relational algebra, r. ramakrishnan and j. gehrke1 2003 ungraded homework p1 reminder :...
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Relational Algebra, R. Ramakrishnan and J. Gehrke 1
2003 Ungraded Homework P1
Reminder: Reserves(sid,bid,day)a. Create a table Together(sid1,sid2,day) that
contains all pairs of sailors that have a reservation for the same boat on the same day.
(Reserves1 (1sid1),Reserves)(Reserves2 (1sid2),Reserves)(Together, sid1,sid2,day(Reserves1 |X| Reserves2))
b. Give the sids of all sailors that have a reservation for 10/13/2003 but not for 10/17/2003.
sid(day=10/13/2003 (Reserves)) sid(day=10/17/2003 (Reserves))
Relational Algebra, R. Ramakrishnan and J. Gehrke 2
2003 Ungraded Homework P1 (cont)
c. Give the name and sid of all sailors that have reservations for all red boats
(Tempsids, sid,bid(Reserves) / bid(color=red (Boats)) )
sid,sname(Tempsids |X| Sailors)
d. Give the name and sid of all sailors that have exactly one reservation for 11/11/2003
(Reserves2 (1sid2,2bid2), day=11/11/2003 Reserves)
(2+R, (sid(day=11/11/2003(Reserves) |X|sid=sid2 and bidbid2
Reserves2))sid,sname((sid(Reserves) 2+R) |X| Sailors))
Relational Algebra, R. Ramakrishnan and J. Gehrke 3
P2 All but one boat problem
SELECT R.sidFROM Reserves RGROUPED BY R.sidHAVING COUNT(DISTINCT R.bid)) + 1= (SELECT count(*) FROM Boats Bo)
Remarks: still needs to be tested if the query runs on ORACLE9i; counting, instead of checking the real bid’s is okay because of referentialintegrity.
Relational Algebra, R. Ramakrishnan and J. Gehrke 4
P3 NFL E/R Design Problem
Design an Entity-Relationship Diagram that models the following objects and relationships in the world of football (NFL): teams, players, games, managers and contracts. Each (NFL-) team has a unique team name, and a city it plays in. Each person being part of the NFL-world has a unique ssn and a name. Additionally, for players their weight, height, position and birth dates are of importance. Players have a contract with at most one team and receive a salary for their services, and teams have at least 24 and at most 99 players under contract. Each team has one to three managers; managers can work for at most 4 teams and receive a salary for each of their employments. Players cannot be managers. A game involves a home-team and visiting-team; additionally, the day of the game, and the score of the game are of importance; teams play each other several times in a season (not on the same day!). Moreover, for each game played we like to know which players participated in the game and how many minutes they played.
Indicate the cardinalities for each relationship type; assign roles (role names) to each
relationship if there are ambiguities! Use sub-types, if helpful to express constraints!
Team
empl.
contr
name
city
Manager
Player
Person
namessn
weight pos
birthd
height
Game
play played-in.
Day
minscore
Home Visit
Sal
Sal
(1,3)
(0,4)
(1,1)
(24,99) (0,1)
(22,*)
(0,*)(0,*)NFLE/RProblem
(0,*)
isa
isa
Scoring: 1. Play relationship a Set: 32. Person/Player/Manager: 33. Weak Game Entity: 34. Played-in: 25. Can Only Play once on a day: 16. Contract: 37. Salary, score, min attribute: 3
Relational Algebra, R. Ramakrishnan and J. Gehrke 6
Using the Default Mappingto Map the E/R Diagram to
theRelational Data Model
Game(home, visit, day, score)
Played_in(home,visit, day, ssn, min)
Player(ssn, birthd, pos,…)
Team(name, city)
Person(ssn, name)
Contract(Team,Player,Salary)
Relational Algebra, R. Ramakrishnan and J. Gehrke 7
SQL-QueriesB1) “Give the dates of all reservations for red
boats” [2]SELECT R.day FROM Reserve R, Boat B WHERE R.bid = B. bid AND B.color = ‘red’B2) “Give the boats (return bid ) that have at least
2 reservations for 5/5/2003” [4]SELECT DISTINCT R1.bidFROM Reserve R1, Reserve R2WHERE R1.day= ‘5/5/03’ AND R2.day =
‘5/5/03’ AND R1.bid=R2.bid AND R1.Sid <> R2.Sid
Relational Algebra, R. Ramakrishnan and J. Gehrke 8
SQL-QueriesB2) “Give the name and sid of all sailors that do
not have any reservations for green boat(“There is no green boat that is reserved by this sailor”)[4]
Wrong:SELECT S.sname, S.sidFROM Sailor S, Reserve R, Boat BWHERE S.sid = R.sid AND R.bid=B.bid AND
not(B.color = ‘green’)
Relational Algebra, R. Ramakrishnan and J. Gehrke 9
SQL-QueriesB2) “Give the name and sid of all sailors that do
not have any reservations for green boat(“There is no green boat that is reserved by this sailor”)[4]
Correct:SELECT S.sname, S.sidFROM Sailor SEXECPT SELECT S.sname S.sidFROM Sailor S, Reserve R, Boat BWHERE S.sid = R.sid AND R.bid=B.bid AND
B.color = ‘green’