RESERVATION.Sid and RESERVATION.Bid are foreign keys referring to SAILOR.Sid and BOAT.Bid, respectively.

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Tyrone Baker
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1 relational schema SAILOR ( Sid: integer, Sname: string, Rating: integer, Age: real ) BOAT ( Bid: integer, Bname: string, Color: string ) RESERVATION ( Sid: integer, Bid: integer, Day: date ) RESERVATION.Sid and RESERVATION.Bid are foreign keys referring to SAILOR.Sid and BOAT.Bid, respectively. sample data SAILOR Sid Sname Rating Age 22 Dustin Brutus Lubber Andy Rusty Horatio Zorba Horatio Art Bob BOAT Bid Bname Color 101 Interlake blue 102 Interlake red 103 Clipper green 104 Marine red RESERVATION Sid Bid Day

2 more SQL queries 1. Find the names of sailors who have no rating (NULL). Sids are included because sailor names are not unique, and we probably want to know the individuals with no rating. SELECT Sname,Sid WHERE Rating IS NULL 2. Find the sids and names of sailors who have reserved both a red boat and a green boat. (two solutions) Note that the following does not work the WHERE is applied to each row of the FROM table in turn, so this is looking for a reservation for a boat that is both red and green at the same time. SELECT DISTINCT Sname,Sid S NATURAL JOIN RESERVATION NATURAL JOIN BOAT B WHERE B.Color='red' AND B.Color='green' A tactic is to instead look for a reservation for a red boat where there is another reservation for the same sailor with a green boat. DISTINCT is needed here because a sailor may have multiple reservations for red boats and so would appear in the results more than once. SELECT DISTINCT Sname,Sid S NATURAL JOIN RESERVATION NATURAL JOIN BOAT B WHERE B.Color='red' AND EXISTS ( SELECT * FROM RESERVATION R2 NATURAL JOIN BOAT B2 WHERE R2.Sid=S.Sid AND B2.Color='green' ) Another strategy is to find all pairs of sailors-reservations-boats, then pick those where the first reservation is for a red boat and the second is for a green boat SELECT DISTINCT S1.Sid,S1.Sname FROM ( SAILOR AS S1 NATURAL JOIN RESERVATION AS R1 NATURAL JOIN BOAT AS B1 ), ( SAILOR AS S2 NATURAL JOIN RESERVATION AS R2 NATURAL JOIN BOAT AS B2 ) WHERE S1.Sid=S2.Sid and B1.Color = 'red' AND B2.Color = 'green'

3 3. Find the sids and names of sailors over the age of 30 who have not reserved a boat. The result is a subset of SAILORS, so that's the FROM. For who have not reserved a boat, we construct the set of sailors who have reserved a boat and then look for sailors not in that set. FROM RESERVATION ) 4. Find the sids and names of sailors over the age of 30 who have not reserved a red boat. This is very similar to #3 in wording, so we try an approach very similar to #3 for the solution: FROM RESERVATION NATURAL JOIN BOAT WHERE Color='red' ) Note that this includes both sailors who have reserved boats of other colors (but not red ones) and sailors who have not reserved any boats. The following excludes sailors who have not reserved any boats by starting from the set of sailors with reservations (in the outer FROM) rather than just the set of sailors: NATURAL JOIN RESERVATION FROM RESERVATION NATURAL JOIN BOAT WHERE Color='red' ) 5. Find the sids and names of the sailor(s) with the highest rating. (two solutions) find sailors whose ratings are greater than or equal to that of every sailor SELECT S.Sid,S.Sname WHERE S.Rating >= ALL ( SELECT Rating ) find sailors whose ratings aren't the highest (because the sailor is paired with another with a higher rating), then pick the sailor(s) not in that group SELECT S.Sid,S.Sname WHERE S.Rating NOT IN ( SELECT S1.Sid S1, SAILOR S2 WHERE S1.Rating < S2.Rating )

4 6. Find the sids and names of the sailor(s) with the second lowest rating. find a sailor where another sailor can be found with a lower rating, but not a second such sailor SELECT * S WHERE EXISTS ( SELECT * S2 WHERE S2.Rating<S.Rating AND NOT EXISTS ( SELECT * S3 WHERE S3.Rating<S.Rating AND S2.Sid <> S3.Sid ) ) 7. List sailors (sid and name) along with the boats (bid and boat name) each sailor hasn't reserved. Strategy: the result is a subset of all combinations of sailors and boats, so find all combinations of sailors and boats and then eliminate those combinations for which there are actually reservations.,bid,bname, BOAT WHERE (Sid,Bid) NOT IN ( SELECT Sid,Bid FROM RESERVATION ) 8. Find the sids and names of sailors who have reserved every boat. Observe that the sailors found in the result set for #7 are exactly those who haven't reserved every boat (because there's at least one boat they haven't reserved). So finding the sailors not in that set would give us a solution WHERE Sid NOT IN ( SELECT Sid, BOAT WHERE (Sid,Bid) NOT IN ( SELECT Sid,Bid FROM RESERVATION ) )

5 9. Find the sids and names of sailors who have reserved every boat named Interlake. This is similar to #8 start with all combinations of sailors and boats named Interlake, then choose those who are not in the set of sailors who have actually reserved boats named Interlake. This results in the set of sailors who haven't reserved a boat named Interlake. The sailors not in this set are what we want if a sailor is not a sailor who hasn't reserved a boat named Interlake, they have reserved all such boats. WHERE Sid NOT IN ( SELECT Sid, BOAT WHERE Bname='Interlake' AND (Sid,Bid) NOT IN ( SELECT Sid,Bid FROM RESERVATION WHERE Bname='Interlake' ) )

RESERVATION.Sid and RESERVATION.Bid are foreign keys referring to SAILOR.Sid and BOAT.Bid, respectively.

RESERVATION.Sid and RESERVATION.Bid are foreign keys referring to SAILOR.Sid and BOAT.Bid, respectively. relational schema SAILOR ( Sid: integer, Sname: string, Rating: integer, Age: real ) BOAT ( Bid: integer, Bname: string, Color: string ) RESERVATION ( Sid: integer, Bid: integer, Day: date ) RESERVATION.Sid