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

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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 SQL queries 1. Find the names of sailors over the age of 20. SELECT Sname WHERE Age > Find the names and ratings of each sailor. SELECT Sname,Rating 3. Find the names of the boats. SELECT Bname FROM BOAT 4. Find all information about sailors over the age of 20. WHERE Age > Find the names of sailors who have reserved boat 103. NATURAL JOIN RESERVATION WHERE Bid=103 There is an important subtlety here sailor names are not unique, so what should the results contain if boat 103 has been reserved by two different sailors with the same name? Without DISTINCT, even the same sailor may appear in the results multiple times if they have reserved boat 103 multiple times probably not what the problem intended. However, with DISTINCT, a name would only be listed once even if there are multiple sailors with that name which does satisfy the problem (all of the names belonging to sailors who have reserved boat 103 are included), but doesn't fully represent all the individuals who have reserved boat 103. To have a list of the individuals who have reserved boat 103, sailor IDs need to be included in order to uniquely identify sailors.,sid NATURAL JOIN RESERVATION WHERE Bid=103

3 6. Find the names of sailors who have reserved a red boat. WHERE Color='red' This has the same issue as #5 with regards to the (non-)uniqueness of sailor names. 7. Find the colors of boats reserved by Lubber. SELECT DISTINCT Color WHERE Sname='Lubber' 8. Find the names of sailors who have reserved at least one boat. The observation here is that RESERVATION lists only those sailors who have reserved at least one boat if a sailor hasn't reserved a boat, there won't be any reservations. So the task is really just to extract sailor names from the reservations. There is the same issue as #5 with regards to the (non-)uniqueness of sailor names. NATURAL JOIN RESERVATION 9. Sailors can be mentored by someone who has a higher rating and is older. Find all possible mentoring pairs. Generate all pairs of sailors, then pick out just those pairs that satisfy the mentored/mentoring definition AS S1, SAILOR AS S2 WHERE S2.Rating > S1.Rating AND S2.Age > S1.Age An alternative is to only combine those pairs of sailors who can be mentored/mentors AS S1 JOIN SAILOR AS S2 ON S2.Rating > S1.Rating AND S2.Age > S1.Age

4 10. Find the names of sailors who have reserved a red boat or a green boat. WHERE Color='red' OR Color='green' This has the same issue as #5 with regards to the (non-)uniqueness of sailor names. 11. Find all sailors whose names start with A. WHERE Sname LIKE 'A%' 12. Find all sailors whose names are five letters long. WHERE Sname LIKE ' ' 13. Find all reservations (date, sailor sids and names, and boat bids and names) for reservations made during September SELECT Day,Sid,Sname,Bid,Bname WHERE Day >= ' ' AND Day <= ' ' 14. Find all reservations (date, sailor sids and names, and boat bids and names) for reservations made on the 10 th of (any) month during the 1990s. SELECT Day,Sid,Sname,Bid,Bname WHERE Day LIKE '199_ 10' 15. Determine the age each sailor will be next year. SELECT Sname,Sid,Age+1 AS NextAge 16. For each sailor, determine the ratio of the sailor's rating to his/her age. SELECT Sname,Sid,Rating/Age AS Ratio

5 17. List the sid, name, and rating of each sailor along with the bids and names of the boats each has reserved, sorted in descending order by rating and, within each rating, in ascending order by boat name. DISTINCT is used because a sailor may have reserved the same boat more than once. SELECT DISTINCT Sid, Sname, Rating, Bid, Bname NATURAL JOIN RESERVATION NATURAL JOIN ORDER BY Rating DESC, Bname ASC