CREC, Dept. of CSE Page 1

LAB MANUAL

For

DATABASE MANAGEMENT SYSTEMS

B.TECH CSE II YEAR I SEMESTER

(JNTUA-R15)

Mrs. S.RUKSANA M.Tech

Assistant Professor

DEPARTMENT OF COMPUTER SCIENCE & ENGINEERING

CHADALAWADA RAMANAMMA ENGINEERING COLLEGE CHADALAWADA NAGAR,

RENIGUNTA ROAD, TIRUPATI (A.P) - 517506

CREC, Dept. of CSE Page 2

JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY ANANTAPUR

B. Tech II - I sem (Common to CSE & IT) T Tu C 3 1 3

(15A05303) DATABASE MANAGEMENT SYSTEMS LABORATORY

Course Objectives:

To create a database and query it using SQL, design forms and generate reports. Understand the significance of integrity constraints, referential integrity constraints, triggers, assertions.

Course Outcomes:

Design databases Retrieve information from data bases Use procedures to program the data access and manipulation Create user interfaces and generate reports

List of Experiments:

1. Practice session: Students should be allowed to choose appropriate DBMS software, install it, configure it and start working on it. Create sample tables, execute some queries, use SQLPLUS features, use PL/SQL features like cursors on sample database. Students should be permitted to practice appropriate User interface creation tool and Report generation tool.

2. A college consists of number of employees working in different departments. In this context, create two tables employee and department. Employee consists of columns empno, empname, basic, hra, da, deductions, gross, net, date-of-birth. The calculation of hra,da are as per the rules of the college. Initially only empno, empname, basic have valid values. Other values are to be computed and updated later. Department contains deptno, deptname, and description columns. Deptno is the primary key in department table and referential integrity constraint exists between employee and department tables. Perform the following operations on the the database:

Create tables department and employee with required constraints.

Initially only the few columns (essential) are to be added. Add the remaining columns separately by using appropriate SQL command

Basic column should not be null

Add constraint that basic should not be less than 5000.

Calculate hra,da,gross and net by using PL/SQL program.

Whenever salary is updated and its value becomes less than 5000 a trigger has to be raised preventing the operation.

The assertions are: hra should not be less than 10% of basic and da should not be less than 50% of basic.

The percentage of hra and da are to be stored separately.

When the da becomes more than 100%, a message has to be generated and with user permission da has to be merged with basic.

Empno should be unique and has to be generated automatically.

If the employee is going to retire in a particular month, automatically a message has to be generated.

The default value for date-of-birth is 1 jan, 1970.

When the employees called daily-wagers are to be added the constraint that salary should be greater than or equal to 5000 should be dropped.

Display the information of the employees and departments with description of the fields.

Display the average salary of all the departments.

Display the average salary department wise.

Display the maximum salary of each department and also all departments put together.

Commit the changes whenever required and rollback if necessary.

Use substitution variables to insert values repeatedly.

Assume some of the employees have given wrong information about date-of-birth. Update the corresponding tables to change the value.

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Find the employees whose salary is between 5000 and 10000 but not exactly 7500.

Find the employees whose name contains ‘en’.

Try to delete a particular deptno. What happens if there are employees in it and if there are no employees.

Create alias for columns and use them in queries.

List the employees according to ascending order of salary.

List the employees according to ascending order of salary in each department.

Use ‘&&’ wherever necessary

Amount 6000 has to be deducted as CM relief fund in a particular month which has to be accepted as input from the user. Whenever the salary becomes negative it has to be maintained as 1000 and the deduction amount for those employees is reduced appropriately.

The retirement age is 60 years. Display the retirement day of all the employees.

If salary of all the employees is increased by 10% every year, what is the salary of all the employees at retirement time.

Find the employees who are born in leap year.

Find the employees who are born on feb 29.

Find the departments where the salary of atleast one employee is more than 20000.

Find the departments where the salary of all the employees is less than 20000.

On first January of every year a bonus of 10% has to be given to all the employees. The amount has to be deducted equally in the next 5 months. Write procedures for it.

As a designer identify the views that may have to be supported and create views.

As a designer identify the PL/SQL procedures necessary and create them using cursors.

Use appropriate Visual programming tools like oracle forms and reports, visual basic etc to create user interface screens and generate reports.

Note: As a designer identify other operations that may be required and add to the above list. The above operations are not in order. Order them appropriately. Use SQL or PL/SQL depending on the requirement.

3. Students may be divided into batches and the following experiments may be given to them to better understand the DBMS concepts. Students should gather the required information, draw ER diagrams, map them to tables, normalize, create tables, triggers, procedures, execute queries, create user interfaces, and generate reports.

Student information system

APSRTC reservation system

Hostel management

Library management

Indian Railways reservation

Super market management

Postal system

Banking system

Courier system

Publishing house system

References:

1. “Oracle Database 11g PL/SQL Programming”, M.Mc Laughlin,TMH.

2. “Learning Oracle SQL and PL/SQL”, Rajeeb C. Chatterjee, PHI.

3. “Introduction to SQL”, Rick F.Vander Lans, Pearson education.

4. “Oracle PL/SQL”, B.Rosenzweig and E.Silvestrova, Pearson education.

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1. Practice session: Students should be allowed to choose appropriate DBMS software, install it, configure it and start working on it. Create sample tables, execute some queries, use SQLPLUS features, use PL/SQL features like cursors on sample database. Students should be permitted to practice appropriate User interface creation tool and Report generation tool.

1. Select installation method

Following two installation methods are available:

Basic Installation – Choose this to perform the full Oracle Database 11g installation with standard configuration options. You can use only filesystem for storage in this option.

Advanced Installation – This gives your full control over the installation, including the ability to choose Automatic Storage Management. Select this option as shown below.

Fig – Select Installation Method (Click on the image to enlarge it)

2. Specify Inventory directory and credentials

Enter the following information:

Enter the inventory directory: /u01/app/oraInventory Specify operating system group name: oinstall

Fig – Specify Inventory directory and credentials (Click on the image to enlarge it)

3. Select Installation Type

Following three installation types are available:

Enterprise Edition – Select this option. Standard Edition Custom

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Fig – Select Installation Type (Click on the image to enlarge it)

4. Specify Install Location

Enter the following information:

Oracle Base: /u01/app/oracle Name: Oracle_Home Path: /u01/app/oarcle/product/11.1.0

Fig – Select Install Location (Click on the image to enlarge it)

5. Product-Specific Prerequisite Checks

In this screen, the Installer verifies that your environment meets all of the minimum requirements for installing and configuring the products that you have chosen to install. You must manually verify and confirm the items that are flagged with warnings and items that require manual checks.

6. Select Configuration Option

Following three configuration options are available:

Create a Database – Select this option.

Configure Automatic Storage Management (ASM) Install Software Only

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Fig – Select Configuration Option (Click on the image to enlarge it)

7. Select a Database Configuration

Following three database configuration options are available.

General Purpose/Transaction Processing – Select this option. Data Warehouse Advance

Fig – Select Database Configuration (Click on the image to enlarge it)

8. Specify Database Configuration

Enter the following information:

Global Database Name: devdb.example.com . An Oracle database is uniquely identified by a Global Database Name, typically of the form “name.domain”

SID: devdb

9. Specify Database Configuration Details

There are four different tabs in this screen as shown below:

Memory Tab o Select the check-box to Enable Automatic Memory Management o Drag the bar to allocate the memory. This will automatically allocate memory size for SGA and PGA

depending on the total database memory allocated. Character Sets Tab. Following three character set options are given in this tab.

o Use the default o Use Unicode (AL32UTF8) o Choose from the list of character sets. Select Database Character Set as Unicode standard UTF-

8AL32UTF8 – Select this option. Security Tab – Just leave the defaults in this tab. Sample Schema Tab – Just leave the defaults in this tab.

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Fig – Select Database Config Details with Memory Tab (Click on the image to enlarge it)

10. Select Database Management Option

Following two management options are available:

Use Grid Control for Database Management o Management Service: Choose an agent.

Use Database Control for Database Management- Select this option. o Select the check-box to Enable Email Notifications o Outgoing mail (SMTP) Server: o Email Address:

Fig – Select Database Management Option (Click on the image to enlarge it)

11. Specify Database Storage Option

Following two storage options are available:

File System – Select this option. o Specify Database File location: /u01/app/oracle/oradata/

Automatic Storage Management. ASM simplifies database storage administration and optimizes database layout for I/O performance.

12. Specify Backup and Recovery Option

Enable automated backup as shown below. If you’ve configured ASM, you can choose to store the automated backups on

ASM. If not, select File System option.

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Fig – Specify Backup and Recovery Options (Click on the image to enlarge it)

13. Specify database schema password

Enter the password for sys, system, sysman, dbsnmp schemas. You have the option of either using the same passwords or different password for all the accounts in this screen.

14. Privileged Operating System Groups

Enter the following information in this screen:

Database Administrator (OSDBA) Group: dba Database Operator (OSOPER) Group: dba ASM administrator (OSASM) Group: dba

15. Oracle Configuration Manager Registration

You can associate your configration information with your metalink account as shown below.

Fig – Oracle Configuration Manager Registration (Click on the image to enlarge it)

16. Final Summary Screen

This will display a summary of all the installation option you’ve selected so far. Click on Install to complete the

installation.

Fig – Installation Summary Screen (Click on the image to enlarge it)

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2. A college consists of number of employees working in different departments. In this context, create two tables employee and department. Employee consists of columns empno, empname, basic, hra, da, deductions, gross, net, date-of-birth. The calculation of hra,da are as per the rules of the college. Initially only empno, empname, basic have valid values. Other values are to be computed and updated later. Department contains deptno, deptname, and description columns. Deptno is the primary key in department table and referential integrity constraint exists between employee and department tables. Perform the following operations on the the database:

2.1 Create tables department and employee with required constraints.

SQL> create table dept(did number primary key, dname varchar(10),dhod varchar(10));

Table created.

SQL> select * from dept;

DID DNAME DHOD

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

101 cse rama

102 cse muni

103 cse munnaiah

104 ece trimurt

105 eee siva

SQL> create table emp1(eid number primary key,ename varchar(10),basic number, hra number, da number, gross

number, net number, dob date,did references dept(did));

Table created.

SQL> select * from emp1;

EID ENAME BASIC HRA DA GROSS NET

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

DOB DID

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

1001 ruks 14000 200 300 200 300

16-JUL-93 101

1002 veena 15000 400 100 200 300

20-JUL-93 102

1003 endira 18000 400 300 400 600

20-JUN-93 103

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2.2 Initially only the few columns (essential) are to be added. Add the remaining columns separately by using appropriate SQL command

SQL> alter table dept add droom number;

Table altered.

2.3 Basic column should not be null

SQL> alter table emp1 modify basic not null;

Table altered.

2.4 Add constraint that basic should not be less than 5000

SQL> alter table emp1 add constraint chk5 check(basic>5000);

Table altered.

2.5 Calculate hra,da,gross and net by using PL/SQL program.

create or replace procedure daprocedure as

cursor c is select * from emp;

ch char(2) :='&ch';

begin

for i in c loop

if i.da=i.basic then

dbms_output.put_line('do u wanna merge da with basic');

if (ch ='y') or (ch ='y') then

update emp set i.da=i.basic where empno=i.empno;

end if;

end if;

end loop;

end;

/ Procedure created

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2.6 Whenever salary is updated and its value becomes less than 5000 a trigger has to be raised preventing the operation.

create or replace trigger trname before update on emp1

for each row

declare

begin

if :new.basic<5000 then

raise_application_error(-20000,'basic cant be less than 5000');

end if;

end;

/

Trigger created.

2.7 The assertions are: hra should not be less than 10% of basic and da should not be less than 50% of basic.

SQL> alter table emp1 add constraint chk6 check(hra>0.01*basic);

Table altered.

SQL> alter table emp1 add constraint chk7 check(da>0.005*basic);

Table altered.

2.8 The percentage of hra and da are to be stored separately.

SQL> alter table emp1 add phra1 number;

Table altered.

SQL> alter table emp1 add pda number;

Table altered.

2.10 Empno should be unique and has to be generated automatically.

declare

cursor c is select * from emp1;

k number:=1;

begin

for i in c

loop

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update emp1 set eid=100||k where eid=i.eid;

k:=k+1;

end loop;

end;

/

2.11 If the employee is going to retire in a particular month, automatically a message has to be generated

declare

cursor c is select months_between(sysdate,dob)/12 from emp;

begin

for i in c loop

if round(i.d)>60 then

dbms_output.put_line('u are going to retire this month');

end if;

end loop;

end;

/

u are going to retire this month

2.12 The default value for date-of-birth is 1 jan, 1970

SQL> alter table emp1 modify dob date default '01-jan-1970';

Table altered.

2.13 When the employees called daily-wagers are to be added the constraint that salary should be greater than or equal to 5000 should be dropped.

SQL> alter table emp1 add constraint chk8 check(basic>=5000);

Table altered.

SQL> alter table emp1 drop constraint chk8;

Table altered.

2.14 Display the information of the employees and departments with description of the fields.

SQL> desc emp1;

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Name Null? Type

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

EID NOT NULL NUMBER

ENAME VARCHAR2(10)

BASIC NOT NULL NUMBER

HRA NUMBER

DA NUMBER

GROSS NUMBER

NET NUMBER

DOB DATE

DID NUMBER

PHRA NUMBER

DOJ DATE

PHRA1 NUMBER

PDA NUMBER

DNAME VARCHAR2(10)

YROFBIRTH NUMBER

SQL> desc dept;

Name Null? Type

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

DID NOT NULL NUMBER

DNAME VARCHAR2(10)

DHOD VARCHAR2(10)

DROOM NUMBER

2.15 Display the average salary of all the departments.

SQL> select avg(basic) from emp1;

AVG(BASIC)

----------

15666.6667

2.16 Display the average salary department wise.

SQL> select avg(basic) from emp1 group by dname;

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AVG(BASIC)

----------

14500

18000

2.17 Display the maximum salary of each department and also all departments put together.

SQL> select max(basic) from emp1 group by dname;

MAX(BASIC)

----------

15000

18000

SQL> select max(basic) from emp1 ;

MAX(BASIC)

----------

18000

2.18 Commit the changes whenever required and rollback if necessary.

SQL> commit;

Commit complete.

SQL> rollback;

Rollback complete.

2.19 Use substitution variables to insert values repeatedly.

SQL> insert into dept values(&did,'&dname','&dhod',&droom);

Enter value for did: 111

Enter value for dname: eee

Enter value for dhod: siva

Enter value for droom: 301

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old 1: insert into dept values(&did,'&dname','&dhod',&droom)

new 1: insert into dept values(111,'eee','siva',301)

1 row created.

2.20 Assume some of the employees have given wrong information about date-of-birth. Update the corresponding

tables to change the value.

SQL> update emp1 set dob='23-feb-2012' where eid=1003;

0 rows updated.

2.21 Find the employees whose salary is between 5000 and 10000 but not exactly 7500.

SQL> select ename from emp1 where basic between 5000 and 10000 and basic not in(7500);

no rows selected.

2.22 Find the employees whose name contains ‘en’.

SQL> select ename from emp1 where ename like '%en%';

ENAME

----------

veena

endira

2.23 Try to delete a particular deptno. What happens if there are employees in it and if there are no employees

SQL> delete from emp1 where did=20;

If the employees are there in the department referential integrity constraint violates and you cannot drop it without dropping corresponding employee entries.

If there are no employees then you can do delete the data.

2.24 Create alias for columns and use them in queries.

SQL> select (basic+500) "bonus" from emp1 ;

bonus

----------

14500

15500

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18500

2.25List the employees according to ascending order of salary.

SQL> select ename from emp1 order by basic asc;

ENAME

----------

ruks

veena

endira

2.26 The retirement age is 60 years. Display the retirement day of all the employees.

SQL> select add_months(dob,60*12) "Retirement Day" from emp1;

Retirement

---------

29-FEB-76

20-JUL-53

20-JUN-53

2.27 If salary of all the employees is increased by 10% every year, what is the salary of all the employees at retirement time.

SQL> select ((basic+basic*0.1)*months_between(sysdate,doj)/12)"salary at retirement" from emp1;

salary at retirement

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

111009.18

133211.016

2.28 Find the employees who are born in leap year.

SQL> select ename FROM emp1 where mod(to_number(to_char(dob,'YYYY')),4)=0 and

mod(to_number(to_char(dob,'YYYY')),100)!=0 or mod(to_number(to_char(dob,'YYYY')),400)=0;

ENAME

----------

Ruks

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2.29 Find the employees who are born on feb 29.

SQL> select ename from emp1 where to_char(dob,'mon dd')='feb 29';

ENAME

----------

Ruks

2.30 Find the departments where the salary of atleast one employee is more than 20000.

SQL> select dname from dept where did in (select e.did from emp1 e,emp1 ee where e.basic>ee.basic);

DNAME

----------

cse

2.31 Find the departments where the salary of all the employees is less than 20000.

SQL> select ename from emp1 where basic>20000;

no rows selected

2.32 As a designer identify the views that may have to be supported and create views.

SQL> create view eview as select ename, eid,did from emp1;

View created.

2.33 As a designer identify the views that may have to be supported and create views. On first January of every year a bonus of 10% has to be given to all the employees. The amount has to be deducted equally in the next 5 months. Write procedure.

create or replace procedure salded as

cursor c is select * from emp1;

begin

for i in c

loop

if to_char(sysdate,'mon dd')='jan 01' then

update emp1 set basic=i.basic+i.basic*0.1 where eid=i.eid;

end if ;

if to_char(sysdate,'mon dd')='feb 01' then

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update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='mar 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='apr 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='may 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='jun 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

end loop;

end;

/

Procedure created.

2.34 As a designer identify the PL/SQL procedures necessary and create them using cursors.

create or replace procedure salded as

cursor c is select * from emp1;

begin

for i in c

loop

if to_char(sysdate,'mon dd')='jan 01' then

update emp1 set basic=i.basic+i.basic*0.1 where eid=i.eid;

end if ;

if to_char(sysdate,'mon dd')='feb 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='mar 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='apr 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

if to_char(sysdate,'mon dd')='may 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

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end if;

if to_char(sysdate,'mon dd')='jun 01' then

update emp1 set basic=i.basic-i.basic*0.02 where eid=i.eid;

end if;

end loop;

end;

/

Procedure created

3. Students may be divided into batches and the following experiments may be given to them to better understand the DBMS concepts. Students should gather the required information, draw ER diagrams, map them to tables, normalize, create tables, triggers, procedures, execute queries, create user interfaces, and generate reports.

3.1 Student information system

Student Result

Unit enrolment

Unit Unit offering has

enrolls

is given

Student has Course

contains

is enrolled

in

Department

Faculty

belongs to

offered

by

runs

Room requirements

Resource

requirements

Room allocations

Staff allocations

comprises Special

consideration application

Grading rules

changes

Assignment result

Exam result

based

on

lodges

needs needs is taught in

is taught

by

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3.2 Library management

NORMALIZATION OF DATABASE:

Database Normalisation is a technique of organizing the data in the database. Normalization is a systematic approach of

decomposing tables to eliminate data redundancy and undesirable characteristics like Insertion, Update and Deletion

Anamolies. It is a multi-step process that puts data into tabular form by removing duplicated data from the relation tables.

Normalization is used for mainly two purpose,

Eliminating reduntant(useless) data.

Ensuring data dependencies make sense i.e data is logically stored.

Problem Without Normalization

Without Normalization, it becomes difficult to handle and update the database, without facing data loss. Insertion,

Updation and Deletion Anamolies are very frequent if Database is not Normalized. To understand these anomalies let us

take an example of Student table.

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S_id S_Name S_Address Subject_opted

401 Adam Noida Bio

402 Alex Panipat Maths

403 Stuart Jammu Maths

404 Adam Noida Physics

Updation Anamoly : To update address of a student who occurs twice or more than twice in a table, we will have to

update S_Address column in all the rows, else data will become inconsistent.

Insertion Anamoly : Suppose for a new admission, we have a Student id(S_id), name and address of a student but if

student has not opted for any subjects yet then we have to insert NULL there, leading to Insertion Anamoly.

Deletion Anamoly : If (S_id) 401 has only one subject and temporarily he drops it, when we delete that row, entire

student record will be deleted along with it.

Normalization Rule

Normalization rule are divided into following normal form.

1. First Normal Form

2. Second Normal Form

3. Third Normal Form

4. BCNF

First Normal Form (1NF)

As per First Normal Form, no two Rows of data must contain repeating group of information i.e each set of column must

have a unique value, such that multiple columns cannot be used to fetch the same row. Each table should be organized into

rows, and each row should have a primary key that distinguishes it as unique.

The Primary key is usually a single column, but sometimes more than one column can be combined to create a single

primary key. For example consider a table which is not in First normal form

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Student Table :

Student Age Subject

Adam 15 Biology, Maths

Alex 14 Maths

Stuart 17 Maths

In First Normal Form, any row must not have a column in which more than one value is saved, like separated with

commas. Rather than that, we must separate such data into multiple rows.

Student Table following 1NF will be :

Student Age Subject

Adam 15 Biology

Adam 15 Maths

Alex 14 Maths

Stuart 17 Maths

Using the First Normal Form, data redundancy increases, as there will be many columns with same data in multiple rows

but each row as a whole will be unique.

Second Normal Form (2NF)

As per the Second Normal Form there must not be any partial dependency of any column on primary key. It means that for

a table that has concatenated primary key, each column in the table that is not part of the primary key must depend upon

the entire concatenated key for its existence. If any column depends only on one part of the concatenated key, then the

table fails Second normal form.

In example of First Normal Form there are two rows for Adam, to include multiple subjects that he has opted for. While

this is searchable, and follows First normal form, it is an inefficient use of space. Also in the above Table in First Normal

Form, while the candidate key is {Student, Subject}, Age of Student only depends on Student column, which is incorrect

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as per Second Normal Form. To achieve second normal form, it would be helpful to split out the subjects into an

independent table, and match them up using the student names as foreign keys.

New Student Table following 2NF will be :

Student Age

Adam 15

Alex 14

Stuart 17

In Student Table the candidate key will be Student column, because all other column i.e Age is dependent on it

New Subject Table introduced for 2NF will be :

Student Subject

Adam Biology

Adam Maths

Alex Maths

Stuart Maths

In Subject Table the candidate key will be {Student, Subject} column. Now, both the above tables qualifies for Second

Normal Form and will never suffer from Update Anomalies. Although there are a few complex cases in which table in

Second Normal Form suffers Update Anomalies, and to handle those scenarios Third Normal Form is there.

Third Normal Form (3NF)

Third Normal form applies that every non-prime attribute of table must be dependent on primary key, or we can say that,

there should not be the case that a non-prime attribute is determined by another non-prime attribute. So this transitive

functional dependency should be removed from the table and also the table must be in Second Normal form. For example,

consider a table with following fields.

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Student_Detail Table :

Student_id Student_name DOB Street city State Zip

In this table Student_id is Primary key, but street, city and state depends upon Zip. The dependency between zip and other

fields is called transitive dependency. Hence to apply 3NF, we need to move the street, city and state to new table,

with Zip as primary key.

New Student_Detail Table :

Student_id Student_name DOB Zip

Address Table :

Zip Street city state

The advantage of removing transtive dependency is,

Amount of data duplication is reduced.

Data integrity achieved.

Boyce and Codd Normal Form (BCNF)

Boyce and Codd Normal Form is a higher version of the Third Normal form. This form deals with certain type of

anamoly that is not handled by 3NF. A 3NF table which does not have multiple overlapping candidate keys is said to be in

BCNF. For a table to be in BCNF, following conditions must be satisfied:

R must be in 3rd Normal Form

and, for each functional dependency ( X -> Y ), X should be a super Key.

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