SQL - Lecture 3 (Aggregation, etc.)

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1 SQL - Lecture 3 (Aggregation, etc.) INFS 614 INFS614 1

2 Example Instances S1 S2 R1 sid bid day /10/ /12/96 sid sname rating age 22 dustin lubber rusty sid sname rating age 28 yuppy lubber guppy rusty INFS614 2

3 SQL: Seen so far Syntax of Basic SQL query: Relation-list (FROM) Target-list (SELECT) Qualification (WHERE) Semantic: Conceptual Evaluation Strategy; String operators: LIKE, NOT LIKE; Set operators: Union (all), Intersect (all), Except (all); Qualification involving sets: IN, EXISTS, UNIQUE, ANY, ALL (and their negated version); Nested queries. INFS614 3

4 Post Processing Processes the result of an SQL query: 1) Order by: sort the result tuples by any column(s), (even columns not appearing in the SELECT clause). 2) Distinct: Duplicate removal in result tuples Example: SELECT Distinct S.sname FROM Sailors S, Reserves R WHERE S.sid=R.sid and R.bid=103 Order by S.sid asc; Aggregates... (later in lecture) INFS614 4

5 Order By: Example SELECT S.sname, S.age FROM Sailors S WHERE S.sname LIKE %U% ORDER BY S.age,S.sname ASC, S.sname DESC Sort order: ASC : Ascending (default) DESC : Descending Listed by priority E.g., ties in S.age ordered by S.sname sid sname rating age 28 yuppy lubber andy guppy rusty zorba sid sid sname rating age age guppy yuppy rusty yuppy guppy lubber INFS614 5

6 Aggregate Operators Significant extension of relational algebra. COUNT (*) COUNT ( [DISTINCT] A) SUM ( [DISTINCT] A) AVG ( [DISTINCT] A) MAX (A) MIN (A) single column INFS614 6

7 Aggregate Operators SELECT COUNT (*) FROM Sailors S COUNT (*) COUNT ( [DISTINCT] A) SUM ( [DISTINCT] A) AVG ( [DISTINCT] A) MAX (A) MIN (A) SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10 single column SELECT COUNT (DISTINCT S.rating) FROM Sailors S WHERE S.sname= Bob SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 SELECT S.sname FROM Sailors S WHERE S.rating= (SELECT MAX(S2.rating) FROM Sailors S2) INFS614 7

8 Find name and age of the oldest sailor(s) The first query is illegal! (We ll look into the reason a bit later, when we discuss GROUP BY.) The third query is equivalent to the second query, and is allowed in the SQL/92 standard, but is not supported in many systems. SELECT S.sname, MAX (S.age) FROM Sailors S SELECT S.sname, S.age FROM Sailors S WHERE S.age = (SELECT MAX (S2.age) FROM Sailors S2) SELECT S.sname, S.age FROM Sailors S WHERE (SELECT MAX (S2.age) FROM Sailors S2) = S.age INFS614 8

9 GROUP BY and HAVING So far, we ve applied aggregate operators to all (qualifying) tuples. Sometimes, we want to apply them to each of several groups of tuples. Consider: Find the age of the youngest sailor for each rating level. In general, we don t know how many rating levels exist, and what the rating values for these levels are! Suppose we know that rating values go from 1 to 10; we can write 10 queries that look like this (!): For i = 1, 2,..., 10: SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i INFS614 9

10 Solution with GROUP BY Query: Find the age of the youngest sailor for each rating level. SELECT S.rating, MIN (S.age) FROM Sailors S GROUP BY S.rating This query: 1) Patitions Sailors S tuples into groups, according to their s.rating value 2) Returns one tuple per partion including: a) the s.rating value, b) the minimum s.age for that partition. INFS614 10

11 Queries With GROUP BY and HAVING SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list [HAVING group-qualification] The grouping-list forms a partition of tuples with the same value for attributes in the list (e.g., same s.rating value). The target-list consists of an (i) attribute list (e.g., s.rating) and (ii) aggregate terms (e.g., MIN (S.age)). The attribute list must be a subset of the grouping-list This is because each target-list attribute must have a single value per group/partition! INFS614 11

12 Conceptual Evaluation The cross-product of relation-list is computed, tuples that fail qualification are discarded, `unnecessary fields are deleted. The remaining tuples are partitioned into groups by the value of attributes in grouping-list. The group-qualification is then applied to eliminate some groups. Any attribute in group-qualification that is not an argument of an aggregate op must also appear in the grouping-list. This is because expressions in group-qualification must have a single value per group! One answer tuple is generated per qualifying group. INFS614 12

13 For each red boat, find the number of reservations for this boat SELECT B.bid, COUNT (*) AS rescount FROM Boats B, Reserves R WHERE R.bid=B.bid AND B.color= red GROUP BY B.bid Grouping over a join of two relations. What do we get if we remove B.color= red from the WHERE clause and add a HAVING clause with this condition?... an error! (B.color is not in the grouping list) INFS614 13

14 For each sailor with more than three reservations, find the number of his reservations SELECT R.sid, COUNT (*) AS rescount FROM Reserves R GROUP BY R.sid HAVING COUNT(*) > 3 Find the age of the youngest sailor for each rating level > 5 which has an average age > 25 SELECT S.rating, MIN(S.age) AS minage FROM Sailors S GROUP WHERE BY S.rating > 5 HAVING GROUP S.rating BY S.rating > 5 AND HAVING AVG(S.age) > 25 INFS614 14

15 Find the age of the youngest sailor older than 18, for each rating with at least 2 (such) sailors older than 18 SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING COUNT (*) > 1 GROUP BY applied S.rating and S.age are mentioned in the SELECT, GROUP BY, HAVING clauses; other attributes discarded sid sname rating age sid sname rating age 22 dustin Dustin lubber lubber andy Andy Zorba horatio Horatio brutus brutus rusty rusty horatio Horatio art Art bob Bob rating age rating The 2nd column of result is unnamed. (Use AS to name it.) INFS Answer relation

16 Find the age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age!) SELECT S.rating, MIN (S.age) Computes total FROM Sailors S number of sailors WHERE S.age > 18 who have rating GROUP BY S.rating equal to S.rating HAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating) HAVING clause includes a correlated subquery. Counts partitions without an S.age>18 predicate What if HAVING clause is replaced by: HAVING COUNT(*) >1 (previous query)? WHERE clause always applied before HAVING clause!! INFS614 16

17 Find the ratings with the minimum average sailor age. Aggregate operations cannot be nested! The following is WRONG: SELECT S.rating FROM Sailors S WHERE AVG(S.age) = (SELECT MIN (AVG (S2.age)) FROM Sailors S2 GROUP BY S2.rating) INFS614 17

18 Continue from previous slide Use nested FROM to compute a temp table of average sailor age, for each rating level (A bit like relational algebra, or views!) SELECT Temp.rating, Temp.avgage FROM (SELECT S.rating, AVG (S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avgage = (SELECT MIN (Temp.avgage) FROM Temp) Any query with a HAVING can be rewritten without a HAVING via the nested FROM INFS614 18

19 Note on Oracle SQL in the labs Do not use the keyword AS to name a temporary table within the FROM clause. INFS614 19

20 Note on Oracle SQL in the labs SELECT Temp.rating, Temp.avgage FROM (SELECT S.rating, AVG (S.age) AS avgage FROM Sailors S GROUP BY S.rating) AS Temp WHERE Temp.avgage = (SELECT MIN (Temp.avgage) FROM Temp) INFS614 20

21 Conclusions (so far ) Post processing on the result of queries is supported. Aggregation is the most complex post processing Group by clause partitions the results into groups Having clause puts condition on groups (just like Where clause on tuples). INFS614 21

22 Null Values Field values in a tuple are sometimes unknown (e.g., a rating has not been assigned) or inapplicable (e.g., no spouse s name). They can also be hidden (for security reasons) SQL provides a single special value null for all such situations. INFS614 22

23 Dealing with the null value Special operators are needed to check if value is/ is not null. Attribute_name IS NULL : TRUE if the attribute s value is null. FALSE otherwise Attribute_name IS NOT NULL : TRUE if the attribute s value is not null. FALSE otherwise Is rating>8 true or false when rating is equal to null? Actually, it s unknown. How about: rating > 8 AND age < 40? Three-valued logic. INFS614 23

24 Three Valued Logic Logical operators AND, OR and NOT using a 3- valued logic (TRUE, FALSE and unknown). NOT T F U F T U AND T F U T T F U F F F F U U F U OR T F U T T T T F T F U U T U U INFS614 24

25 Null Values: Some issues The presence of null complicates many issues: WHERE clause eliminates tuples for which the qualification does not evaluate to TRUE. Important in nested queries involving EXISTS or UNIQUE SELECT S1.sname FROM SAILORS S1 WHERE EXISTS ( SELECT * FROM SAILORS S2 WHERE S1.sname = S2.sname AND AND S1.SID < S2.SID AND S1.RATING <> S2.RATING) sid sname rating age 22 dustin lubber horatio null andy horatio Two tuples in SQL are duplicates in a relation if corresponding attributes have the same value or both contains null. INFS614 25

26 Issues with the null value: Summary WHERE and HAVING clause eliminates rows (groups) for which the qualification does not evaluate to true (i.e., evaluate to false or unknown). Aggregate functions ignore null values (except count(*)). Distinct treats all null values as the same. The arithmetic operations +, -, *, / return null if one of the arguments is null. INFS614 26

27 Outer joins sid sname rating age 22 dustin lubber rusty (left outer-join) sid bid day /10/ /12/96 = sid sname rating age bid day 22 dustin /10/96 31 lubber Null Null 58 rusty /12/96 Outer joins are generators of null values! INFS614 27

28 Oracle Notation: Select * From Sailor S left/right/full outer join Reserves R on S.sid = R.sid; What do these return?? Select S.sid, count(r.bid) From Sailor S left outer join Reserves R on S.sid = R.sid; Group by s.sid; Select S.sid, count(*) From Sailor S left outer join Reserves R on S.sid = R.sid; Group by s.sid; INFS614 28

29 SELECT statement functions Values can be transformed before aggregation: e.g.: SELECT sum(s.a/2) FROM S; The decode function (Oracle specific): decode(attribute, domain value1, output value1, domain value2, output value2,, default output): SELECT sum(decode(major, INFS, 1, 0)) as Total_IS_Stu, sum(decode(major, INFS, 0, 1)) as Total_NonIS_Stu FROM Student ; Calculating GPA from letter grades? INFS614 29

30 Examples Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) INFS614 30

31 Query 1 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) List the classes (Class-no) currently taken by students whose names start with 'T'. SELECT distinct e.class-no FROM Enrollment e, Student s WHERE e.student-ssn = s.ssn AND s.s-name LIKE 'T%' INFS614 31

32 Query 2 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) List the students (SSN) who are currently taking exactly one class. SELECT e.student-ssn FROM Enrollment e GROUP BY e.student-ssn HAVING 1=count(*) INFS614 32

33 Query 3 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) Give the percentage of the students (among all students) who are currently taking courses offered by ISE (D-code='ISE'). SELECT count(distinct e.student-ssn)/count(distinct s.ssn) *100.0 as Percent FROM Enrollment e, Class c, Student s WHERE e.class-no=c.class-no and c.d-code='ise'; INFS614 33

34 Query 4 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) List the faculty members (F-Name) who teach 2 or more classes. List these faculty members by the number of classes they teach (ascending order). SELECT f.f-name FROM Faculty f, Class c WHERE f.ssn=c.instructor-ssn GROUP BY f.ssn, f.f-name HAVING count(distinct c.class-no)>=2 ORDER BY count(distinct c.class-no), F-Name; INFS614 34

35 Query 5 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) List the students (SSN and Name) along with the number of classes they are taking. If a student is not taking any class, the student should also be listed (with 0 as the number of classes he/she is taking). The list should be ordered by the number of classes (in an ascending order), and in case of a tie, by the SSN of the students. SELECT s.ssn, s.s-name, count(distinct Class-no) FROM Student s, Enrollment e WHERE s.ssn = e.student-ssn (+) GROUP BY s.ssn, s.s-name ORDER BY count(distinct Class-no), Ssn INFS614 35

36 Query 6 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) List the faculty members (F-Name) who teach more than twice as many classes as Professor Smith (F-Name='Smith') is teaching. (Note that if Professor Smith is not teaching anything, then any professor who teaches at least one class will satisfy the query.) SELECT f.f-name FROM Faculty f, Class c WHERE f.ssn=c.instructor-ssn GROUP BY f.ssn, f.f-name HAVING count(distinct c.class-no) > (SELECT 2*count(distinct c2.class-no) FROM Faculty f2, Class c2 WHERE f2.f-name='smith' and f2.ssn=c2.instructor-ssn) INFS614 36

37 Query 7 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) Find the number of departments which do not have a chairman (Chair_ssn is 'NULL'). SELECT count(d.d-code) FROM Department d WHERE d.chair-ssn is NULL INFS614 37

38 Query 8 Department (D-code, D-Name, Chair-SSn) Course (D-code, C-no, Title, Units) Prereq (D-code, C-no, P-code, P-no) Class (Class-no, D-code, C-no, Instructor-SSn) Faculty (Ssn, F-Name, D-Code, Rank) Student (Ssn, S-Name, Major, Status) Enrollment (Class-no, Student-Ssn) Transcript (Student-Ssn, D-Code, C-no, Grade) For each department (i.e., student s Major), give the number of graduate students (status='grad') and the number of other students (status <> 'Grad'). The two numbers must be shown in the same row as the department code. Hint: use decode. SELECT Major, sum(decode(status, 'Grad', 1,0)) as Grad, sum(decode(status, 'Grad', 0,1)) as NoGrad FROM Student GROUP BY Major INFS614 38

39 Query 9 For each department (D-code), give the highest rank(s) of the professors in the department along with the number of the faculty with that highest rank. The output contains one row for each department. Assume Full>Associate>Assistant, i.e., lexicographic order is fine. Note that some department may not have professors in some ranks (e.g., a department may not have full, or associate or assistant professors). SELECT d.d-code as dept, d.maxrank, count(f.ssn) as num FROM (SELECT D-code, max(rank) as maxrank FROM Faculty GROUP BY D-code) AS d, Faculty f WHERE d.d-code=f.d-code AND f.rank=d.maxrank GROUP BY d.d-code, d.maxrank ORDER BY d.d-code INFS614 39

40 Query 10 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)" Find the names of all students who are enrolled in two classes that meet at the same time" 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)" INFS614 40

41 Examples of Oracle s Rollup/Cube INFS614 41

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