Relational Model Concepts
Relational Model Concepts
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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 Structure
Employee
Empno
101
102
103
Ename
Anu
Raja
Rani
Salary
40000
50000
35000
Dno
10
20
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 operations
Additional operations
Extended operations
Basic operations
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Select ()
Project ()
Union (U)
Rename ()
Set difference (-)
Cartesian product (x)
Additional operations
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Set intersection ()
Natural join
Division ()
Assignment ()
Extended operations
• Aggregate operations
• Outer join
Basic operations
The 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 name
Example
1) dno=10 (emp)
2) ename=‘raja’(emp)
Note :
Display eno,ename,salar dno (all heading)
Project operation
The required attributes alone can be listed from a
relation (selects certain columns from the table)
Syntax
 <select condition> (R)
Note :
R is table name
Example
1) 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/tR 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)
Example
R IXI S =  r.A, r.B, r.C, r.D, s.E (r.B =s.B and r.D=s.D (RXS))
equal
r.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
All combinations of
tuples satisfying the joint tuples are included in
condition appear in the the result.
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 Algebra
person_name (company_name="First Bank Corporation” (works))
SQL Query
SQL> Select person_name from works where company_name=“First Bank
Corporation”;
employee (person_name, street, city)
works (person_name, company_name, salary)
Company (company_name,city)
manages (person_name, manager_name)
b) Find the names and cities of residence of all employees
who work for First Bank Corporation
Relation Algebra
person_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 ;
employee (person_name, street, city)
works (person_name, company_name, salary)
Company (company_name,city)
manages (person_name, manager_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 Algebra
person_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 ;
employee (person_name, street, city)
works (person_name, company_name, salary)
Company (company_name,city)
manages (person_name, manager_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 Algebra
person_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;
employee (person_name, street, city)
works (person_name, company_name, salary)
Company (company_name,city)
manages (person_name, manager_name)
e)
Find the names of all employees who live in the same city and on
the same street as do their managers.
Relation Algebra
person_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;
employee (person_name, street, city)
works (person_name, company_name, salary)
Company (company_name,city)
manages (person_name, manager_name)
f) Find the names of all employees in this database who
do not work for First Bank Corporation
Relation Algebra
person_name (employee.person_name<> First Bank Corporation” (Works))
SQL Query
SQL> Select person_name from works where company_name<> “First
Bank Corporation”;