Car Maintenance ExampleSolution

TablesREGISTRATION

CAR_MAKE

CAR_MODEL

CAR_COLOUR MODEL_YEAR

LINCENCE_NO

3679MR82 Toyota Corolla Blue 2006 1967fr89768

E-TS865 Nissan Micro Red 2004 1973Smith121

PE57UVP Peugeot 407 Blue 2007 1990byt3212

PISE567 Volkswagen

Eos Lime 2006 DF-678-WV

ROMA482 Volkswagen

Golf GT Black 2007 AQ-123-AV

Z-BA975 Peugeot 207 Black 2007 1980vrt7312PART_NO PART_NAME PART_COST12390 Paint

sealantsGhȻ14.95

12391 Wiper GhȻ19.9512392 Brake pads GhȻ24.9912393 Brake Discs GhȻ49.5412395 Spark Plugs GhȻ0.9912396 Airbag GhȻ24.9512397 Tyres GhȻ25.00

CAR

PART

Tables(cont’d)

INSPECTION_CODE REGISTRATION INSPECTION_DATE EVALUATION

100036 PE57UVP 10/05/2008 FAIL100390 ROMA482 01/09/2008  106750 E-TS865 01/03/2006 PASS122456 Z-BA975 03/10/2008 FAIL145678 PISE567 30/09/2007 PASS200450 E-TS865 21/02/2005 PASS200456 E-TS865 01/04/2007 FAIL

MAINTENANCE_RECORD

INSPECTION_CODE PART_NO106750 12396106750 12397100036 12393200450 12391100036 12397200450 12392200456 12397

REPAIR

Questions

1. Display all the part names and their

prices where the cost of the part is

greater than GhȻ20.00

2. List the car registration and model

details and part numbers for all cars

where the model year is 2007, where an

inspection was carried out after

01/03/2008, which resulted in a part

being required for the repair.

Question 1: Solution

PART_NAME PART_COST

Brake pads

GhȻ24.99

Brake Discs

GhȻ49.54

Airbag GhȻ24.95

Tyres GhȻ25.00

Π𝑝𝑎𝑟𝑡𝑛𝑎𝑚𝑒¿

Question 2: Solution

STEP1:•The first part of the query states that we

need the attributes REGISTRATION and CAR_MODEL from the CAR relation.

•Further, we are only interested in cars whose MODEL_YEAR is 2007

Question 2: Solution

STEP1:

REGISTRATION

CAR_MODEL

PE57UVP 407

ROMA482 Golf GT

Z-BA975 207

Π 𝑟𝑒𝑔𝑖𝑠𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ,𝑐𝑎𝑟𝑚𝑜𝑑𝑒𝑙(𝜎𝑚𝑜𝑑𝑒𝑙 𝑦𝑒𝑎𝑟=2007

(𝐶𝐴𝑅))

Question 2: Solution

STEP2:•Next, the query requires information about

inspections which were carried out after 01/03/2008. Information about inspections is stored in the MAINTENANCE_RECORD relation.

•Since the query is not asking for specific attributes, we assume that the values of all attributes in the MAINTENANCE_RECORD is required.

•Further, we must restrict the query to only tuples where the INSPECTION_DATE > 01/03/2008.

Question 2: Solution

STEP2:𝜎 𝑖𝑛𝑠𝑝𝑒𝑐𝑡𝑖𝑜𝑛𝑑𝑎𝑡𝑒>01/03/2008

(𝑀𝐴𝐼𝑁𝑇𝐸𝑁𝐴𝑁𝐶𝐸𝑅𝐸𝐶𝑂𝑅𝐷 )

INSPECTION_CODE

REGISTRATION

INSPECTION_DATE

EVALUATION

100036 PE57UVP 10/05/2008 FAIL

100390 ROMA482

01/09/2008  

122456 Z-BA975 03/10/2008 FAIL

Question 2: Solution

STEP3:•Next, perform a NATURAL JOIN on the

query results from STEPs 1 & 2 with the common column (REGISTRATION) in both the CAR and MAINTENANCE_RECORD relations.

•STEP3a:▫In performing the NATURAL JOIN first

perform the Cartesian Product on CAR and MAINTENANCE_RECORD relations. Prefix each attribute in the CAR and MAINTENANCE_RECORD relations with C and M respectively.

Question 2: Solution

STEP3a:Cartesian Product: X CAR

M.INSPECTION_CODE

M.REGISTRATION

M.INSPECTION_DATE

M.EVALUATION

C.REGISTRATION

C.CAR_MODEL

100036 PE57UVP 10/05/2008 FAIL PE57UVP 407

100036 PE57UVP 10/05/2008 FAIL ROMA482 Golf GT

100036 PE57UVP 10/05/2008 FAIL Z-BA975 207

100390 ROMA482 01/09/2008   PE57UVP 407

100390 ROMA482 01/09/2008   ROMA482 Golf GT

100390 ROMA482 01/09/2008   Z-BA975 207

122456 Z-BA975 03/10/2008 FAIL PE57UVP 407

122456 Z-BA975 03/10/2008 FAIL ROMA482 Golf GT

122456 Z-BA975 03/10/2008 FAIL Z-BA975 207

Question 2: SolutionSTEP3b:

SELECT only the rows for which the REGISTRATION values are equal

M.INSPECTION_CODE

M.REGISTRATION

M.INSPECTION_DATE

M.EVALUATION

C.REGISTRATION

C.CAR_MODEL

100036 PE57UVP

10/05/2008 FAIL PE57UVP 407

100390 ROMA482

01/09/2008   ROMA482 Golf GT

122456 Z-BA975 03/10/2008 FAIL Z-BA975 207

Question 2: SolutionSTEP3c:

Perform a PROJECT on either C.REGISTRATION or M.REGISTRATION to the result in STEP3b

and drop the prefix’s C or M in the final relation.

INSPECTION_CODE REGISTRATION INSPECTION_DATE EVALUATION CAR_MODEL

100036 PE57UVP 10/05/2008 FAIL 407

100390 ROMA482 01/09/2008   Golf GT

122456 Z-BA975 03/10/2008 FAIL 207

Question 2: Solution

STEP4:

•The query further requires that the

information be restricted for cars where a

part was needed for a repair. This

information can be found in the REPAIR

relation by looking for a PART_NO in the

REPAIR relation, which corresponds to a

specific INSPECTION_CODE in the

MAINTENANCE_RECORD relation.

Question 2: SolutionSTEP4:Perform a NATURAL JOIN with the common

column (INSPECTION_CODE) in both the REPAIR and Query Result of STEP3c.

INSPECTION_CODE REGISTRATION INSPECTION_DATE

EVALUATION CAR_MODEL

PART_NO

100036 PE57UVP 10/05/2008 FAIL 407 12393

100036 PE57UVP 10/05/2008 FAIL 407 12397

QueryResult

Question 2: SolutionSTEP5:

Finally, the original query requested that we list “the car registration, model details and part numbers.”This requires that we perform a PROJECT on the query result of STEP4.

REGISTRATION CAR_MODEL PART_NO

PE57UVP 407 12393

PE57UVP 407 12397

Π 𝑟𝑒𝑔𝑖𝑠𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ,𝑐𝑎𝑟𝑚𝑜𝑑𝑒𝑙 ,𝑝𝑎𝑟𝑡𝑛𝑜(𝑄𝑢𝑒𝑟𝑦𝑅𝑒𝑠𝑢𝑙𝑡 )