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MIS3150 Data and Information Management

Query Languages –SQL

Arijit Sengupta

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Structure of this semester

Database Fundamentals

Relational Model

Normalization

ConceptualModeling Query

Languages

AdvancedSQL

0. Intro 1. Design

Newbie Users ProfessionalsDesigners

Java DB Applications –

JDBC

4. Applications

MIS3150

2. Querying

Developers

Transaction Management

DataMining

3. AdvancedTopics

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Today’s Buzzwords

• Query Languages• Formal Query Languages• Procedural and Declarative Languages• Relational Algebra• Relational Calculus• SQL• Aggregate Functions• Nested Queries

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Objectives

At the end of the lecture, you should • Get a formal as well as practical perspective on query

languages• Have a background on query language basics (how they

came about)• Be able to write simple SQL queries from the specification• Be able to look at SQL queries and understand what it is

supposed to do• Be able to write complex SQL queries involving nesting• Execute queries on a database system

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Set Theory Basics

• A set: a collection of distinct items with no particular order

• Set description: { b | b is a Database Book} {c | c is a city with a population of over a

million} {x | 1 < x < 10 and x is a natural number}

• Most basic set operation:Membership: x S (read as x belongs to S if

x is in the set S)

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Other Set Operations

• Addition, deletion (note that adding an existing item in the set does not change it)

• Set mathematics:Union R S = { x | x R or x S} Intersection R S = { x | x R and x S}Set Difference R – S = { x | x R and x S}Cross-product R x S = { <x,y> | x R and y S}

• You can combine set operations much like arithmetic operations: R – (S T)

• Usually no well-defined precedence

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Relational Query Languages

• Query languages: Allow manipulation and retrieval of data from a database.

• Relational model supports simple, powerful QLs: Strong formal foundation based on logic. Allows for much optimization.

• Query Languages != programming languages! QLs not expected to be “Turing complete”. QLs not intended to be used for complex calculations. QLs support easy, efficient access to large data sets.

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Formal Relational Query Languages

Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation:

Relational Algebra: More operational, very useful for representing execution plans.

Relational Calculus: Lets users describe what they want, rather than how to compute it. (Non-operational, declarative.)

Understanding Algebra & Calculus is key to understanding SQL, query processing!

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Structured Query Language

• Need for SQLOperations on Data TypesDefinition ManipulationOperations on SetsDeclarative (calculus) vs. Procedural (algebra)

• Evolution of SQLSEQUEL ..SQL_92 .. SQL_93SQL Dialects

Does SQL treat Relations as ‘Sets’?

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Preliminaries

• A query is applied to relation instances, and the result of a query is also a relation instance. Schemas of input relations for a query are fixed

(but query will run regardless of instance!) The schema for the result of a given query is also

fixed! Determined by definition of query language constructs.

• Positional vs. named-field notation: Positional notation easier for formal definitions,

named-field notation more readable. Both used in SQL

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Example Instances

sid sname GPA age

22 dustin 3.5 25.0

31 lubber 3.8 25.558 rusty 4.0 23.0

sid sname GPA age28 yuppy 3.9 24.031 lubber 3.8 25.544 guppy 3.5 25.558 rusty 4.0 23.0

sid cid semester

22 101 Fall 9958 103 Spring 99

R1

S1 S2

• Students, Registers, Courses relations for our examples.

cid cname dept

101 Database CIS103 Internet ECI

C1

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• 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”.)

Relational Algebra

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Projection

sname GPA

yuppy 3.9lubber 3.8guppy 3.5rusty 4.0

)2(, Sgpasnameage

24.025.523.0

)2(Sage

• 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??) Note: real systems typically

don’t do duplicate elimination unless the user explicitly asks for it. (Why not?)

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Vertical Slices

• ProjectionSpecifying Elements

No Specification

List all information about Students

select *

fromSTUDENT;

(Student)

Conditional

List IDs, names, and addresses of all students

select StudentID, name, address

from STUDENT;

StudentID, name, address (Student)

Algebra: projection

<A1,A2,...Am> (R)

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Does SQL treat Relations as ‘Sets’?

What are the different salaries we pay to our employees?

select salary

from EMPLOYEE;

OR is the following better?

select DISTINCT salary

from EMPLOYEE;

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Selection

)2(8.3

Sgpa

sid sname GPA age28 yuppy 3.9 35.058 rusty 4.0 35.0

sname GPAyuppy 3.9rusty 4.0

))2(8.3

(, Sgpagpasname

• Selects rows that satisfy selection condition.

• No duplicates in result! (Why?)

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

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

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Horizontal Slices

• Restriction Specifying Conditions

Unconditional

List all students

select *

fromSTUDENT;

(Student)

Conditional

List all students with GPA > 3.0

select *

from STUDENT

where GPA > 3.0;

GPA > 3.0 (Student)

Algebra: selection

or restriction (R)

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Specifying Conditions

List all students in ...select * from STUDENTwhere city in (‘Boston’,’Atlanta’);

List all students in ...select * from STUDENTwhere zip not between 60115 and 60123;

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Pattern Matching

‘%’ any string with n characters, n>=0‘_’ any single character. x exact sequence of string x.

List all CIS 3200 level courses.select * from COURSEwhere course# like ? ;

List all CIS courses.select * from COURSEwhere course# like ‘CIS%’;

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Missing or Incomplete Information

•List all students whose address or telephone number is missing:

select *

from STUDENT

where Address is null or GPA is null;

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Horizontal and Vertical

Query:List all student ID, names and addresses who have GPA > 3.0 and date of birth before Jan 1, 1980.

select StudentID, Name, Addressfrom STUDENTwhere GPA > 3.0 and DOB < ‘1-Jan-80’order by Name DESC;

Algebra: StudentID,name, address ( GPA > 3.0 and DOB < ‘1-Jan-80’ (STUDENT))Calculus: {t.StudentID, t.name, t.address | t Student t.GPA > 3.0

t.DOB < ‘1-Jan-80’}

Order by sorts result in descending (DESC) order. Note: The default order is ascending (ASC) as in:

order by Name;

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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 gpa age

22 dustin 3.5 25.031 lubber 3.8 25.558 rusty 4.0 23.044 guppy 3.5 25.528 yuppy 3.9 24.0

sid sname gpa age31 lubber 3.8 25.558 rusty 4.0 23.0

S S1 2

S S1 2sid sname gpa age

22 dustin 3.5 25.0

S S1 2

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Union

List students who live in Atlanta or GPA > 3.0

select StudentID, Name, DOB, Address

from STUDENT

where Address = ‘Atlanta’

union

select StudentID, Name, DOB, Address

from STUDENT

where GPA > 3.0;

Can we perform a Union on any two Relations ?

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Union Compatibility

Two relations, A and B, are union-compatible

ifA and B contain a same number of attributes, andThe corresponding attributes of the two have the same domains Examples

CIS=Student (ID: Did; Name: Dname; Address: Daddr; Grade: Dgrade);

Senior-Student (SName: Dname; S#: Did; Home: Daddr; Grade: Dgrade);

Course (C#: Dnumber; Title: Dstr; Credits: Dnumber)

Are CIS-Student and Senior-Student union compatible?

Are CIS-Student and Course union compatible?

What happens if we have duplicate tuples?

What will be the column names in the resulting Relation?

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Union, Intersect, Minus

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ UNIONselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ UNIONselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ INTERSECTselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ INTERSECTselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ MINUSselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

select CUSTNAME, ZIPfrom CUSTOMERwhere STATE = ‘MA’ MINUSselect SUPNAME, ZIPfrom SUPPLIERwhere STATE = ‘MA’ ORDER BY 2;

B

A

B

A

B

AA

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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 GPA Age (sid) cid semester

22 dustin 3.5 25.0 22 101 Fall 99

22 dustin 3.5 25.0 58 103 Spring 99

31 lubber 3.8 25.5 22 101 Fall 99

31 lubber 3.8 25.5 58 103 Spring 99

58 rusty 4.0 23.0 22 101 Fall 99

58 rusty 4.0 23.0 58 103 Spring 99

Renaming operator:

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Joins

• Condition Join:

• Result schema same as that of cross-product.• Fewer tuples than cross-product, might be able

to compute more efficiently• Sometimes called a theta-join.

R c S c R S ( )

(sid) sname GPA age (sid) cid Semester

22 dustin 3.5 25.0 58 103 Spring 9931 lubber 3.8 25.5 58 103 Spring 99

S RS sid R sid

1 11 1

. .

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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 GPA age cid semester

22 dustin 3.5 25.0 101 Fall 9958 rusty 4.0 23.0 103 Spring 99

S Rsid

1 1

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Find names of students who have taken course #103

Solution 2:)Re,1(

103gistersTemp

cid

)1,2( StudentsTempTemp sname Temp( )2

Solution 3:))Re(

103( Studentsgisters

cidsname

))Re((103

Studentsgisterscidsname

Solution 1:

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Connecting/Linking Relations

List information about all students and the classes they are taking

What can we use to connect/link Relations?Join: Connecting relations so that relevant tuples can be retrieved.

ID Name ***s1 Jose ***s2 Alice ***s3 Tome ****** *** ***

Emp# ID C# ***

e1 s1 BA 201 ***

e3 s2 CIS 300 ***

e2 s3 CIS 304 ***

*** *** ***

Student

Class

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Join

CartesianProduct

Student: 30 tuples Class: 4 tuples

Total Number of Tuples in the Cartesian Product. ? (match each tuple of student to every tuple of class)

Select tuples having identical Student Ids.Expected number of such Tuples: Join Selectivity

R1 R2

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Join Forms

• General Join FormsEquijoinOperator Dependent

• Natural Join

• Outer JoinLeftRight

Fullselect s.*, c.*from STUDENT s, CLASS cwhere s.StudentID = c.SID (+);

select s.*, c.*from STUDENT s, CLASS cwhere s.StudentID = c. SID;

=x > y

<>...

R1 R2

R1 R2

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Find names of students who have taken a CIS course

• Information about departments only available in Courses; so need an extra join:

)Re)''

(( StudentsgistersCoursesCISdeptsname

A more efficient solution:

))Re)''

((( StudentsgisCoursesCISdeptcidsidsname

A query optimizer can find this given the first solution!

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Find students who have taken an MIS or a CS course

• Can identify all MIS or CS courses, then find students who have taken one of these courses:

))''''

(,1( CoursesCSdeptMISdept

Temp

)Re1( StudentsgisTempsname

Can also define Temp1 using union! (How?)

What happens if is replaced by in this query?

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Find students who have taken a CIS and an ECI Course

• Previous approach won’t work! Must identify students who have taken CIS courses, students who have taken ECI courses, then find the intersection (note that sid is a key for Students):

))Re)''

((,1( gisCoursesCISdeptsid

Temp

))21(( StudentsTempTempsname

))Re)''

((,2( gisCoursesECIdeptsid

Temp

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Relational Calculus

• Comes in two flavours: Tuple relational calculus (TRC) and Domain relational calculus (DRC).

• Calculus has variables, constants, comparison ops, logical connectives and quantifiers. TRC: Variables range over (i.e., get bound to) tuples. DRC: Variables range over domain elements (= field

values). Both TRC and DRC are simple subsets of first-order logic.

• Expressions in the calculus are called formulas. An answer tuple is essentially an assignment of constants to variables that make the formula evaluate to true.

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Find students with GPA > 3.7 who have taken a CIS Course

7.3.| GPAtStudentstt

sidtsidrgisrr ..Re

''... CISdeptccidrcidcCoursescc

TRC:

7.3,,,|,,, GStudentsAGNIAGNI

IIrgisSCrIrSCrIr Re,,,,

'',,,, CISDCrCCoursesDCNCDCNC

DRC:

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Find students who have taken all CIS courses

StudentsAGNIAGNI ,,,|,,,

''^,,,, CISDCoursesDCNCDCNC

CCrIrIgisSCrIrSCrIr Re,,,,

DRC:

Studentstt|

''.^ CISdeptcCoursescc

cidccidrsidtsidrgisrr ....Re

TRC:

How will you do this with Relational Algebra?

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Monotonic and Non-Monotonic Queries

• Monotonic queries: queries for which the size of the results either increase or stay the same as the size of the inputs increase. The result size never decreases

• Non-monotonic queries: queries for which it is possible that the size of the result will DECREASE when the size of the input increases

• Examples of each?

• Which of the algebra operations is non-monotonic?• What does this signify?

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Summaries and Aggregates

Calculate the average GPA select avg. (GPA)from STUDENT,

Find the lowest GPA select min (GPA) as minGPAfrom STUDENT,

How many CIS majors? select count (StudentId)from STUDENTwhere major=‘CIS’;

Discarding duplicates select avg (distinct GPA)STUDENTwhere major=‘CIS’(is this above query correct?)

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Aggregate Functions

COUNT (attr) - a simple count of values in attrSUM (attr) - sum of values in attrAVG (attr) - average of values in attrMAX (attr) - maximum value in attrMIN (attr) - minimum value in attr

Take effect after all the data is retrieved from the databaseApplied to either the entire resulting relation or groupsCan’t be involved in any query qualifications (where clause)

Would the following query be permitted?

select StudentId

from STUDENT

where GPA = max (GPA);

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Grouping Results Obtained

Show all students enrolled in each course.

select cno, StudentID

from REGISTRATION

group by cno; Is this grouping OK?Calculate the average GPA of students by county.

select county, avg (GPA) as CountyGPA

from STUDENT

group by county;Calculate the enrollment of each class.

select cno, year , term, count (StudentID) as enroll

from REGISTRATION

group by cno, year, term;

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Selections on Groups

Show all CIS courses that are full.

select cno, count (StudentID)

from REGISTRATION

group by cno

having count (StudentID) > 29;

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Grouping Results after Join

Calculate the average GPA of each class

select course#, avg (GPA)from STUDENT S, CLASS Cwhere S.StudentID = C.SIDgroup by course#,

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Nesting Queries

SELECT attribute(s)

FROM relation(S)

WHERE attr [not] {in | comparison operator | exists }

( query statement(s) );

List names of students who are taking “BA201”

select Name

from Student

where StudentID in

( select StudentID

from REGISTRATION

where course#=‘BA201’);

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Sub Queries

List all students enrolled in CIS coursesselect namefrom STUDENTwhere StudentId in

(select StudentIdfrom REGISTRATIONwhere cno like ‘CIS%’);

List all students enrolled in CIS coursesselect namefrom STUDENTwhere StudentId in

(select StudentIdfrom REGISTRATIONwhere cno like ‘CIS%’);

List all courses taken by Student (Id 1011) select cnamefrom COURSEwhere cnum = any

(select cnofrom REGISTRATIONwhere StudentId = 1011);

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Sub Queries

Who received the highest grade in CIS 8140select StudentIdfrom REGISTRATIONwhere cnum = ‘CIS 8140’ and

grade >=all(select gradefrom REGISTRATIONwhere cno = ‘CIS 8140’);

Who received the highest grade in CIS 8140select StudentIdfrom REGISTRATIONwhere cnum = ‘CIS 8140’ and

grade >=all(select gradefrom REGISTRATIONwhere cno = ‘CIS 8140’);

List all students enrolled in CIS courses.select namefrom STUDENT Swhere exists

(select *from REGISTRATIONwhere StudentId = S.StudentId

and cno like ‘CIS%’);

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Relational Views

• Relations derived from other relations.• Views have no stored tuples.• Are useful to provide multiple user views.

View 1 View 2 View N

BaseRelation 1

BaseRelation 2

•What level in the three layer model do views belong?•Which kind of independence do they support?

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View Creation

Create View view-name [ ( attr [ , attr ] ...) ]

AS subquery

[ with check option ] ;

DROP VIEW view-name;

Create a view containing the student ID, Name, Age and GPA for those who are qualified to take 300 level courses, i.e., GPA >=2.0.

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View Options

• With Check Option enforces the query condition for insertion or update

To enforce the GPA >=2.0 condition on all new student tuples inserted into the view

• A view may be derived from multiple base relations

Create a view that includes student IDs, student names and their instructors’ names for all CIS 300 students.

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View Retrieval

Queries on views are the same as that on base relations.

Queries on views are expanded into queries on their base relations.

select Name, Instructor-Name

from CIS300-Student

where Name = Instructor-Name;

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View: Update

Update on a view actually changes its base relation(s)!

update Qualified-Student

set GPA = GPA-0.1

where StudentID = ‘s3’;

insert into Qualified-Student

values ( ‘s9’, ‘Lisa’, 4.0 )

insert into Qualified-Student

values ( ‘s10’, ‘Peter’, 1.7 )

Why are some views not updateable?

What type of views are updateable?

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Non-monotonic queries – again!

• Need to use either MINUS or NOT EXISTS!

• Find courses where no student has gpa over 3.5

• Find students who have taken all courses that Joe has taken

• How would you solve these?

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Summary

• SQL is a low-complexity, declarative query language

• The good thing about being declarative is that internally the query can be changed automatically for optimization

• Good thing about being low-complexity? No SQL query ever goes into an infinite loopNo SQL query will ever take indefinite amount of

space to get the solution

• Can be used for highly complex problems!