query languages (basic operator)

8/3/2019 Query Languages (basic operator)

1/21

Query Languages

Language in which user requests information from the database.

Categories of languages

Procedural

Non-procedural, or declarative

Pure languages:

Relational algebra

Tuple relational calculus

Domain relational calculus


2/21

Relational Algebra

It is a Procedural language

It consist of set of operation which takes one or two relation asinput and produce relation as output.

Six basic operators

select: - unary

project: - unary

rename: -unary

union: - binary

set difference: - binary

Cartesian product: x - binary

The operators take one or two relations as inputs and producea new relation as a result.


3/21

Additional Operations

We define additional operations that do not add any power to the

relational algebra, but that simplify common queries.

Set intersection

Natural join

Division

Assignment


4/21

Select Operation

Notation: p(r)

pis called the selection predicate Defined as:

Each term is one of:

op or

where opis one of: =, , >, .


5/21

Select Operation Example

Relation rA B C D

1

5

12

23

7

7

3

10

A=B ^ D > 5(r)A B C D

1

23

7

10


6/21

Project Operation

Notation:

where A1, A2are attribute names and ris a relation name.

The result is defined as the relation of kcolumns obtained byerasing the columns that are not listed

Duplicate rows removed from result, since relations are sets

Example: To eliminate the branch_nameattribute of account

account_number, balance(account)

)(,,, 21 rkAAA


7/21

Project Operation Example

Relation r: A B C

10

20

30

40

1

1

1

2

A C

1

1

1

2

=

A C

1

1

2

A,C (r)


8/21

Find those customers who live in xyz :

customer_name(cutomer-city = xyz(account))

Composition of Operations


9/21

Union Operation

Notation: rs

Defined as:

r s= {t| tror ts}

For rsto be valid.

1. r,smust have the samearity (same number of attributes)

2. The attribute domains must be compatible (example: 2nd

column of rdeals with the same type of values as does the 2nd

column of s)

Example: to find all customers with either an account or a loan

customer_name(depositor) customer_name(borrower)


10/21

Union Operation Example

Relations r, s:

r s:

A B

1

2

1

A B

2

3

rs

A B

1

21

3


11/21

Set Difference Operation

Notation r s

Defined as:

r s = {t| trand t s}

Set differences must be taken between compatiblerelations.

rand smust have the same arity

attribute domains of rand smust be compatible


12/21

Set Difference Operation Example

Relations r, s:

r s:

A B

1

2

1

A B

2

3

r

s

A B

11


13/21

Cartesian-Product Operation

Notation rx s

Defined as:

rx s= {t q| t rand qs}

Assume that attributes of r(R) and s(S) are disjoint. (That is, R

S= ). If attributes of r(R) and s(S) are not disjoint, then renaming must

be used.


14/21

Cartesian-Product Operation Example

Relations r, s:

rx s:

A B

1

2

A B

111

12222

C D

101020

1010102010

E

aab

baabb

C D

10102010

E

aabbr

s


15/21

Composition of Operations Can build expressions using multiple operations

Example: A=C(r x s)

r x s

A=C(r x s)

A B

111

12222

C D

101020

1010102010

E

aab

baabb

A B C D E

122

101020

aab


16/21

Rename Operation

Allows us to name, and therefore to refer to, the results of

relational-algebra expressions. Allows us to refer to a relation by more than one name.

Example:

x(E)

returns the expression Eunder the name X

If a relational-algebra expression Ehas arity n, then

returns the result of expression Eunder the name X, and with the

attributes renamed to A1 , A2, ., An.

)(),...,,( 21E

nAAAx


17/21

Banking Example

branch (branch_name, branch_city, assets)

customer (customer_name, customer_street, customer_city)

account (account_number, branch_name, balance)

loan (loan_number, branch_name, amount)

depositor (customer_name, account_number)

borrower(customer_name, loan_number)


18/21

Example Queries

Find all loans of over $1200

Find the loan number for each loan of an amount greater than

$1200

amount> 1200 (loan)

loan_number(amount> 1200 (loan))

Find the names of all customers who have a loan, an account, or both,

from the bank

customer_name(borrower) customer_name(depositor)


19/21

Example Queries

Find the names of all customers who have a loan at the Perryridge branch.

Query 1

customer_name (branch_name = Perryridge(

borrower.loan_number = loan.loan_number (borrower x loan)))


20/21

Example Queries

Find the names of all customers who have a loan at the Perryridgebranch but do not have an account at any branch of the bank.

customer-name(branch-name = Perryridge

(borrower.loan-number = loan.loan-number(borrower x loan))) customer-name(depositor)


21/21

Formal Definition

A basic expression in the relational algebra consists of either one of the

following: A relation in the database

A constant relation

Let E1 and E2 be relational-algebra expressions; the following are all

relational-algebra expressions:

E1E2

E1E2

E1 x E2

p(E1), Pis a predicate on attributes in E1

s(E1), Sis a list consisting of some of the attributes in E1

x(E1), x is the new name for the result of E1

query languages (basic operator)

Documents