Database Management System - GNUopen.gnu.ac.kr/lecslides/2017-2-DB/DBLec02_RelAlgebra... · 2017. 9. 7. · Database Management System Lecture 2 Relational Algebra and SQL * Some

DatabaseManagementSystem

Lecture2

RelationalAlgebraandSQL

*SomematerialsadaptedfromR.Ramakrishnan,J.Gehrke

Today’sAgenda• RelationalAlgebra

• ComplexSQL

2DatabaseManagementSystem

RelationalAlgebra


RelationalDBandAlgebra• SQL

• PracticaldefinitionofrelationalDB• OperatesonTables(bags)

• Operations• Keywords

• Statements:SELECT,FROM,WHERE,…

• RelationalAlgebra

• MathematicaldefinitionofRelationalDB• OperatesonRelations(Sets)

• Operations• set-basedoperations

• Intersection,Union,...

• Thedefaultistoproduceabagofrowsasaqueryresult

• Wantaset,useDISTINCT


DescribingarelationalDBmathematically• Twoingredients• Arelationisasetoftuples

• Definequeryoperatorsasasetfunctions


Recap:CrossproductwithSet• LetA={a,b,c}andB={1,2}

• Crossproductinsettheoryisdefinedasorderedpairs(2-tuples)whereeachpairconsistsofanelementfromAandB

• HowaboutA={a,b,c},B={1,2},andC={α,β}?

A× B={(a,1),(b,1),(c,1),(a,2),(b,2),(c,2)}


DefiningRelations

• Anyinstanceoftherelationisalwaysasubset(⊆) of attributes

• name × sal × num × status

• Eachrelationinstanceisasubsetofthecrossproductofitsdomains

• oneelementofarelationiscalledtuple

• Arelationisalwaysasetbydefinition

Person(name,salary,num,status)name={allpossiblestringsof30characters}salary={realnumbersbetween0and100,000,000}num ={integerbetween0and9999}status={“a”,“b”}


Recap:SetTheory

• Whatdothesereturn?• A∩B

• A ∪ B

• A– B

• A× B

A={1,3,5,7}B={1,2,3,4}


RelationalAlgebrahasAdditionalOperations

• Introducingnewoperators(Cforcondition,Lforattributelist,Rforrenamingspecification)• A⋈cB

• A÷B

• 𝝈c(A)

• 𝜋 L(A)

• 𝜌 R(A)

A={1,3,5,7}B={1,2,3,4}


RelationalAlgebraasaQueryLanguage• Wedon’tnormallyuserelationalalgebradirectly• Productsdon’tallowyoutowriterelationalalgebraqueries

• But,itisusedinternallyinaDBMStorepresentaqueryplan

• Itisalsooftenusedintheoreticalworkondatabases• (althoughfragmentsoffirstorderlogicarefrequentlyusedaswell...)


RelationalAlgebraQueriesw/outOperators• WhatdoesthefollowingSQLqueryreturn?

• Answer:Student(Itiscalledidentityfunction)

• Arelationnamebyitselfisavalidrelationalalgebraquery

• Listingtherelationnamejustreturnsthetuplesintherelation

SELECT*FROMStudent;

StudentJohnCusackWill Smith

Student


RelationalAlgebra:Selectionoperator(𝝈)

• Therelationalalgebraquery

• IssimilartotheSQLquery

Number Owner Balance Type7003001 JaneSmith 1,000,000 Savings

7003003 AlfredHitchcock 4,400,200 Savings

7003005 TakumiFujiwara 2,230,000 Checking

7003007 Brian Mills 1,200,000 Savings

Account

𝝈 Balance<3000 (Account)

SELECT*FROMAccountWHEREBalance<3,000,000;


RelationalAlgebra:Selectionoperator(𝝈)• Select(𝝈)isaunaryoperator:

• Itisalwaysappliedtoasinglerelation

𝝈 :R→R

𝝈 Balance<3000 (Account)

Selectoperator Relationorrelationalalgebraexpression

thepredicate(condition)AttributeComparator(≥,>,=,≠,<,≤)Attribute|Constant


Exercises• 𝝈 Balance<3,000,000 (Account)

• 𝝈 Number<7003005 (Account)

• 𝝈 Balance=Number (Account)

• 𝝈 Type=“checking” (𝝈 Balance<3,000,000 (Account))





Account


RelationalAlgebra:ProjectionOperator(𝜋)

• Therelationalalgebraquery:

• IssimilartotheSQLquery





Account

𝜋 Number,Owner (Account)

SELECTNumber,OwnerFROMAccount;


RelationalAlgebra:Projectionoperator(𝜋)• Projection(𝜋)isaunaryoperator:

• Itisalwaysappliedtoasinglerelation

𝜋 :R→R

𝜋 Number,Owner (Account)

Projectionoperator Relationorrelationalalgebraexpression

Listofattributestokeep


Example





7003009 AlfredHitchcock 3,400,200 Checking

Account

𝜋 Owner (Account)SELECTNumberFROMAccount;

Vs.

OwnerJaneSmith

AlfredHitchcock

TakumiFujiwara

Brian Mills

Number7003001

7003003

7003005

7003007

7003009

• Relationsarealwayssets• Queryanswerisasetofnames• andJ.Smithappearsjustonce

intheanswer


CombiningSelectandProject• Areanyoftheseequivalent?

𝜋 Owner(𝝈 Balance<3,000,000 (Account))𝝈 Balance<3,000,000(𝜋 Owner,Balance (Account))𝜋 Owner(𝝈 Balance<3,000,000(𝜋 Owner,Balance(Account)))𝝈 Type=“checking” (𝝈 Balance<3,000,000(𝜋 Owner,Balance(Account)))





7003009 AlfredHitchcock 3,400,200 Checking

Account


RelationalAlgebra:CrossProductoperator(×)• Usedinthebasicdefinitionofarelation• “Aninstanceofarelationisasubsetofthecrossproductofitsdomains”

• Isalsoanoperatorintherelationalalgebra


Example• Supposewehavefollowingtworelations

• Thecrossproductproduceseverypossiblecombinationsofteacherandcourses

Teacher(TID,Tname)Course(CID,Cname)

TID Tname

101 Emma Thompson

105 BillyElliot

110 JohnWaine

TeacherCID Cname

346 HowtoAct

491 HowtoThink

Course

TID Tname CID Cname

101 Emma Thompson 346 HowtoAct

101 Emma Thompson 491 HowtoThink

105 BillyElliot 346 HowtoAct

105 BillyElliot 491 HowtoThink

110 JohnWaine 346 HowtoAct

110 JohnWaine 491 HowtoThink

TeacherXCourse


SELECT*FROMTeacher,Course;

RelationalAlgebra:Joinoperator(⋈)• Join()isabinaryoperator

• Itisalwaysappliedtoatworelationsandreturnsone⋈:R× R→R

Account⋈Number=Account Deposit

relationorrelationalalgebraexpression

Relationorrelationalalgebraexpression

thejoinpredicate(condition)Attributecomparator(≥,>,=,≠,<,≤) Attribute


RelationalAlgebra:Joinoperator(⋈)

• Therelationalalgebraquery

• isequivalentto

Account

Number Owner Balance Type

Accnt TxID Date Amount

Deposit

Account⋈Number=Accnt(Deposit)

𝝈 Number=Accnt (Account× Deposit)


RelationalAlgebra:Joinoperator(⋈)• Thejoinoperatorisdefinedforconvenience

• Anyquerywithajoincanalwaysberewrittenintocrossproductfollowedbyselection

R1⋈a1=a2R2≡𝝈 a1=a2(R1× R2)


NotesonJoin• EachsimpleBooleanpredicateinthejoinconditionmustcompareanattributefromonerelationtoanattributeintheotherrelation

• type=“checking”isnotajoincondition

• ifyouhaveajoinwithNOcondition,thenitisjustacrossproduct

Account⋈Number=Account^type=“checking”Deposit


Examples

• S⋈advisor=fid(F)

• S⋈S.age <F.age (F)

• Themostcommonjoiniscalledaequi-join(forequalitycondition)

sid name advisor age

101 Bill 301 20

102 John 302 20

103 Edward 301 19

104 Albert 301 19

105 Thompson 302 19

SinstanceofStudentfid name age

301 Morrison 45

302 Groot 37

FinstanceofFaculty

R1⋈A1=A2R2


select*fromStudentass,Facultyasfwheres.advisor =f.fid;

select*fromStudentass,Facultyasfwheres.age <f.age;

SQLstatementtoanrelationalAlgebraexpression

• SELECT-FROM-WHEREqueriesaresometimesdescribedasequivalenttotheSelect-Project-Join(SPJ)subsetofrelationalalgebra

SELECTDISTINCTattributesFROMT1,T2,…WHEREconditions

?𝜋 attributes(𝝈 conditions(T1× T2× … ))


ComplexSQL


MoreSQLqueryconstructs

SELECTFROMWHERE

………

(SELECT… FROM…WHERE…)

UNION(SELECT… FROM…WHERE…)

ORDERBYGROUPBYHAVING

………

1.2.3.

4.

5.

1.Extensions:SUM,COUNT,MIN,AVG,etc2.ExtensionsincludevariouskindsofJOINs3.Additionalcomparators,e.g.EXISTS,IN,ANY

4.OperatorsthattakestwoormorecompleteSQLqueriesasarguments,e.g.,UNIONandINTERSECT

5.Severaladditionalclauses,e.g.,ORDERBY,GROUPBY,andHAVING



SELECTFROMWHERE

………




………

1.2.3.

4.

5.





SampleDatabase• Let’sconsiderthefollowingDBfortheexamples

• WearegoingtootherDBstimetotime

Customer(Number,Name,Address,Crating,Camount,Cbalance,Salesperson)

Salesperson(Number,Name,Address,Office)

foreignkeycustomer.Salesperson ->Salesperson.Number


SELECT(1/4)• AggregateOperators:COUNT,SUM,MIN,MAX,andAVG

• IfoneaggregateoperatorappearsintheSELECTclause• ALLOFTHEENTRIESintheselectclauseMUSTBEANAGGREGATEOPERATOR

• UnlessthequeryincludesaGROUPBYclause(moreonlater)

SELECTMIN(Cbalnace),MAX(Cbalance),AVG(Cbalance)FROMCustomer;

SELECTMIN(Cbalnace),MAX(Cbalance),AVG(Cbalance)FROMCustomerWHEREage>35;


Stoptothink• Whatwould/shouldthequeryresultbe?

• Isitallowed?

SELECTName,Crating,AVG(Cbalance)FROMCustomer;


SELECT (2/4)• Whatisthedifferencebetweenthesetwoqueries?

• Whenwillthesetwoqueriesreturnthesameanswer?• orwhataretheconditionsforittohappen

SELECTCOUNT(Name)FROMCustomer;

SELECTDISTINCTNameFROMCustomer;

Vs.


SELECT (3/4)• WhatistheimplicationofusingDISTINCT• WhencomputingtheSUMorAVGofanattribute?

• WhencomputingtheMINorMAXofanattribute?

SUM(DISTINCT(AGE))Vs.SUM(age)

MIN(DISTINCT(AGE))Vs.MIN(age)

TheSUMorAVGwillbecomputedonlydistinctvalues

NoDifference:theresultdoesnotdependonwhetherornotduplicatesareremoved


SELECT (4/4)• SELECTclauselistcanalsoincludesimplearithmeticexpressionsusing+,-,*,/

SELECT(Camount – Cbalance)ASAvailableCredit,NameFROMCustomerWHERECamount >0



SELECTFROMWHERE

………




………

1.2.3.

4.

5.





FROM:SyntacticSugarsandnewoperators• ThereareanumberofjointypesthatcanbeexpressedinFROMclause• Innerjoin(theregularjoin)

• Crossjoin

• naturaljoin

• leftouterjoin

• rightouterjoin

• fullouterjoin

syntacticsugarsthatcanbeexpressedusingSELECT-FROM-WHEREqueries

Newoperators


FROM• Thesetwoqueriesareequivalent

SELECTC.Name,S.NameFROMCustomerCJOIN SalespersonSON C.Salesperson =S.NumberWHEREC.Crating <6;

1.

2. SELECTC.Name,S.NameFROMCustomerC,SalespersonSWHEREC.Salesperson =S.Number ANDC.Crating <6;

𝜋 C.Name,S.Name(𝝈C.CreditRating <6(Customer⋈C.Salesperson =S.Number Salesperson))

𝜋 C.Name,S.Name(𝝈C.CreditRating <6^C.Salesperson=S.Number(Customer× Salesperson))


FROM:JOINwithUSINGclause• JOINwithUSINGclausewhenattributesinthe2tableshavethesamename

• TheseTwoqueriesareequivalent

• USINGclausedoesn’tneed(andcan’thave)acorrelationname

Course(CNumber,CName,Description)Teacher(TNumber,TName,Phone)Offering(CNumber,TNumber,Time,Days,Room)

SELECTC.CNumber,C.CName,RoomFROMCourseCJOIN OfferingUSING(CNumber);

SELECTC.CNumber,C.Name,RoomFROMCourseCJOIN OfferingOON C.CNumber=O.CNumber;


FROM:BasicJoin≡(INNER)JOIN• FortheINNERJOIN

• Thequeryresultincludesall“matches”butexcludes• customerrowsthatdonothaveaSalesperson

• Salespersonrowsthatarenotassignedtoanycustomers

• Thekeyword“INNER”isoptional• abovequeryisequivalentto

SELECTC.Name,S.NameFROMCustomerCINNERJOINSalespersonSON C.Salesperson =S.Number;

SELECTC.Name,S.NameFROMCustomerCJOINSalespersonSON C.Salesperson =S.Number;


FROM:crossproduct≡CROSSJOIN• Thefollowingqueriesareequivalent

SELECT*FROMCustomer,Salesperson;

SELECT*FROMCustomerCROSSJOINSalesperson;


FROM:Equi-Jioin vs.Natual Join(1/3)• Whenthejoinisbasedonequalityofattributes,wealwayshavetwoidenticalattributesintheresult

Name DeptID

Smith 1

James 2

Brown 3

Johnson 1

Robert

FacultyDeptID DeptName

1 Engineering

2 Communications

3 Marketing

Department

SELECT*FROMFacultyFINNERJOINDepartmentDONF.DeptID =D.DeptID;F.Name F.DeptID D.DeptID D.DeptName

Smith 1 1 Engineering

Johnson 1 1 Engineering

James 2 2 Communication

Brown 3 3 Markeing

Equi-Join


FROM:Equi-Jioin vs.Natual Join(1/3)• Equi-JoinwiththeUSINGconstruct:applicablewithcolumnshavingsamename

Name DeptID

Smith 1

James 2

Brown 3

Johnson 1


1 Engineering

2 Communications

3 Marketing

Department

SELECT*FROMFacultyFINNERJOINDepartmentDUSING(DeptID);Name DeptID DeptName

Smith 1 Engineering

Johnson 1 Engineering

James 2 Communication

Brown 3 Markeing

Equi-JoinwithUSINGconstruct


FROM:Equi-Jioin vs.Natual Join(3/3)• NATURALJOIN:Equi-Joinwithonlyonecolumnforeachequallynamedcolumns

Name DeptID

Smith 1

James 2

Brown 3

Johnson 1


1 Engineering

2 Communications

3 Marketing

Department

SELECT*FROMFacultyNATURALJOINDepartment;

Name DeptID DeptName

Smith 1 Engineering

Johnson 1 Engineering

James 2 Communication

Brown 3 Markeing

NATURALJOINIf you don’t specify which attributes to join on, natural join will join on all attributes with the same name


FROM:moreonNATURALJOIN(1/2)• NATURALJOINislikea“macro”thatjoinstableswithanequalityconditionforallattributeswiththesamename

• NATURALJOINdropsoneofduplicatecolumnsautomatically

Course(CNumber,CName,Description)

Teacher(TNumber,TName,Phone)

Offering(CNumber,TNumber,Time,Days,Room)


FROM:moreonNATURALJOIN (2/2)• Listthecourseandteachernameforallcourseofferings

• ThisquerycanbeexpressedwiththeNATURALJOINorwithanINNERJOIN• Thesetwoqueriesareequivalent

• Theyareequivalentbecausethejoinattributeshavethesameattributenames

• Butisitalwaysuseful?

SELECTCName,TNameFROMCourseC,OfferingO,TeachingTWHEREC.CNumber =O.CNumber ANDO.TNumber =T.Tnumber

SELECTCName,TNameFROMCourseNATURALJOINOfferingNATURALJOINTeacher;


FROM:INNERJOINVs.OUTERJOIN(1/2)• FortheINNERJOIN

• thequeryresultdoesnotinclude(p.40)• acustomerthatdoesnothaveasalesperson

• asalespersonthatisnotassignedtoanycustomers

SELECTC.Name,S.NameFROMCustomerCINNERJOINSalespersonSONC.Salesperson =S.Number

47

Number Name Address Crating Camount Cbalance Salesperson

1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

Customer

Number Name Address Office

55 Miller 5th Str. 101

77 Khan 7th Str. 102

83 Dunham 8th Str. 103

Salesperson


FROM: INNERJOINVs.OUTERJOIN(2/2)• AnINNER(regular)JOINincludesonlythosecustomersthathavesalespersons(onlythematches)

• ALEFTOUTERJOINwillincludeallmatchesplusall– customersthatdonothaveaSalesperson

• ARIGHTOUTERJOINwillincludeallmatchesplusall– salespersonsthatarenotassignedtoanycustomers

• AFULLOUTERJOINwillincludeallofthese

48

SELECTC.Name,S.NameFROMCustomerasCINNERJOINSalespersonasS

ONC.Salesperson =S.Number;


FROM:LEFTOUTERJOIN

49


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

Customer





Salesperson

INNERJOIN onC.Salesperson =S.Number gives:

LEFTOUTERJOINonC.Salesperson =S.Number gives:

1 Smith 1st Str. 700 10,000 9,000 55 55 Miller 5th Str. 101

2 Jones 2nd Str. 700 8,000 4,000 77 77 Khan 7th Str. 102



3 Mills 3rd Str. 700 11,000 8,000 NULL NULL NULL NULL NULL


FROM:RIGHTOUTERJOIN

50


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

Customer





Salesperson


RIGHTOUTERJOINonC.Salesperson =S.Number gives:





NULL NULL NULL NULL NULL NULL NULL 83 Dunham 8th Str. 103


FROM:FULLOUTERJOIN

51


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

Customer





Salesperson


RIGHTOUTERJOINonC.Salesperson =S.Number gives:





3 Mills 3rd Str. 700 11,000 8,000 NULL NULL NULL NULL NULL

NULL NULL NULL NULL NULL NULL NULL 83 Dunham 8th Str. 103


*notsupportedinmysql

FROM:aformofsubquery• YoucanputacompletequeryexpressionintheFROMclause• alsoknownasnestedqueriesorsubqueries

• Parenthesesareimportant

52

SELECT...FROMEmployeeE,(SELECT...FROM...WHERE...)WHERE...


RelationalAlgebraOperators


Eightstandardrelationalalgebraoperators• 𝜋 project Wehaveseenalready

• 𝝈 select Wehaveseenalready

• ∪ union Fromsettheory

• ∩ intersect Fromsettheory

• – difference Fromsettheory

• × crossproduct Wehaveseenalready

• ⋈ join Wehaveseenalready

• ÷ divide

• 𝜌 renaming

54

canonlyusedwithunion-compatiblerelations


Union-compatiblerelations• Tworelationsareunion-compatibleif• havesamenumberofattributes

• havesamedomains

• Example

55

Checking(CNum: int, COwner: string, CBalance: int)

Savings(SNum: int, SOwner: string, SBalance: int)


Example: ∪ union

56

Cnum Cowner Cbalance101 Smith 1000

102 Mills 2000

104 Jones 1000

105 Schwab 3000

Checking

Snum Sowner Sbalance103 Smith 5000

Savings

Checking∪SavingsCnum Cowner Cbalance101 Smith 1000

102 Mils 2000

104 Jones 1000

105 Schwab 3000

103 Smith 5000

notethatattributesarefromthefirstrelationinthequery


SELECTCNum,COwner,CBalanceFROMCheckingUNIONSELECTSNum,SOwner,SBalanceFROMSavings;

Example: ∩ intersection

57

Cnum Cowner Cbalance101 Smith 1000

102 Mils 2000

104 Jones 1000

105 Schwab 3000

Checking

Snum Sowner Sbalance103 Smith 5000

Savings

Checking∩ Savings

𝜋Cowner(Checking)∩𝜋 Sowner(Savings)

?Itisempty– notuplesappearinbothrelations

Smith– theonlyownerinSavingsAcount


Example:– difference• FindalltuplesthatareintheCheckingrelationbutarenotintheSavingsrelation

• EveryoneinCheckingexceptSmith

58

CheckingAccount −SavingsAccount

𝜋 COwner(CheckingAccount)−𝜋 SOwner(SavingsAccount)


*notsupportedinmysql

WorkaroundfordifferenceoperationexamplequerySELECT*FROMpLEFTOUTERJOINqONp.id =q.id WHEREq.id ISNULL


SELECTFROMWHERE

………




………

1.2.3.

4.

5.





UNIONandINTERSECTION• TwocompletequerieswithUNIONinbetween

60

• TwocompletequerieswithINTERSECTinbetween

(SELECTC.NameFROMCustomerCWHEREC.Name LIKE“B%”)UNION(SELECTS.NameFROMSalespersonSWHERES.Name LIKE“B%”);

(SELECTC.NameFROMCustomerC)INTERSECT(SELECTS.NameFROMSalespersonS);

• TwocompletequerieswithEXCEPT(i.e.,DIFFERENCE)inbetween• MySQLdoesn’tsupportEXCEPT

(SELECT C.NameFROM Customer C) EXCEPT(SELECT S.NameFROM Salesperson S);


ALLinUNION,INTERSECT,andEXCEPT• Ifyoudon’tspecifyALL,theresultiscomputedonsets• Eliminateduplicatesfromfirstoperand

• Eliminateduplicatesfromsecondoperand

• Computeoperation

• Eliminateduplicatesfromresult

• Notethedifferenceandchosewisely• UNIONVs. UNIONALL

• INTERSECTVs. INTERSECTALL

• EXCEPTVs.EXCEPTALL



SELECTFROMWHERE

………




………

1.2.3.

4.

5.





GROUPBY• AnySQLquerycanhavetheanswer“grouped”• oneoutputrowforeachgroup

63

SELECTSalesperson,COUNT(*)FROMCustomer;

SELECTSalesperson,COUNT(*)FROMCustomerGROUPBYSalesperson;


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

CustomerSalesperson COUNT(*)

55 1

77 1

NULL 1


GROUPBY

64



1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

4 Bill 4th Str. 700 13,000 5,000 55

5 Jane 5th Str. 800 3,000 3,000 55

6 Harley 8th Str. 700 2,000 8,000 20

7 Khale 9th Str. 900 6,000 1,000 77

Customer


Example:GROUPBY

65


1.Makegroupsresultingin4Groups2.Evaluate“SELECTSalesperson,Count(*)”foreachgroup


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

4 Bill 4th Str. 700 13,000 5,000 55

5 Jane 5th Str. 800 3,000 3,000 55

6 Harley 8th Str. 700 2,000 8,000 20

7 Khale 9th Str. 900 6,000 1,000 77


55 3

NULL 1

77 2

20 1


SQLHAVING• HAVINGclausespecifiesapredicateevaluatedagainsteachgroup

• AgroupisintheresultifitsatisfiestheHAVINGcondition

66

SELECTSalesperson,COUNT(*)FROMCustomerGROUPBYSalespersonHAVINGCOUNT(*)>1;


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 55

3 Mills 3rd Str. 700 11,000 8,000 NULL


55 2


Example:GROUPBY

67

SELECTSalesperson,COUNT(*)FROMCustomerGROUPBYSalespersonHAVINGCOUNT(*)>1;

1.Makegroupsresultingin4Groups2.CheckifCOUNT(*)>1holds3.Evaluate“SELECTSalesperson,Count(*)”foreachgroup


1 Smith 1st Str. 700 10,000 9,000 55

2 Jones 2nd Str. 700 8,000 4,000 77

3 Mills 3rd Str. 700 11,000 8,000 NULL

4 Bill 4th Str. 700 13,000 5,000 55

5 Jane 5th Str. 800 3,000 3,000 55

6 Harley 8th Str. 700 2,000 8,000 20

7 Khale 9th Str. 900 6,000 1,000 77


55 3

NULL 1

77 2

20 1

Salesperson COUNT(*)

55 3

77 2


NoteonGROUPBY,HAVING• Theonlyattributethatcanappearina“grouped”queryare• thegroupingattributes

• aggregateoperatorsthatareappliedtothegroup

• Thus,thefollowingisnotlegal

• Becausether canbemorethanonenameforeachgroup

68

SELECTNameFROMCustomerGROUPBYSalesperson;


Exercise

• WriteSQLqueriesforthefollowing• Averagenumberofwinsandlossesacrossteams

• Averagenumberofwinsandlossesperconference

• Battingaverageforeachplayer,wherebattingaverageisthenumberofhitsdividedbyatbats

69

Team(Name,Games,Wins,Losses,Conference)

Player(Name,Hits,AtBats,HomeRuns,Team)Player.Team ->Team.Name


ORDERBY• Sorttheresultofaquery

70

SELECTNumber,Name,SalespersonFROMCustomerORDERBYName;

Number Name … Salesperson

1 Smith … 55

2 Jones … 77

3 Mills … NULL

4 Bill … 55

5 Jane … 55

6 Harley … 20

7 Khale … 77

Customer


4 Bill … 55

6 Harley … 20

5 Jane … 55

2 Jones … 77

7 Khale … 77

3 Mills … NULL

1 Smith … 55

Customer



71

SELECTNumber,Name,SalespersonFROMCustomerORDERBYNameDESC;


1 Smith … 55

2 Jones … 77

3 Mills … NULL

4 Bill … 55

5 Jane … 55

6 Harley … 20

7 Khale … 77

Customer


1 Smith … 55

3 Mills … NULL

7 Khale … 77

2 Jones … 77

5 Jane … 55

6 Harley … 20

4 Bill … 55

Customer



72

SELECTNumber,Name,SalespersonFROMCustomerORDERBYName,Salesperson;


1 Smith … 55

2 Jones … 77

3 Mills … NULL

4 Bill … 55

5 Jane … 55

6 Harley … 20

7 Khale … 77

8 Bill … 20

Customer


8 Bill … 20

4 Bill … 55

6 Harley … 20

5 Jane … 55

2 Jones … 77

7 Khale … 77

3 Mills … NULL

1 Smith … 55

Customer


Subqueries• Itcanbeusedinthewhereclause(inadditiontotheFROMclause)

• Innerqueryreturns• Asinglevaluethatrepresentsmaxcreditrating

• Outerqueryreturns• Thenameandnumberofthecustomerwiththehighestcreditratings

73

SELECTC1.Number,C1.NameFROMCustomerC1WHEREC1.CRating=(SELECTMAX(C2.Crating)

FROMCustomerC2); Innerquery

Outerquery


Example

1. FROMclauseinouterquery2. TakearowfromtheCustomertable3. CheckiftherowsatisfiestheWHEREclause4. Evaluatetheinnerquery(result:800)5. EvaluateifCratingisequaltotheresult

74


FROMCustomerC2);


1 Smith 1st Str. 200 10,000 9,000 55

2 Jones 2nd Str. 800 8,000 4,000 55

3 Mills 3rd Str. 700 11,000 8,000 NULL

Customer


Subqueries• Subqueriescanbeusedinthewhereclause(inadditiontothefromclause)

• SixComparators:=, >, < >=, <=, <> (not equal) • innerquerymustreturnasinglevalue

• Iftheinnerquerydoesnotmentionanyattributesfromtheouterquery(C1notmentionedintheinnerquery)• Thenyouonlyneedtoevaluatetheinnerqueryonce

• Theinner(sub)queryisNOTcorrelated

75


FROMCustomerC2);


Subqueries:SOME/ALLcomparison

• ForSOME,theexpressionmustbetrueforatleastonerow inthesubqueryanswer• “ANY”isequivalenttoSOME

• Whatdoesthisqueryreturn?

76

SELECTS.NameFROMSalespersonSWHERES.Name =SOME(SELECTC.Salesperson

FROMCustomerCWHEREC.CRating =700);

The name of each salespeople that has a customer with a credit rating of 700


Subqueries:SOME/ALLcomparison

• ForALL,theexpressionmustbetrueforallrowsinthesubqueryanswer


77

SELECTS.NameFROMSalespersonSWHERES.Name =ALL(SELECTC.Salesperson

FROMCustomerCWHEREC.CRating =700);

Thenameofthesalespersonthathasallthecustomerswitharatingof700(ifsuchasalespersonexists)


Subqueries:IN/NOTINcomparison(1/4)

• WithIN,theattributematchesatleastonevaluereturnedfromthesubquery• Sameas“=SOME”

78

SELECTC1.Number,C1.NameFROMCustomerC1WHEREC1.NameIN(SELECTName

FROMSalesperson);



• WithNOTIN,theattributematchesnone ofthevaluesreturnedfromthesubquery• Sameas“<>ALL”

79

SELECTC1.Number,C1.NameFROMCustomerC1WHEREC1.Name NOTIN(SELECTName

FROMSalesperson);


Subqueries:IN/NOTINcomparison(3/4)• Aretheseequivalent?

• DoweneedtouseDISTINCTforthesetobeequivalent?

• Isthesubquerycorrelated?

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SELECTS.Number,S.NameFROMSalespersonSWHERES.Number IN (SELECTC.Salesperson

FROMCustomerC);

SELECTDISTINCT S.Number,S.NameFROMSalespersonS,CustomerCWHERES.Number =S.Salesperson;



• Becausethesubquerymentionsanattributefromatableintheouterquery• Thesubquerymustbe(re-)evaluatedforeachrowintheouterquery(eachtimetheWHEREclauseisevaluated)

• Correlatedsubqueriescanbeveryexpensive!

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SELECTS.Number,S.NameFROMSalespersonSWHERES.Number IN(SELECTC.Salesperson

FROMCustomerCWHEREC.Name =S.Name);


Subqueries:EXISTS/NOTEXISTS(1/2)

• Iftheanswertothesubqueryisnotempty...thentheEXISTSpredicatereturnsTRUE• Isthissubquerycorrelated?


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SELECTC.NameFROMCustomerCWHEREEXISTS (SELECT*

FROMSalespersonSWHERES.Number =C.Salesperson AND

S.Name =C.Name);


Subqueries:EXISTS/NOTEXISTS(2/2)

• Fourpredicatescanbeappliedtoasubquery• EXISTS :isthesubqueryanswernon-empty?

• NOTEXISTS:isthesubqueryanswerempty?

• UNIQUE :doesthesubqueryreturnjustonerow?

• NOTUNIQUE:doesthesubqueryreturnmultiplerows?

83

SELECTC.NameFROMCustomerCWHEREEXISTS (SELECT*

FROMSalespersonSWHERES.Number =C.Salesperson AND

S.Name =C.Name);


MissingRelationalAlgebraOperatorDivide


DivideOperator(p.54)• Supposewehaveaextratableinourdatabase

• Howdowefindcustomersthathaveatleastoneaccountofeachaccounttype?

85





Account

TypeChecking

Savings

AccountTypes

𝜋Owner,Type(Account) ÷ AccountTypesFindaccountownerswhohaveALLtypesofaccounts


ForNextWeek• Review– Quizonthematerial• Ch.4to4.2• Ch.5.5

• Readingassignments• Ch.2-2.5• Ch.3.5

• Besureyouunderstand• Aggregateoperations• howjoinoperates• setoperators• GROUPBY,HAVING,ORDERBY,Subqueries


Database Management System - GNUopen.gnu.ac.kr/lecslides/2017-2-DB/DBLec02_RelAlgebra... · 2017. 9. 7. · Database Management System Lecture 2 Relational Algebra and SQL * Some

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