## Questions Based on SQL – Part 6

Question 1 : Consider the following relations: Student(snum: integer, sname: string, major: string, level: string, age: integer) Class(name: string, meets at: string, room: string, fid: integer) Enrolled(snum: integer, cname: string) Faculty(fid: integer, fname: string, deptid: integer) The meaning of these relations is straightforward; for example, Enrolled has one record per student-class pair such that the student is enrolled in the class. Write the following queries in SQL. No duplicates should be printed in any of the answers.

1. Find the names of all Juniors (level = JR) who are enrolled in a class taught by I. Teach.

Solution : SELECT DISTINCT S.Sname FROM Student S, Class C, Enrolled E, Faculty F WHERE S.snum = E.snum AND E.cname = C.name AND C.fid = F.fid AND F.fname = ‘I.Teach’ AND S.level = ‘JR’

2. Find the age of the oldest student who is either a History major or enrolled in a course taught by I. Teach.

Solution : SELECT MAX(S.age) FROM Student S WHERE (S.major = ‘History’) OR S.snum IN (SELECT E.snum FROM Class C, Enrolled E, Faculty F WHERE E.cname = C.name AND C.fid = F.fid AND F.fname = ‘I.Teach’ )

3. Find the names of all classes that either meet in room R128 or have five or more students enrolled.

Solution : SELECT C.name FROM Class C WHERE C.room = ‘R128’ OR C.name IN (SELECT E.cname FROM Enrolled E GROUP BY E.cname HAVING COUNT (*) >= 5)

4. Find the names of all students who are enrolled in two classes that meet at the same time.

Solution : SELECT DISTINCT S.sname FROM Student S WHERE S.snum IN (SELECT E1.snum FROM Enrolled E1, Enrolled E2, Class C1, Class C2 WHERE E1.snum = E2.snum AND E1.cname <> E2.cname AND E1.cname = C1.name AND E2.cname = C2.name AND C1.meets at = C2.meets at)

5. Find the names of faculty members who teach in every room in which some class is taught.

Solution : SELECT DISTINCT F.fname FROM Faculty F WHERE NOT EXISTS (( SELECT * FROM Class C ) EXCEPT (SELECTC1.room FROM Class C1 WHERE C1.fid = F.fid ))

6. Find the names of faculty members for whom the combined enrollment of the courses that they teach is less than five.

Solution : SELECT DISTINCT F.fname FROM Faculty F WHERE 5 > (SELECT COUNT (E.snum) FROM Class C, Enrolled E WHERE C.name = E.cname AND C.fid = F.fid)

7. For each level, print the level and the average age of students for that level.

Solution : SELECT S.level, AVG(S.age) FROM Student S GROUP BY S.level

8. For all levels except JR, print the level and the average age of students for that level.

Solution : SELECT S.level, AVG(S.age) FROM Student S WHERE S.level <> ‘JR’ GROUP BY S.level

9. For each faculty member that has taught classes only in room R128, print the faculty member’s name and the total number of classes she or he has taught.

Solution : SELECT F.fname, COUNT(*) AS CourseCount FROM Faculty F, Class C WHERE F.fid = C.fid GROUP BY F.fid, F.fname HAVING EVERY ( C.room = ‘R128’ )

10. Find the names of students enrolled in the maximum number of classes.

Solution : SELECT DISTINCT S.sname FROM Student S WHERE S.snum IN (SELECT E.snum FROM Enrolled E GROUP BY E.snum HAVING COUNT (*) >= ALL (SELECT COUNT (*) FROM Enrolled E2 GROUP BY E2.snum ))

11. Find the names of students not enrolled in any class.

Solution : SELECT DISTINCT S.sname FROM Student S WHERE S.snum NOT IN (SELECT E.snum FROM Enrolled E )

12. For each age value that appears in Students, find the level value that appears most often. For example, if there are more FR level students aged 18 than SR, JR, or SO students aged 18, you should print the pair (18, FR).

Solution : SELECT S.age, S.level FROM Student S GROUP BY S.age, S.level, HAVING S.level IN (SELECT S1.level FROM Student S1 WHERE S1.age = S.age GROUP BY S1.level, S1.age HAVING COUNT (*) >= ALL (SELECT COUNT (*) FROM Student S2 WHERE s1.age = S2.age GROUP BY S2.level, S2.age))

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Questions On SQL – Part 5 | Relational Algebra |

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