relational model concepts. the relational model represents the database as a collection of...

Relational Model Concepts

Relational Model Concepts

The relational model represents the database as a collection of relations. Each relation resembles a table of values.

• A table is called a relation.

• A row is called a tuple.

• A column header is called an attribute.

• The data type describing the type that can appear in each column is called a domain.

• A domain (D) is a set of atomic values.

Basic StructureEmployee Empno Ename Salary Dno 101 Anu 40000 10 102 Raja 50000 20 103 Rani 35000 30

Employee - Relation name

Empno, ename, salary, dno – Attribute

All records are Tuples

Query Language

• Categories– Procedural

• Example Relational Algebra

– Non Procedural• Example Relational Calculus

Relation Algebra• It is procedural language• It consists of set of operations that take

one or more relations as input and produces new relation as output.

The operation can be divided into three parts

Basic operationsAdditional operationsExtended operations

Basic operations

• Select ()

• Project ()

• Union (U)

• Rename ()

• Set difference (-)

• Cartesian product (x)

Additional operations

• Set intersection ()

• Natural join

• Division ()

• Assignment ()

Extended operations

• Aggregate operations

• Outer join

Basic operationsThe Select operation ()

– It is used to select a subset of the tuples (records) from a relation that satisfy a selection condition.

Syntax <select condition> (R)

Note :R is table nameExample

1) dno=10 (emp)

2) ename=‘raja’(emp)Note :Display eno,ename,salar dno (all heading)

Project operationThe required attributes alone can be listed from a

relation (selects certain columns from the table)Syntax

<select condition> (R)Note :R is table nameExample1) eno,salary (emp)

display eno, salary (two column)2) ename dno=10 (emp)Note :Display ename column only

The union operation (U)

• RUS is all tuples that are either in R or in S. Duplicate tuples are eliminated.

Syntax

RUS

Example

ename (emp1) U ename (emp2)

The Set difference operation ( - )

• It is used to find tuples that are in one relation but are not in another.

• Syntax

R - S

• Example

ename (emp) U ename (empit)

Cartesian Product operation (X)• It allows us to combine information from

any 2 relations.

Syntax

R X S = { tq/tR and q s}

• Example

dno=10 (emp X dept)

Rename operation ()

• Allow us to name the relational algebra expressions

• Allow us to refer to a relation by more than one name.

Example

temp (eno,salary (emp))

Join operation

• R=(A,B,C,D) s=(E,B,D)– (A,B,C,D,E Column heading)

R IXI S=(A,B,C,D,E)

ExampleR IXI S = r.A, r.B, r.C, r.D, s.E (r.B =s.B and r.D=s.D (RXS))

equalr.B =s.B and r.D=s.D (R IXI S))

• Division operation ( r s)

R=(A1,A2,A3, …… Am, B1,B2,B3,…….Bn)

S=(B1,B2,…..Bn)

The result of r s is

r s =(A1,A2,A3, …… Am)

Assignment Operation ( )• It can be assigned to a temporary relation

variable (S).

Example

Temp ( ) eno,salary (emp)

Temp1 ( ) eno,salary salary>10000 (emp)

Aggregate functions

• Aggregate function takes a collection of values and returns a single values as a result.

AVG – average value

MIN – minimum value

MAX – Maximum value

SUM – Sum of values

COUNT – number of record

Syntax

G Aggregate_function (attribute_name) (relation name)

Example

G Sum(salary)(emp)

dno G Sum(salary)(emp)

Find sum of salary for employees to each department.

Difference between join (IXI) and Cartesian product (X)

Join (IXI) Cartesian product (X)

Only combinations of tuples satisfying the joint condition appear in the result.

All combinations of tuples are included in the result.

employee (person_name, street, city)works (person_name, company_name, salary)Company (company_name,city)manages (person_name, manager_name)Consider the relation database, where the primary keys are underlined.

Give an expression in relational algebra, SQL query of the following

a) Find the names of all employees who work for First Bank Corporation

b) Find the names and cities of residence of all employees who work for First Bank Corporation

c) Find the names, street address, cities of residence of all employees who work for bank and earn more than Rs. 10,000 per annum.

d) Find the names of all employees in this database who live in the same city as the company for which they work.

e) Find the names of all employees who live in the same city and on the same street as do their managers.

f) Find the names of all employees in this database who do not work for First Bank Corporation

employee (person_name, street, city)works (person_name, company_name, salary)Company (company_name,city)manages (person_name, manager_name)

a) Find the names of all employees who work for First Bank Corporation

Relation Algebraperson_name (company_name="First Bank Corporation” (works))

SQL Query

SQL> Select person_name from works where company_name=“First Bank Corporation”;


b) Find the names and cities of residence of all employees who work for First Bank Corporation

Relation Algebraperson_name, city (company_name="First Bank Corporation” ^

employee.person_name=works.person_name(Employee X works))

SQL Query

SQL> Select person_name, city from employee, works where company_name=“First Bank Corporation” and employee.person_name=works.person_name ;


c) Find the names, street address, cities of residence of all employees who work for bank and earn more than Rs. 10,000 per annum.

Relation Algebraperson_name,street,city (company_name="First Bank Corporation” ^ Salary>10000 ^

employee.person_name=works.person_name(Employee X works))

SQL Query

SQL> Select person_name, street, city from employee, works where company_name=“First Bank Corporation” and salary>10000 and employee.person_name=works.person_name ;


d) Find the names of all employees in this database who live in the same city as the company for which they work.

Relation Algebraperson_name (employee.city=company.city ^ employee.person_name

=works.person_name^company.company_name=works.company_name(Employee X works X company))

SQL Query

SQL> Select person_namefrom employee, works, company where employee.city=company.city and works.comapany_name = company.company_name;


e) Find the names of all employees who live in the same city and on the same street as do their managers.

Relation Algebraperson_name (employee.person_name=manges.person_name ^ company.city=employee.city (Employee X

manages X company))

SQL Query

SQL> Select person_name from employee ,manages , company where employee.person_name=manages.person_name And company.city=employee.city;


f) Find the names of all employees in this database who do not work for First Bank Corporation

Relation Algebraperson_name (employee.person_name<> First Bank Corporation” (Works))

SQL Query

SQL> Select person_name from works where company_name<> “First Bank Corporation”;

relational model concepts. the relational model represents the database as a collection of...

Documents