L02-06-SQL - v180129

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L04:SQL

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Announcements!

• PollsonPiazza.Openfor2days• Outlinetoday:- practicingmorejoinsandspecifyingkeyandFKconstraints- nestedqueries

• Nexttime:"witnesses"(traditionallystudentsfindthistopicthemostdifficult)

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QueriesviaSQLhavemultiplewords:Ifyoumasterthisstructureyouknow50%aboutSQLQueries

SELECT …FROM …WHERE …GROUP BY …HAVING …ORDER BY …

• Listofattributes tobeincludedinfinalresult(alsocalledprojection!("*"selectsallattributes)

• Indicatesthetable(s)fromwhichdataistoberetrieved

• Listsacomparisonpredicate,whichrestrictstherowsreturnedbythequery,e.g.“price<20”ordifferentjoin conditions

• Groups rowsthathaveonemorecommonvaluestogetherintoasmallersetofrows

• Acomparisonpredicate usedtorestricttherowsresultingfromtheGROUPBY clause

• Identifieswhichcolumns areusedtosorttheresultingdata,plusthedirection eachcolumnissortedby(ascendingordescending)

Note1 :SQLisgenerallycaseinsensitive,e.g.SELECT

=Select=select

Note2 :Thewordsalwaysappearinthisorder– youCANNOTreorderthem

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HowtospecifyForeignKeyconstraints

• Supposewehavethefollowingschema:

• Andwewanttoimposethefollowingconstraint:- ‘Onlybonafidestudentsmayenrollincourses’i.e.astudentmustappearinthe

Studentstabletoenrollinaclass

student_id aloneisnotakey- whatis?

sid name gpa

101 Bob 3.2

123 Mary 3.8

student_id cid grade

123 564 A

123 537 A+

Students Enrolled

Wesaythatstudent_id isaforeignkey thatreferstoStudents

Students(sid: string, name: string, gpa: float)Enrolled(student_id: string, cid: string, grade: string)

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DeclaringPrimaryKeys

CREATE TABLE Students(sid CHAR(20) PRIMARY KEY,name CHAR(20),gpa REAL

)


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DeclaringPrimaryKeys

CREATE TABLE Students(sid CHAR(20),name CHAR(20),gpa REAL,PRIMARY KEY (sid)

)


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DeclaringForeignKeys

CREATE TABLE Enrolled(student_id CHAR(20),cid CHAR(20),grade CHAR(10),

)


PRIMARY KEY (student_id, cid),FOREIGN KEY (student_id) REFERENCES Students(sid)

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AnexampleofSQLsemantics

SELECT R.AFROM R, SWHERE R.A = S.B

A13

B C2 33 43 5

A B C1 2 31 3 41 3 53 2 33 3 43 3 5

CrossProduct

A B C3 3 43 3 5

A33

ApplyProjectionApply

Selections/Conditions

Output

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NotethesemanticsofajoinSELECT R.AFROM R, SWHERE R.A = S.B

Recall:Crossproduct(AXB)isthesetofalluniquetuplesinA,B

Ex:{a,b,c}X{1,2}={(a,1),(a,2),(b,1),(b,2),(c,1),(c,2)}

=Filtering!

=Returningonlysome attributes

Rememberingthisorderiscriticaltounderstandingtheoutputofcertainqueries(seelateron…)

1. Takecrossproduct:𝑋 = 𝑅×𝑆

2. Applyselections/conditions:𝑌 = 𝑟, 𝑠 ∈ 𝑋 𝑟. 𝐴 = 𝑟. 𝐵}

3. Applyprojections togetfinaloutput:𝑍 = (𝑦. 𝐴, )𝑓𝑜𝑟𝑦 ∈ 𝑌

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Note:wesay“semantics”not“executionorder”

• Theprecedingslidesshowwhatajoinmeans

• NotactuallyhowtheDBMSexecutesitunderthecovers

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Practicing more Joins

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Quiz4

SELECT DISTINCT cNameFROM WHERE

Product (pName, price, category, manufacturer)Company (cName, stockPrice, country)

Q: Find all US companies that manufacture at least twodifferent products.

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Quiz4

Q: Find all US companies that manufacture at least twodifferent products.


SELECT DISTINCT cNameFROM Product P1, Product P2, CompanyWHERE country = 'USA'

and P1.manufacturer = cNameand P2.manufacturer = cNameand P1.pName <> P2.pName <>

C

P1

P2

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Quiz4PName Price Category Manufacturer

Gizmo $19.99 Gadgets GizmoWorks

... ... ... ...

P1

CompanyCName StockPrice Country

GizmoWorks 25 USA

... ... ...

Cname

GizmoWorks

PName Price Category Manufacturer

... ... ... ...

Powergizmo $29.99 Gadgets GizmoWorks

P2

SELECT DISTINCT cNameFROM Product P1, Product P2, CompanyWHERE country = 'USA'

and P1.manufacturer = cNameand P2.manufacturer = cNameand P1.pName <> P2.pName

<>

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Quiz5

Q: Find all US companies that manufacture a product below $20 and a product above $15.


SELECT DISTINCT cNameFROM Product as P1, Product as P2, CompanyWHERE country = 'USA'

and P1.price < 20and P2.price > 15and P1.manufacturer = cNameand P2.manufacturer = cName


Gizmo 19.99 Gadgets GizmoWorks

Powergizmo 29.99 Gadgets GizmoWorks

SingleTouch 149.99 Photography Canon

MultiTouch 203.99 Household Hitachi

Product


GizmoWorks 25 USA

Canon 65 Japan

Hitachi 15 Japan

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Quiz5

Q: Find all US companies that manufacture a product below $20 and a product above $15.


P.price > 15

C

P1

P2 P.price < 20

Note that we did not specify any condition that P1 and P2 need to be distinct. An alternative interpretation is "...and another product above..."

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Quiz5PName Price Category Manufacturer


... ... ... ...

P1


GizmoWorks 25 USA

... ... ...

Cname

GizmoWorks



... ... ... ...

P2

SELECT DISTINCT cNameFROM Product as P1, Product as P2, CompanyWHERE country = 'USA'

and P1.price < 20and P2.price > 15and P1.manufacturer = cNameand P2.manufacturer = cName

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Quiz6 302

?SELECT countryFROM Product, CompanyWHERE manufacturer = cName

and category = 'Gadgets'




SingleTouch $149.99 Photography Canon

MultiTouch $203.99 Household Hitachi

Product CompanyCName StockPrice Country

GizmoWorks 25 USA

Canon 65 Japan

Hitachi 15 Japan

Q:Findallcountriesthathavecompaniesthatmanufacturesomeproductinthe‘Gadgets’category!

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Quiz6




SingleTouch $149.99 Photography Canon

MultiTouch $203.99 Household Hitachi

Product CompanyCName StockPrice Country

GizmoWorks 25 USA

Canon 65 Japan

Hitachi 15 Japan

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Q:Findallcountriesthathavecompaniesthatmanufacturesomeproductinthe‘Gadgets’category!

SELECT countryFROM Product, CompanyWHERE manufacturer = cName

and category = 'Gadgets'

Country

USA

USA

Joins can introduce duplicates -> remember to use DISTINCT!

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Nested queries(Subqueries)

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High-levelnoteonnestedqueries

• WecandonestedqueriesbecauseSQLiscompositional:

- Everything(inputs/outputs)isrepresentedasmultisets- theoutputofonequerycanthusbeusedastheinputtoanother(nesting)!

• Thisisextremelypowerful!

• High-levelidea:subqueriesreturnrelations(yetsometimesjustvalues)

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Subqueries=Nestedqueries

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

(SELECT ...FROM ...WHERE ... )

Outerblock

Innerblock

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Subqueries

important!

• AsubqueryisaSQLquerynestedinsidealargerquery• Suchinner-outerqueriesarecallednestedqueries• Asubquerymayoccurina:- SELECTclause- FROMclause- WHEREclause- HAVINGclause

• Ruleofthumb:avoidwritingnestedquerieswhenpossible;keepinmindthatsometimesit’simpossible

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1.SubqueriesinSELECT

What happens if the subquery returns more than one city ?Runtime error

Q: For each product return the city where it is manufactured!

SELECT P.pname, (SELECT C.cityFROM Company2 CWHERE C.cid = P.cid)

FROM Product2 P

Product2 (pname, price, cid)Company2 (cid, cname, city)

® "Scalar subqueries"

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Q: For each product return the city where it is manufactured!

SELECT P.pname, C.cityFROM Product2 P, Company2 CWHERE C.cid = P.cid

"unnesting the query" Whenever possible, don't use nested queries

SELECT P.pname, (SELECT C.cityFROM Company2 CWHERE C.cid = P.cid)

FROM Product2 P


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Better: we can unnestby using a GROUP BY:

Q: Compute the number of products made by each company!

SELECT C.cname, ( SELECTcount (*)FROM Product2 PWHERE P.cid = C.cid)

FROM Company2 C


SELECT C.cname, count(*)FROM Company2 C, Product2 PWHERE C.cid=P.cidGROUP BY C.cname

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2.SubqueriesinFROMclause

Q: Find all products whose prices are > 20 and < 30!

SELECT X.pname FROM ( SELECT *

FROM Product2 as PWHERE price >20 ) as X

WHERE X.price < 30

SELECT pnameFROM Product2WHERE price > 20 and price < 30

unnesting


XPName Price cid

Powergizmo $29.99 1

MultiTouch $203.99 3

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Subqueries in WHERE clause

IN, ANY, ALL

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3.SubqueriesinWHEREWhat do these queries compute?

SELECT aFROM RWHERE a IN

(SELECT * from U)?

305Ra12

SELECT aFROM RWHERE a < ANY

(SELECT * from U)

SELECT aFROM RWHERE a < ALL

(SELECT * from U)

Ua234

?

?

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3.SubqueriesinWHEREWhat do these queries compute?

SELECT aFROM RWHERE a IN

(SELECT * from U)

Since 2 is in the set(2, 3, 4)

a2

305Ra12

SELECT aFROM RWHERE a < ANY

(SELECT * from U)

a12

SELECT aFROM RWHERE a < ALL

(SELECT * from U)

a1

Ua234

Since 1 and 2 are <than at least one ("any") of 2, 3 or 4

Since 1 is < than each ("all") of 2, 3, and 4

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SomethingtrickyaboutNestedQueries

SELECT c.cityFROM Company c,

Product pr, Purchase p

WHERE c.name = pr.makerAND pr.name = p.productAND p.buyer = ‘Joe B’

Arethesequeriesequivalent?

Bewareofduplicates!

SELECT c.cityFROM Company cWHERE c.name IN (SELECT pr.makerFROM Purchase p, Product prWHERE p.name = pr.product

AND p.buyer = ‘Joe B‘)

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SomethingtrickyaboutNestedQueries

SELECT DISTINCT c.cityFROM Company c,

Product pr, Purchase p

WHERE c.name = pr.makerAND pr.name = p.productAND p.buyer = ‘Joe B’

Arethesequeriesequivalent?

SELECT DISTINCT c.cityFROM Company cWHERE c.name IN (SELECT pr.makerFROM Purchase p, Product prWHERE p.name = pr.product

AND p.buyer = ‘Joe B‘)

Nowtheyareequivalent(bothusesetsemantics)

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Correlatedsubqueries

• Inallpreviouscases,thenestedsubqueryintheinnerselectblockcouldbeentirelyevaluatedbeforeprocessingtheouterselectblock.

• Thisisnolongerthecaseforcorrelatednestedqueries.• WheneveraconditionintheWHEREclauseofanestedqueryreferencessomecolumnofatabledeclaredintheouterquery,thetwoqueriesaresaidtobecorrelated.

• Thenestedqueryisthenevaluatedonceforeachtuple(orcombinationoftuples)intheouterquery.

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CorrelatedQueries(UsingExternalVars inInternalSubquery)

SELECT DISTINCT titleFROM Movie AS mWHERE year <> ANY(

SELECT yearFROM MovieWHERE title = m.title)

Movie(title, year, director, length)

Notealso:thiscanstillbeexpressedassingleSFWquery…

Findmovieswhosetitleappearsinmorethanoneyear.

Notethescopingofthevariables!

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ComplexCorrelatedQuery

SELECT DISTINCT x.name, x.makerFROM Product AS xWHERE x.price > ALL(

SELECT y.priceFROM Product AS yWHERE x.maker = y.maker

AND y.year < 1972)

Findproducts(andtheirmanufacturers)thataremoreexpensivethanallproductsmadebythesamemanufacturerbefore1972

Product(name, price, category, maker, year)

Canbeverypowerful(alsomuchhardertooptimize)

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3.SubqueriesinWHERE(existential)

Existential quantifiers $

Using IN:

Q: Find all companies that make some products with price < 25!

SELECT DISTINCT C.cname FROM Company2 CWHERE C.cid IN ( 1, 2 )


PName Price cid

Gizmo $19.99 1

Powergizmo $29.99 1

SingleTouch $14.99 2


cid CName City

1 GizmoWorks Oslo

2 Canon Osaka

3 Hitachi Kyoto

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Using IN:


SELECT DISTINCT C.cname FROM Company2 CWHERE C.cid IN ( SELECT P.cid

FROM Product2 PWHERE P.price < 25)


PName Price cid

Gizmo $19.99 1

Powergizmo $29.99 1



cid CName City

1 GizmoWorks Oslo

2 Canon Osaka

3 Hitachi Kyoto

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"Set membership"

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Using EXISTS:


SELECT DISTINCT C.cname FROM Company2 CWHERE EXISTS ( SELECT *

FROM Product2 PWHERE C.cid = P.cid

and P.price < 25)


PName Price cid

Gizmo $19.99 1

Powergizmo $29.99 1



cid CName City

1 GizmoWorks Oslo

2 Canon Osaka

3 Hitachi Kyoto

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"Test for empty relations"

Correlated subquery

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Using ANY (also some):


SELECT DISTINCT C.cname FROM Company2 CWHERE 25 > ANY ( SELECT price

FROM Product2 PWHERE P.cid = C.cid)


SQLlite does not support "ANY" L

PName Price cid

Gizmo $19.99 1

Powergizmo $29.99 1



cid CName City

1 GizmoWorks Oslo

2 Canon Osaka

3 Hitachi Kyoto

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"Set comparison"

Correlated subquery

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Now, let's unnest:


SELECT DISTINCT C.cname FROM Company2 C, Product2 PWHERE C.cid = P.cid

and P.price < 25

Existential quantifiers are easy ! J


PName Price cid

Gizmo $19.99 1

Powergizmo $29.99 1



cid CName City

1 GizmoWorks Oslo

2 Canon Osaka

3 Hitachi Kyoto

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3.SubqueriesinWHERE(universal)

Universal quantifiers "

Q: Find all companies that make only products with price < 25!

Q: Find all companies for which all products have price < 25!

Universal quantifiers are more complicated ! L(Think about the companies that should not be returned)

same as:


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3.SubqueriesinWHERE(existnot->universal)

2. Find all companies s.t. all their products have price < 25!

1. Find the other companies: i.e. they have some product ³ 25!

SELECT DISTINCT C.cname FROM Company2 CWHERE C.cid IN ( SELECT P.cid

FROM Product2 PWHERE P.price >= 25)

SELECT DISTINCT C.cname FROM Company2 CWHERE C.cid NOT IN ( SELECT P.cid

FROM Product2 PWHERE P.price >= 25)

Q: Find all companies that make only products with price < 25!315

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Using NOT EXISTS:

SELECT DISTINCT C.cname FROM Company2 CWHERE NOT EXISTS ( SELECT *

FROM Product2 PWHERE C.cid = P.cid

and P.price >= 25)




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Using ALL:



SELECT DISTINCT C.cname FROM Company2 CWHERE 25 > ALL ( SELECT price

FROM Product2 PWHERE P.cid = C.cid)


SQLlite does not support "ALL" L

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QuestionforDatabaseFans&Friends

• Howcanweunnesttheuniversalquantifierquery?

Thistopicgoesbeyondthecourseobjectives;onlyfor

thosewhoarereallyinterested

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Queriesthatmustbenested

• Definition:AqueryQismonotoneif:- Wheneverweaddtuplestooneormoreofthetables…- …theanswertothequerycannotcontainfewertuples

• Fact:allunnestedqueriesaremonotone- Proof:usingthe"nestedforloops"semantics

• Fact:Querywithuniversalquantifierisnotmonotone- Addonetupleviolatingthecondition.Then"all"returnsfewertuples

• Consequence:wecannotunnestaquerywithauniversalquantifier

Thistopicgoesbeyondthecourseobjectives;onlyfor

thosewhoarereallyinterested

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Thedrinkers-bars-beersexample

Find drinkers that frequent some bar that serves some beer they like.

Find drinkers that frequent only bars that serve some beer they like.

Find drinkers that frequent only bars that serve only beer they like.

x: $y. $z. Frequents(x, y)ÙServes(y,z)ÙLikes(x,z)

x: "y. Frequents(x, y)Þ ($z. Serves(y,z)ÙLikes(x,z))

x: "y. Frequents(x, y)Þ "z.(Serves(y,z) Þ Likes(x,z))

Find drinkers that frequent some bar that serves only beers they like.

x: $y. Frequents(x, y)Ù"z.(Serves(y,z) Þ Likes(x,z))

Likes(drinker, beer)Frequents(drinker, bar)Serves(bar, beer)

Challenge: write these in SQL.Solutions: http://queryviz.com/online/

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BasicSQLSummary

• SQLprovidesahigh-leveldeclarativelanguageformanipulatingdata(DML)

• TheworkhorseistheSFWblock

• Setoperatorsarepowerfulbuthavesomesubtleties

• Powerful,nestedqueriesalsoallowed.

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WITH clause

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WITHclause:temporaryrelationsSELECT pname, priceFROM Product2WHERE price =

(SELECT max(price)FROM Product2)

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WITH max_price(value) as(SELECT max(price)FROM Product2)

SELECT pname, priceFROM Product2, max_priceWHERE price = value

Product (pname, price, cid)

TheWITHclausedefinesatemporaryrelationthatisavailableonlytothequeryinwhichitoccurs.Sometimeseasiertoread.Veryusefulforqueriesthatneedtoaccessthesameintermediateresultmultipletimes

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WITHclause:temporaryrelationsSELECT pname, priceFROM Product2WHERE price =

(SELECT max(price)FROM Product2)

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WITH max_price as(SELECT max(price) as valueFROM Product2)

SELECT pname, priceFROM Product2, max_priceWHERE price = value

Product (pname, price, cid)

TheWITHclausedefinesatemporaryrelationthatisavailableonlytothequeryinwhichitoccurs.Sometimeseasiertoread.Veryusefulforqueriesthatneedtoaccessthesameintermediateresultmultipletimes

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Witnesses

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Motivation:Whatarethesequeriessupposedtoreturn?

PName Price cidGizmo 15 1SuperGizmo 20 1iTouch1 300 2iTouch2 300 2

Product2cid cname city1 GizmoWorks Oslo2 Apple MountainView

Company2

Findforeachcompanyid,themaximumpriceamongstitsproducts ?

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Company2

cid mp1 202 300

Findforeachcompanyid,themaximumpriceamongstitsproducts

Findforeachcompanyid,theproductwithmaximumpriceamongstitsproducts ?

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Company2

cid mp pname1 20 SuperGizmo2 300 iTouch12 300 iTouch2

cid mp1 202 300

Findforeachcompanyid,themaximumpriceamongstitsproducts

Findforeachcompanyid,theproductwithmaximumpriceamongstitsproducts(Recallthat"groupbycid"canjustgiveusonesingletuplepercid)

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Witnesses:simple(1/4)

Q:Findthemostexpensiveproduct +itsprice315

Product2 (pname, price, cid)

(Findingthemaximumpricealonewouldbeeasy)

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SELECT max(P1.price)FROM Product2 P1

Butwewantthe"witnesses,"i.e.theproduct(s)withthemaxprice.Howdowedothat?

OurPlan:• 1.Computemaxpriceinasubquery

Q:Findthemostexpensiveproduct +itsprice

Product2 (pname, price, cid)315

1.


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SELECT P2.pname, P2.priceFROM Product2 P2

OurPlan:• 1.Computemaxpriceinasubquery• 2.Computeeachproductanditsprice...

Q:Findthemostexpensiveproduct +itsprice


SELECT max(P1.price)FROM Product2 P1

Butwewantthe"witnesses,"i.e.theproduct(s)withthemaxprice.Howdowedothat?

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1.

2.


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SELECT P2.pname, P2.priceFROM Product2 P2WHERE P2.price =

(SELECT max(P1.price)FROM Product2 P1)

OurPlan:• 1.Computemaxpriceinasubquery• 2.Computeeachproductanditsprice...

andcomparethepricewiththemaxprice



Q:Findthemostexpensiveproduct +itsprice315

Documents

L02-06-SQL - v180129