOOKCATEGORY
(CATEGORYID
INTEGER
PRIMARY KEY,
DESCRIPTION
CREATE TABLE CCATALOG
(BOOKID INTEGER PRIMARY KEY,
TITLE
AUTHORID
INTEGER
NOT NULL,
PUBLISHERID
INTEGER
NOT NULL,
CATEGORYID
INTEGER
NOT NULL,
YEAROFPUBLISH INTEGER
NOT NULL,
PRICE
INTEGER
NOT NULL,
FOREIGN KEY (AUTHORID) REFERENCES AAUTHOR(AUTHORID),
FOREIGN KEY (PUBLISHERID) REFERENCES PPUBLISHER(PUBLISHERID),
FOREIGN KEY (CATEGORYID) REFERENCES BOOKCATEGORY(CATEGORYID));
CREATE TABLE OORDERDETAILS
(ORDERNO INTEGER
PRIMARY KEY,
BOOKID
INTEGER
NOT NULL,
QUANTITY INTEGER
NOT NULL,
FOREIGN KEY (BOOKID) REFERENCES CCATALOG(BOOKID));
INSERT INTO AAUTHOR VALUES (1,'NAVATHE','ARLINGTON','USA');
INSERT INTO AAUTHOR VALUES (2,'RAGHU RAMAKRISHNAN','CALIFORNIA','USA');


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INSERT INTO AAUTHOR VALUES (3,'DHAMDHERE','MUMBAI','INDIA');
INSERT INTO AAUTHOR VALUES (4,'BJARNE','NEW JERSY','USA');
INSERT INTO AAUTHOR VALUES (5,'TANENBAUM','AMSTERDAM','NETHERLAND');
INSERT INTO PPUBLISHER VALUES (1,'JOHN WILEY','NEW YORK','USA');
INSERT INTO PPUBLISHER VALUES (2,'PEARSON','BANGALORE','INDIA');
INSERT INTO PPUBLISHER VALUES (3,'O REILLY','NEW JERSY','USA');
INSERT INTO PPUBLISHER VALUES (4,'TMH','CALCUTTA','INDIA');
INSERT INTO PPUBLISHER VALUES (5,'JOHN WILEY','NEW DELHI','INDIA');
INSERT INTO BOOKCATEGORY VALUES (1,'DATABASE MANAGEMENT');
INSERT INTO BOOKCATEGORY VALUES (2,'OPERATING SYSTEMS');
INSERT INTO BOOKCATEGORY VALUES (3,'C++');
INSERT INTO BOOKCATEGORY VALUES (4,'COMPUTER NETWORKS');
INSERT INTO BOOKCATEGORY VALUES (5,'C');
INSERT INTO CCATALOG VALUES (1,'FUNDAMENTALS OF DBMS',1,2,1,2004,500);
INSERT INTO CCATALOG VALUES (2,'PRINCIPLES OF DBMS',2,1,1,2004,400);
INSERT INTO CCATALOG VALUES (3,'OPERATING SYSTEMS',3,4,2,2004,200);
INSERT INTO CCATALOG VALUES (4,'C++ BIBLE',4,5,3,2003,500);
INSERT INTO CCATALOG VALUES (5,'COMPUTER NETWORKS',5,3,4,2002,250);
INSERT INTO CCATALOG VALUES (6,'FUNDAMENTALS OF C',1,2,5,2004,700);
INSERT INTO CCATALOG VALUES (7,'OPERATING SYSTEMS 2',3,2,2,2001,600);
INSERT INTO OORDERDETAILS VALUES (1,1,1);
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.
SELECT
FROM
WHERE

*
AAUTHOR A
EXISTS
(SELECT
FROM
WHERE

A1.AUTHORID,COUNT(A1.AUTHORID)
AAUTHOR A1,CCATALOG C
A1.AUTHORID=C.AUTHORID AND
A.AUTHORID=A1.AUTHORID AND
C.YEAROFPUBLISH > 2000 AND
C.PRICE > (SELECT
AVG(PRICE)
FROM
CCATALOG)
GROUP BY
A1.AUTHORID
HAVING
COUNT(A1.AUTHORID) >=2);

iv. Find the author of the book which has maximum sales.
SELECT
FROM

DISTINCT A.NAME
AAUTHOR A, CCATALOG C, OORDERDETAILS ODM


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WHERE


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A.AUTHORID=C.AUTHORID AND ODM.BOOKID=C.BOOKID AND
EXISTS
(SELECT
OD.BOOKID,SUM(OD.QUANTITY)
FROM
OORDERDETAILS OD
WHERE
OD.BOOKID=ODM.BOOKID
GROUP BY
BOOKID
HAVING
SUM(OD.QUANTITY)>= ALL
(SELECT
SUM(QUANTITY)
FROM
OORDERDETAILS
GROUP BY
BOOKID));

v. Demonstrate how you increase the price of books published by a specific publisher by 10%.
UPDATE
SET
WHERE

CCATALOG
PRICE = (1.1) * PRICE
AUTHORID = (SELECT
AUTHORID
FROM
AAUTHOR
WHERE
NAME = 'NAVATHE');
vii. Create suitable front end for querying and displaying the results
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-street: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 and the foreign keys
CREATE TABLE BBRANCH
(BRANCHNAME
VARCHAR(30)
BRANCHCITY
VARCHAR(30)
ASSETS
NUMBER(10,2)

PRIMARY KEY,
NOT NULL,
NOT NULL);

CREATE TABLE BBANKACCOUNT
(ACCNO
NUMBER(5)
PRIMARY KEY,
BRANCHNAME
VARCHAR(30)
NOT NULL,
BALANCE
NUMBER(10,2),
FOREIGN KEY (BRANCHNAME) REFERENCES BBRANCH (BRANCHNAME));
CREATE TABLE BBANKCUSTOMER
(CUSTOMERNAME VARCHAR(30)
CUSTOMERSTREET VARCHAR(30)
CUSTOMERCITY
VARCHAR(30)

PRIMARY KEY,
NOT NULL,
NOT NULL);

CREATE TABLE DDEPOSITOR
NOT NULL,
ACCNO
NUMBER(5)
NOT NULL,
PRIMARY KEY (CUSTOMERNAME, ACCNO),
FOREIGN KEY (CUSTOMERNAME) REFERENCES BBANKCUSTOMER (CUSTOMERNAME),
FOREIGN KEY (ACCNO) REFERENCES BBANKACCOUNT (ACCNO) ON DELETE CASCADE);


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CREATE TABLE LLOAN
(LOANNUMBER
INTEGER
PRIMARY KEY,
BRANCHNAME
VARCHAR(30)
NOT NULL,
AMOUNT
NUMBER(10,2)
NOT NULL,
FOREIGN KEY (BRANCHNAME) REFERENCES BBRANCH (BRANCHNAME));
CREATE TABLE BBORROWER
NOT NULL,
LOANNUMBER
INTEGER
NOT NULL,
PRIMARY KEY (CUSTOMERNAME, LOANNUMBER),
FOREIGN KEY (CUSTOMERNAME) REFERENCES BBANKCUSTOMER (CUSTOMERNAME),
FOREIGN KEY (LOANNUMBER) REFERENCES LLOAN (LOANNUMBER));
ii. Enter at least five tuples for each relation
INSERT INTO BBRANCH VALUES('CHAMRAJPET','BANGALORE',50000);
INSERT INTO BBRANCH VALUES('RESIDENCY ROAD','BANGALORE',10000);
INSERT INTO BBRANCH VALUES('M G ROAD','BANGALORE',100000);
INSERT INTO BBRANCH VALUES('CP','DELHI',100000);
INSERT INTO BBRANCH VALUES('JANTARMANTAR','DELHI',100000);
INSERT INTO BBANKACCOUNT VALUES(1,'CHAMRAJPET',2000);
INSERT INTO BBANKACCOUNT VALUES(2,'RESIDENCY ROAD',5000);
INSERT INTO BBANKACCOUNT VALUES(3,'M G ROAD',6000);
INSERT INTO BBANKACCOUNT VALUES(4,'CP',9999);
INSERT INTO BBANKACCOUNT VALUES(5,'JANTARMANTAR',999);
INSERT INTO BBANKACCOUNT VALUES(6,'M G ROAD',999);
INSERT INTO BBANKACCOUNT VALUES(9,'CP',10000);
INSERT INTO BBANKACCOUNT VALUES(11,'JANTARMANTAR',9999);
INSERT INTO BBANKCUSTOMER VALUES('ANNE','BULL TEMPLE ROAD','BANGALORE');
INSERT INTO BBANKCUSTOMER VALUES('DANNY','BANNERGATTA ROAD','BANGALORE');
INSERT INTO BBANKCUSTOMER VALUES('TOM','J C ROAD','BANGALORE');
INSERT INTO BBANKCUSTOMER VALUES('NICK','CP','DELHI');
INSERT INTO BBANKCUSTOMER VALUES('ROVER','JANTARMANTAR','DELHI');
INSERT INTO DDEPOSITOR VALUES('ANNE',1);
INSERT INTO DDEPOSITOR VALUES('DANNY',2);
INSERT INTO DDEPOSITOR VALUES('TOM',3);
INSERT INTO DDEPOSITOR VALUES('NICK',4);
INSERT INTO DDEPOSITOR VALUES('ROVER',5);
INSERT INTO DDEPOSITOR VALUES('DANNY',10);
INSERT INTO LLOAN VALUES(1,'CHAMRAJPET',1000);
INSERT INTO LLOAN VALUES(2,'RESIDENCY ROAD',2000);
INSERT INTO LLOAN VALUES(3,'M G ROAD',3000);


-13-

RNSIT, Bengaluru



V Semester

INSERT INTO LLOAN VALUES(4,'CP',4000);
INSERT INTO LLOAN VALUES(5,'JANTARMANTAR',5000);
INSERT INTO BBORROWER VALUES('ANNE',1);
INSERT INTO BBORROWER VALUES('ANNE',2);
INSERT INTO BBORROWER VALUES('TOM',3);
INSERT INTO BBORROWER VALUES('NICK',4);
INSERT INTO BBORROWER VALUES('ROVER',5);
iii. Find all the customers who have at least two accounts at the Main branch.
SELECT
*
FROM
BBANKCUSTOMER C
WHERE EXISTS
(SELECT
DP.CUSTOMERNAME, COUNT (DP.CUSTOMERNAME)
FROM
DDEPOSITOR DP,BBANKACCOUNT BA
WHERE
DP.ACCNO=BA.ACCNO AND
C.CUSTOMERNAME=DP.CUSTOMERNAME AND
BA.BRANCHNAME='RESIDENCY ROAD'
GROUP BY DP.CUSTOMERNAME
HAVING
COUNT(DP.CUSTOMERNAME)>=2);
iv. Find all the customers who have an account at all the branches located in a specific city.
SELECT
FROM
WHERE

*
BBANKCUSTOMER BC
NOT EXISTS
(SELECT
BRANCHNAME
FROM
BBRANCH
WHERE
BRANCHCITY='DELHI'
MINUS
SELECT
BA.BRANCHNAME
FROM
DDEPOSITOR D,BBANKACCOUNT BA
WHERE
D.ACCNO=BA.ACCNO AND
BC.CUSTOMERNAME=D.CUSTOMERNAME );

v. Demonstrate how you delete all account tuples at every branch located in a specific city.
DELETE FROM
WHERE

BBANKACCOUNT
BRANCHNAME IN
(SELECT BRANCHNAME
FROM BBRANCH
WHERE BRANCHCITY='BANGALORE');


-14-

RNSIT, Bengaluru

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