Relational Algebra and My SQL

Relational Algebra and My SQLProf. Sin Min Lee

Deparment of Computer Science

San Jose State University

Functional Dependency

•For all t1, t2 element of R if t1[A] = t2[A] => t1[B] = t2[B]

A, B R(….) , we say FD: AB if

Definition: A R(….) is a key if FD: A R

Definition: A R(….) is a candidate key if FD: AR

and not proper subset of A is a key.

4. Find all the candidate keys for the following table ( Mid1 Study Guide)

R(A B C D) 1 2 3 4 2 2 3 5 3 2 5 1 1 2 5 6 S (A B C D) 1 2 3 4 2 2 3 5 3 2 5 1 1 2 5 6

Queries: Q1:

AR (…)A is not a key

Q2:BR (…)B is not a key

Q3:CR (…)C is not a key

Q4:DR (…)Yes-D is a key

Q5:ABR (…)AB together cannot form the key

Q6:ACR (…)Yes- AC together cannot form the key

Q7:BCR (…)BC together cannot form the key

Relational Algebra Basic operations:

Selection ( ) Selects a subset of rows from relation. Projection ( ) Deletes unwanted columns from relation. Cross-product ( ) Allows us to combine two relations. Set-difference ( ) Tuples in reln. 1, but not in reln. 2. Union ( ) Tuples in reln. 1 and in reln. 2.

Additional operations: Intersection, join, division, renaming: Not essential, but (very!) useful.

Since each operation returns a relation, operations can be composed! (Algebra is “closed”.)

Projectionsname rating

yuppy 9lubber 8guppy 5rusty 10

sname rating

S,

( )2

age

35.055.5

age S( )2

Deletes attributes that are not in projection list.

Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation.

Projection operator has to eliminate duplicates! (Why??, what are the consequences?) Note: real systems typically

don’t do duplicate elimination unless the user explicitly asks for it.

Selection

rating

S82( )

sid sname rating age28 yuppy 9 35.058 rusty 10 35.0

sname ratingyuppy 9rusty 10

sname rating rating

S,

( ( ))82

Selects rows that satisfy selection condition.

Schema of result identical to schema of (only) input relation.

Result relation can be the input for another relational algebra operation! (Operator composition.)

Union, Intersection, Set-Difference

All of these operations take two input relations, which must be union-compatible: Same number of fields. `Corresponding’ fields

have the same type. What is the schema of

result?

sid sname rating age

22 dustin 7 45.031 lubber 8 55.558 rusty 10 35.044 guppy 5 35.028 yuppy 9 35.0

sid sname rating age31 lubber 8 55.558 rusty 10 35.0

S S1 2

S S1 2

sid sname rating age

22 dustin 7 45.0S S1 2

Cross-Product Each row of S1 is paired with each row of R1. Result schema has one field per field of S1 and

R1, with field names `inherited’ if possible.Conflict: Both S1 and R1 have a field called

sid.

( ( , ), )C sid sid S R1 1 5 2 1 1

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 22 101 10/ 10/ 96

22 dustin 7 45.0 58 103 11/ 12/ 96

31 lubber 8 55.5 22 101 10/ 10/ 96

31 lubber 8 55.5 58 103 11/ 12/ 96

58 rusty 10 35.0 22 101 10/ 10/ 96

58 rusty 10 35.0 58 103 11/ 12/ 96 Renaming operator:

Joins Condition Join:

Result schema same as that of cross-product. Fewer tuples than cross-product. Filters tuples not

satisfying the join condition. Sometimes called a theta-join.

R c S c R S ( )

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 58 103 11/ 12/ 9631 lubber 8 55.5 58 103 11/ 12/ 96

S RS sid R sid

1 11 1

. .

Joins Equi-Join: A special case of condition join where

the condition c contains only equalities.

Result schema similar to cross-product, but only one copy of fields for which equality is specified.

Natural Join: Equijoin on all common fields.

sid sname rating age bid day

22 dustin 7 45.0 101 10/ 10/ 9658 rusty 10 35.0 103 11/ 12/ 96

)11(,..,,..,

RSsidbidagesid

Division Not supported as a primitive operator, but useful for

expressing queries like: Find sailors who have reserved all boats.

Precondition: in A/B, the attributes in B must be included in the schema for A. Also, the result has attributes A-B. SALES(supId, prodId); PRODUCTS(prodId); Relations SALES and PRODUCTS must be built using

projections. SALES/PRODUCTS: the ids of the suppliers supplying

ALL products.

Examples of Division A/B

sno pnos1 p1s1 p2s1 p3s1 p4s2 p1s2 p2s3 p2s4 p2s4 p4

pnop2

pnop2p4

pnop1p2p4

snos1s2s3s4

snos1s4

snos1

A

B1B2

B3

A/B1 A/B2 A/B3

Expressing A/B Using Basic Operators

Division is not essential op; just a useful shorthand. (Also true of joins, but joins are so common that

systems implement joins specially. Division is NOT implemented in SQL).

Idea: For SALES/PRODUCTS, compute all products such that there exists at least one supplier not supplying it. x value is disqualified if by attaching y value from B,

we obtain an xy tuple that is not in A.))Pr)((( SalesoductsSales

sidsidA

The answer is sid(Sales) - A

EQUALITY JOIN, NATURAL JOIN, JOIN, SEMI-JOIN Equality join connects tuples from two relations that match on

certain attributes. The specified joining columns are kept in the resulting relation. ∏name(бdname=‘toy’(Emp Dept)))

Natural join connects tuples from two relations that match on the specified common attributes ∏name(бdname=‘toy’(Emp Dept)))

How is an equality join between Emp and Dept using dno different than a natural join between Emp and Dept using dno? Equality join: SS#, name, age, salary, Emp.dno,

Dept.dno, … Natural join: SS#, name, age, salary, dno, dname, …

Join is similar to equality join using different comparison operators A S op = {=, ≠, ≤, ≥, <, >} att op att

(dno)

(dno)

EXAMPLE JOIN

Equality Join, (Emp Dept)))

SS# Name Age Salary dno

1 Joe 24 20000 2

2 Mary 20 25000 1

3 Bob 22 27000 1

4 Kathy 30 30000 2

5 Shideh 4 4000 1

EMP

dno dname floor mgrss#

1 Toy 1 5

2 Shoe 2 1

Dept

(dno)

SS# Name Age Salary EMP.dno Dept.dno dname floor mgrss#

1 Joe 24 20000 2 2 Shoe 2 1

2 Mary 20 25000 1 1 Toy 1 5

3 Bob 22 27000 1 1 Toy 1 5

4 Kathy 30 30000 2 2 Shoe 2 1

5 Shideh 4 4000 1 1 Toy 1 5

EXAMPLE JOIN

Natural Join, (Emp Dept)))


1 Joe 24 20000 2

2 Mary 20 25000 1

3 Bob 22 27000 1

4 Kathy 30 30000 2

5 Shideh 4 4000 1

EMP


1 Toy 1 5

2 Shoe 2 1

Dept

(dno)

SS# Name Age Salary dno dname floor mgrss#

1 Joe 24 20000 2 Shoe 2 1

2 Mary 20 25000 1 Toy 1 5

3 Bob 22 27000 1 Toy 1 5

4 Kathy 30 30000 2 Shoe 2 1

5 Shideh 4 4000 1 Toy 1 5

EXAMPLE JOIN

Join, (Emp ρx(Emp))))


1 Joe 24 20000 2

2 Mary 20 25000 1

3 Bob 22 27000 1

4 Kathy 30 30000 2

5 Shideh 4 4000 1

EMP


1 Toy 1 5

2 Shoe 2 1

Dept

Salary > 5 * salary

SS# Name Age Salary dno x.SS# x.Name x.Age x.Salary x.dno

2 Mary 20 25000 1 2 Shideh 4 4000 1

3 Bob 22 27000 1 3 Shideh 4 4000 1

4 Kathy 30 30000 2 4 Shideh 4 4000 1

EQUALITY JOIN, NATURAL JOIN, JOIN, SEMI-JOIN (Cont…)

Example: retrieve the name of employees who earn more than Joe: ∏name(Emp (sal>x.sal)бname=‘Joe’(ρ x(Emp)))

Semi-Join selects the columns of one relation that joins with another. It is equivalent to a join followed by a projection: Emp (dno)Dept ≡∏SS#, name, age, salary, dno(Emp

Dept)

JOIN OPERATORS Condition Joins: - Defined as a cross-product followed by a selection:

R ⋈c S = σc(R S) ( is called the bow-tie)⋈where c is the condition.

- Example:Given the sample relational instances S1 and R1

The condition join S ⋈S1.sid<R1.sid R1 yields

Equijoin:Special case of the condition join where the join condition consists solely of

equalities between two fields in R and S connected by the logical AND operator ( ).∧

Example: Given the two sample relational instances S1 and R1

The operator S1 R.sid=Ssid R1 yields

SQL SQL (Structured Query Language) is

the standard language for commercial DBMSs SEQUEL (Structured English QUEry Language)

was originally defined by IBM for System R standardization of SQL began in the 80s current standard is SQL-99 SQL is more than a query language it includes a DDL, DML and

administration commands SQL is an example of a transform-oriented language. A language designed to use relations to transform inputs into required

outputs.

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Basic structure of an SQL Basic structure of an SQL queryquery22

GeneralStructure

SELECT, ALL / DISTINCT, *,AS, FROM, WHERE

Comparison IN, BETWEEN, LIKE "% _"

Grouping GROUP BY, HAVING,COUNT( ), SUM( ), AVG( ), MAX( ), MIN( )

Display Order ORDER BY, ASC / DESC

LogicalOperators

AND, OR, NOT

Output INTO TABLE / CURSORTO FILE [ADDITIVE], TO PRINTER, TO SCREEN

Union UNION

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fieldfield typetype widthwidth contentscontentsid numeric 4 student id numbername character 10 namedob date 8 date of birthsex character 1 sex: M / Fclass character 2 classhcode character 1 house code: R, Y, B, Gdcode character 3 district coderemission logical 1 fee remissionmtest numeric 2 Math test score

22 The Situation:Student ParticularsThe Situation:Student Particulars

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General StructureGeneral StructureII

SELECTSELECT [[ALL / DISTINCTALL / DISTINCT] ] expr1expr1 [ [ASAS col1col1], ], expr2expr2 [ [ASAS col2col2] ] ;;

FROMFROM tablenametablename WHEREWHERE conditioncondition

SELECT ...... FROM ...... WHERE ......SELECT ...... FROM ...... WHERE ......

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General StructureGeneral StructureII The query will select rows from the source tablename and output the result in table form.

Expressions expr1, expr2 can be : (1) a column, or (2) an expression of functions and fields.



And col1, col2 are their corresponding column names in the output table.

3.2 SQL

SELECT a1, ..., an

FROM R1, R2, …, Rm

WHERE Con1, …,Conk

This means: π a1, ..., an

( σ Con1( … (σ Conk

( R1 × R2 × … × Rm))…))

Example:

Find the SSN and tax for each person.

SELECT SSN, Tax FROM Taxrecord, Taxtable WHERE wages + interest + capital_gain = income

AS – keyword used to rename relations

Two SQL expressions can be combined by:

INTERSECT

UNION

MINUS – set difference

Example:

Find the names of the streets that intersect.

SELECT S.NAME, T.NAME

FROM Streets AS S, Streets AS T

WHERE S.X = T.X and

S.Y = T.Y

Example: Assume we have the relations:

Broadcast ( Radio, X , Y )

Town ( Name, X, Y )

Find the parts of Lincoln, NE that can

be reached by at least one Radio station.

(SELECT X, Y

FROM Town

WHERE Name = “Lincoln”)

INTERSECT

(SELECT X, Y

FROM Broadcast)

Example:

Find the SSN and tax for each person.

πSSN,Tax σwages+interest+capital_gain = income Taxrecord × Taxtable

Example:

Find the area of Lincoln reached by a radio station.

( πX,Y ( σName=“Lincoln” Town ) ) ( πX,Y Broadcast )

Another way of connecting SQL expressions

is using the IN keyword.

SELECT ……..

FROM ……..

WHERE a IN ( SELECT b

FROM …..

WHERE ….. )

SQL with aggregation – SELECT aggregate_function FROM ……. WHERE ……

aggregate_function –

Max (c1a1 + ……..+ cnan) where ai are attributes

Min (c1a1 + ……..+ cnan) and ci are constants

Sum(a) where a is an attribute that is Avg(a) constant in each constraint tuple Count(a)

Example:

Package(Serial_No, From, Destination, Weight)

Postage (Weight , Fee)

Find the total postage of all packages sent

from Omaha.

SELECT Sum(Fee)

FROM Package, Postage

WHERE Package.Weight = Postage.Weight AND

Package.From = “ Omaha “

GROUP BY –

SELECT a1, …, an, aggregate_function FROM ….. WHERE ……

GROUP BY a1, ..., ak

• Evaluates basic SQL query• Groups the tuples according to different values of

a1,..,ak

• Applies the aggregate function to each group separately

• {a1, …, ak} {a1, …, an}

Example:

Find the total postage sent out from each city.

SELECT Package.From, Sum(Postage.Fee)

FROM Package, Postage

WHERE Package.Weight = Postage.Weight

GROUP BY Package.From

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General StructureGeneral StructureII DISTINCT will eliminate duplication in the output

while ALL will keep all duplicated rows.

condition can be : (1) an inequality, or (2) a string comparison using logical operators AND, OR, NOT.



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General StructureGeneral StructureIIBefore using SQL, open the student file:

USE studentUSE student

eg. 1eg. 1 List all the student records.List all the student records.

SELECT * FROM student

id name dob sex class mtest hcode dcode remission9801 Peter 06/04/86 M 1A 70 R SSP .F.9802 Mary 01/10/86 F 1A 92 Y HHM .F.9803 Johnny 03/16/86 M 1A 91 G SSP .T.9804 Wendy 07/09/86 F 1B 84 B YMT .F.9805 Tobe 10/17/86 M 1B 88 R YMT .F.: : : : : : : : :

Result

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General StructureGeneral StructureIIeg. 2eg. 2 List the names and house code of 1A students.List the names and house code of 1A students.

SELECT name, hcode, class FROM student ;

WHERE class="1A"

Class

11AA

11AA

11AA

11BB

11BB

::

Class

11AA

11AA

11AA

11BB

11BB

::

class="1A"

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General StructureGeneral StructureII

name hcode classPeter R 1AMary Y 1AJohnny G 1ALuke G 1ABobby B 1AAaron R 1A: : :

Result

eg. 2eg. 2 List the names and house code of 1A students.List the names and house code of 1A students.

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General StructureGeneral StructureIIeg. 3eg. 3 List the residential district of the Red House List the residential district of the Red House

members.members.

SELECT DISTINCT dcode FROM student ;

WHERE hcode="R"

dcodeHHMKWCMKKSSPTSTYMT

Result

Data Manipulation

Select: query data in the database Insert: insert data into a table Update: updates data in a table Delete: delete data from a table

Source: Database Systems Connolly/Begg

Retrieve all columns and all rows

SELECT firstColumn,…,lastColumn

FROM tableName;

SELECT *

FROM tableName;

Use of Distinct

SELECT DISTINCT columnName

FROM tableName;

columnName

A

B

C

D

columnName

A

A

B

B

C

D

Calculated fields

SELECT columnName/2

FROM tableName

price

5.00

3.00

6.00

price

10.00

6.00

12.00

Comparison Search Condition

= equals

< > is not equal to (ISO standard)

!= “ “ “ “ (allowed in some dialects)

< is less than

> is greater than

<= is less than or equal to

>= is greater than or equal to


Comparison Search Condition

An expression is evaluated left to right.

Subexpressions in brackets are evaluated first.

NOTs are evaluated before ANDs and ORs.

ANDs are evaluated before ORs.


Range Search ConditionSELECT columnNameFROM tableNameWHERE columnName BETWEEN 20

AND 30;

SELECT columnNameFROM tableNameWHERE columnName >= 20 AND columnName <= 30;

Set membership search condition

SELECT columnName

FROM tableName

WHERE columnName

IN (‘name1’, ‘name2’);

SELECT columnName

FROM tableName

WHERE columnName = ‘name1’

OR columnName = ‘name2’;

Pattern matching symbols

% represents any sequence of zero

or more characters (wildcard).

_ represents any single character


Pattern match search condition‘h%’ : begins with the character h .

‘h_ _ _’ : four character string beginning with the character h.

‘%e’ : any sequence of characters, of length at least 1, ending with the character e.

‘%CS157B%’ : any sequence of characters of any length containing CS157B


Pattern match search condition

LIKE ‘h%’

begins with the character h .

NOT LIKE ‘h%’

does not begin with the character h.


Pattern match search condition

To search a string that includes a

pattern-matching character

‘15%’

Use an escape character to represent

the pattern-matching character.

LIKE ‘15#%’ ESCAPE ‘#’


NULL search condition

DOES NOT WORK

comment = ‘ ’

comment != ‘ ’

DOES WORK

comment IS NULL

comment IS NOT NULL

Sorting

The ORDER BY clause consists of list of column

identifiers that the result is to be sorted on, separated by commas.

Allows the retrieved rows to be ordered by ascending (ASC) or descending (DESC) order


Sorting

Column identifier may be A column name A column number (deprecated)


Sorting

SELECT type, rent

FROM tableName

ORDER BY type, rent ASC;


type rent

Apt

Apt

Flat

Flat

450

500

600

650

type rent

Flat

Apt

Flat Apt

650 450

600 500

Aggregate Functions

COUNT returns the number … SUM returns the sum … AVG returns the average … MIN returns the smallest … MAX returns the largest …

value in a specified column.


Use of COUNT( * )

How many students in CS157B?

SELECT COUNT( * ) AS my count

FROM CS157B

my count

40

GROUP BY clause

When GROUP BY is used, each item in the SELECT list must be single-valued per group.

The SELECT clause may contain only Column names Aggregate functions Constants An expression involving combinations of the

above


GroupingSELECT dept, COUNT(staffNo) AS my count

SUM(salary)

FROM tableName

GROUP BY dept

ORDER BY dept

dept my count

Salary

A

B

C

2

2

1

300.00300.00 200.00

dept staffNo Salary

A

B

C

A

B

1

1

1

2

2

200.00

200.00

200.00

100.00

100.00

Restricting Grouping

HAVING clause is with the GROUP BY clause. filters groups into resulting table. includes at least one aggregate

function. WHERE clause

filters individual rows into resulting table.

Aggregate functions cannot be used.


SELECT dept, COUNT(staffNo) AS my count, SUM(salary) AS my sum

FROM StaffGROUP BY deptHAVING COUNT(staffNo) > 1ORDER BY dept;


dept my count

my sum

A

B

2

2

300.00 300.00

dept staffNo Salary

A

B

C

A

B

1

1

1

2

2

200.00

200.00

200.00

100.00

100.00

Subqueries

SELECT columnNameA

FROM tableName1

WHERE columnNameB = (SELECT columnNameB

FROM tableName2

WHERE condition);


result from inner SELECT applied as a condition for the outer SELECT

Subquery with Aggregate Function

SELECT fName, salary –

( SELECT AVG(salary)

FROM Staff ) AS salDiff

FROM Staff

WHERE salary > ( SELECT AVG(salary)

FROM Staff );


List all staff whose salary is greater than the average salary,show by how much their salary is greater than the average.

Nested Subqueries: Use of IN

SELECT propertyFROM PropertyForRentWHERE staff IN(

SELECT staffFROM StaffWHERE branch = (

SELECT branch FROM Branch WHERE street = ‘112 A St’));


Selects branch at 112 A St


SELECT property

FROM PropertyForRent

WHERE staff IN(

SELECT staff

FROM Staff

WHERE branch = ( branch ) );


Select staff members who works at branch.


SELECT property

FROM PropertyForRent

WHERE staff IN( staffs who works

at branch on ‘112 A St’);


Since there are more than one row selected, “=“ cannot be used.

Use of ANY/SOME

SELECT name, salary

FROM Staff

WHERE salary > SOME( SELECT salary

FROM Staff

WHERE branch = ‘A’ );


Result:{2000,3000,4000}

Result: {list of staff with salary greater than 2000.}

Use of ALL

SELECT name, salary

FROM Staff

WHERE salary > ALL( SELECT salary

FROM Staff

WHERE branch = ‘A’ );


Result:{2000,3000,4000}

Result: {list of staff with salary greater than 4000.}

Use of Any/Some and All

If the subquery is empty: ALL returns true ANY returns false

ISO standard allows SOME to be

used interchangeably with ANY.


Relational Algebra and My SQL

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