数据库系统基础教程第二章答案Word文档下载推荐.doc
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数据库系统基础教程第二章答案Word文档下载推荐.doc
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balance
ForrelationCustomersandthefirsttuple,thecomponentsare:
Robbieà
firstName
Banksà
lastName
901-222à
idNo
12345à
account
Exercise2.2.1d
ForrelationAccounts,arelationschemais:
Accounts(acctNo,type,balance)
ForrelationCustomers,arelationschemais:
Customers(firstName,lastName,idNo,account)
Exercise2.2.1e
Anexampledatabaseschemais:
Accounts(
acctNo,
type,
balance
)
Customers(
firstName,
lastName,
idNo,
account
Exercise2.2.1f
Asuitabledomainforeachattribute:
acctNoà
Integer
typeà
String
balanceà
firstNameà
lastNameà
idNoà
String(becausethereisahyphenwecannotuseInteger)
accountà
Exercise2.2.1g
AnotherequivalentwaytopresenttheAccountrelation:
acctNo
balance
type
34567
25
savings
23456
1000
checking
12345
12000
AnotherequivalentwaytopresenttheCustomersrelation:
idNo
firstName
lastName
account
805-333
Lena
Hand
901-222
Robbie
Banks
Exercise2.2.2
Examplesofattributesthatarecreatedforprimarilyservingaskeysinarelation:
UniversalProductCode(UPC)usedwidelyinUnitedStatesandCanadatotrackproductsinstores.
SerialNumbersonawidevarietyofproductstoallowthemanufacturertoindividuallytrackeachproduct.
VehicleIdentificationNumbers(VIN),auniqueserialnumberusedbytheautomotiveindustrytoidentifyvehicles.
Exercise2.2.3a
Wecanorderthethreetuplesinanyof3!
=6ways.Also,thecolumnscanbeorderedinanyof3!
=6ways.Thus,thenumberofpresentationsis6*6=36.
Exercise2.2.3b
Wecanorderthethreetuplesinanyof5!
=120ways.Also,thecolumnscanbeorderedinanyof4!
=24ways.Thus,thenumberofpresentationsis120*24=2880
Exercise2.2.3c
Wecanorderthethreetuplesinanyofm!
ways.Also,thecolumnscanbeorderedinanyofn!
ways.Thus,thenumberofpresentationsisn!
m!
Exercise2.3.1a
CREATETABLEProduct(
maker CHAR(30),
model CHAR(10)PRIMARYKEY,
typeCHAR(15)
);
Exercise2.3.1b
CREATETABLEPC(
modelCHAR(30),
speedDECIMAL(4,2),
ramINTEGER,
hdINTEGER,
priceDECIMAL(7,2)
Exercise2.3.1c
CREATETABLELaptop(
screenDECIMAL(3,1),
Exercise2.3.1d
CREATETABLEPrinter(
colorBOOLEAN,
typeCHAR(10),
Exercise2.3.1e
ALTERTABLEPrinterDROPcolor;
Exercise2.3.1f
ALTERTABLELaptopADDodCHAR(10)DEFAULT‘none’;
Exercise2.3.2a
CREATETABLEClasses(
classCHAR(20),
typeCHAR(5),
countryCHAR(20),
numGunsINTEGER,
boreDECIMAL(3,1),
displacementINTEGER
Exercise2.3.2b
CREATETABLEShips(
nameCHAR(30),
launchedINTEGER
Exercise2.3.2c
CREATETABLEBattles(
dateDATE
Exercise2.3.2d
CREATETABLEOutcomes(
shipCHAR(30),
battleCHAR(30),
resultCHAR(10)
Exercise2.3.2e
ALTERTABLEClassesDROPbore;
Exercise2.3.2f
ALTERTABLEShipsADDyardCHAR(30);
Exercise2.4.1a
R1:
=σspeed≥3.00(PC)
R2:
=πmodel(R1)
model
1005
1006
1013
Exercise2.4.1b
=σhd≥100(Laptop)
=Product(R1)
R3:
=πmaker(R2)
maker
E
A
B
F
G
Exercise2.4.1c
=σmaker=B(ProductPC)
=σmaker=B(ProductLaptop)
=σmaker=B(ProductPrinter)
R4:
=πmodel,price(R1)
R5:
=πmodel,price(R2)
R6:
=πmodel,price(R3)
R7:
=R4R5R6
price
1004
649
630
1049
2007
1429
Exercise2.4.1d
=σcolor=trueANDtype=laser(Printer)
=πmodel(R1)
3003
3007
Exercise2.4.1e
=σtype=laptop(Product)
=σtype=PC(Product)
=πmaker(R1)
=πmaker(R2)
=R3–R4
Exercise2.4.1f
=ρPC1(PC)
R2:
=ρPC2(PC)
R3:
=R1(PC1.hd=PC2.hdANDPC1.model<
>
PC2.model)R2
R4:
=πhd(R3)
hd
250
80
160
Exercise2.4.1g
=R1(PC1.speed=PC2.speedANDPC1.ram=PC2.ramANDPC1.model<
=πPC1.model,PC2.model(R
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