1. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke1 Database Management Systems Chapter 5 SQL

2. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke2 Example Instances R1 S1 S2  We will use these instances of the Sailors and Reserves relations in our examples.  If the key for the Reserves relation contained only the attributes sid and bid, how would the semantics differ?

3. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke3 Aggregate Operators  In addition to simply retrieving data, we often want to perform some computation or summarization. COUNT (*) : The number of rows COUNT ( [ DISTINCT ] A) : The number of (unique) values in the A column. SUM ( [ DISTINCT ] A) : The sum of all (unique) values in the A column. AVG ( [ DISTINCT ] A) : The average of all (unique) values in the A column. MAX (A) : The maximum value in the A column. MIN (A) : The minimum value in the A column. single column

4. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke4 Aggregate Operation Examples SELECT COUNT (*) FROM Sailors S; SELECT COUNT (DISTINCT S.rating) FROM Sailors S WHERE S.sname=‘Bob’; Q: Count the number of sailors. Q: Count the number of different ratings for sailors named ‘Bob’.

5. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke5 Aggregate Operation Examples SELECT AVG (S.age) FROM Sailors S WHERE S.rating=10; SELECT AVG ( DISTINCT S.age) FROM Sailors S WHERE S.rating=10 ; Q: Find the average age of all sailors with a rating of 10. Q: Find the average age of all sailors with different ages with a rating of 10.

6. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke6 Aggregate Operation Examples SELECT S.sname, S.age FROM Sailors S WHERE S.age = (SELECT MAX(S2.age) FROM Sailors S2); Q: Find the name and age of the oldest sailor. SELECT S.sname, MAX (S.age) FROM Sailors S; Illegal in SQL, if the select clause uses an aggregate operation, then it must use only aggregate operations.

7. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke7 Aggregate Operation Examples SELECT S.sname, FROM Sailors S WHERE S.age > (SELECT MAX(S2.age) FROM Sailors S2 WHERE S2.rating=10); Q: Find the names of sailors who are older than the oldest sailor with a rating of 10.

8. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke8 Motivation for Grouping  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 (!): SELECT MIN (S.age) FROM Sailors S WHERE S.rating = i; For i = 1, 2,..., 10:

9. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke9 Queries With GROUP BY and HAVING  The select-list contains (i) attribute names (ii) terms with aggregate operations (e.g., MIN ( S.age )).  The attribute list (i) must be a subset of grouping-list. Intuitively, each answer tuple corresponds to a group, and these attributes must have a single value per group. (A group is a set of tuples that have the same value for all attributes in grouping-list.) SELECT [DISTINCT] select-list FROM relation-list WHERE qualification GROUP BY grouping-list HAVING group-qualification Q: Find the age of the youngest sailor for each rating level. SELECT S.rating, MIN(S.age) FROM Sailors S GROUP BY S.rating; If the select clause uses an aggregate operation, then it must use only aggregate operations unless the query contains a GROUP BY clause!

10. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke10 Q: Find age of the youngest sailor with age 18, for each rating with at least 2 such sailors SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 GROUP BY S.rating HAVING COUNT (*) > 1; Answer relation: Sailors instance:

11. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke11 Conceptual Evaluation  The cross-product of relation-list is computed, tuples that fail qualification are discarded, ` unnecessary’ fields are deleted, and 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. Expressions in group-qualification must have a single value per group !  One answer tuple is generated per qualifying group.

12. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke12 Q: Find age of the youngest sailor with age 18, for each rating with at least 2 sailors between 18 and 60. SELECT S.rating, MIN (S.age) AS minage FROM Sailors S WHERE S.age >= 18 AND S.age <= 60 GROUP BY S.rating HAVING COUNT (*) > 1; Answer relation: Sailors instance:

13. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke13 Q: For each red boat, find the number of reservations for this boat  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?  Wrong. Only columns that appear in the GROUP BY clause can appear in the HAVING clause, unless they appear as arguments to an aggregate operator in the HAVING clause. SELECT B.bid, COUNT (*) AS scount FROM Boats B, Reserves R WHERE R.bid=B.bid AND B.color=‘red’ GROUP BY B.bid;

14. Database Management Systems 3ed, R. Ramakrishnan and J. Gehrke14 Q: Find age of the youngest sailor with age > 18, for each rating with at least 2 sailors (of any age)  Shows HAVING clause can also contain a subquery.  Compare this with the query where we considered only ratings with 2 sailors over 18!  What if HAVING clause is replaced by:  HAVING COUNT(*) >1  CORRECT ? SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age > 18 GROUP BY S.rating HAVING 1 < (SELECT COUNT (*) FROM Sailors S2 WHERE S.rating=S2.rating);