1 Section 5 - Grouping Data u The GROUP BY clause allows the grouping of data u Aggregate functions are most often used with the GROUP BY clause u GROUP BY divides a table into sets, then Aggregate functions return summary values for those sets.

2 GROUP BY Syntax u SELECT select_list FROM table_list [WHERE conditions] GROUP BY group_by_list;

3 Example u SELECT pub_id, COUNT(title) FROM titles GROUP BY pub_id; u All items in the Select list that are not in the Group By list must generate a single value for each group

4 Groups within Groups u You may nest Groups within other groups by separating the columns with commas u Example: SELECT pub_id, type, COUNT(type) FROM titles GROUP BY pub_id, type;

5 Restrictions u Again: Each item in the SELECT list must produce a single value u Wrong: SELECT pub_id, type, COUNT(type) FROM titles GROUP BY pub_id;

6 More Restrictions u You can NOT use expressions in the GROUP BY clause Wrong: SELECT pub_id, SUM(price) FROM titles GROUP BY pub_id, SUM(price);

7 No Column Numbers u Unlike the ORDER BY clause, you cannot use the column select list position number in the GROUP BY clause Wrong: SELECT pub_id, SUM(price) FROM titles GROUP BY 1;

8 Multiple Summaries u To see the summary values for a publisher and for the type of books within that publisher you will need two SELECT statements u SELECT pub_id, SUM(price) FROM titles GROUP BY pub_id; u SELECT pub_id, type, SUM(price) FROM titles GROUP BY pub_id, type;

9 Exercise u Display a list of the authors and the state they live in. Sort the list by the author’s last name within state

10 Discussion u SELECT au_lname, au_fname, state FROM authors ORDER BY state, au_lname; We don't need a Group By for this statement because no summary information was asked for

11 Exercise u Display a list of states and the number of authors that are from each state. Also, show how many different cities are in each state. Sort in state order.

12 Discussion u SELECT state, count(*), count(distinct city) FROM authors GROUP BY state ORDER BY state, au_lname; This gets us the number of authors per state and the number of distinct cities in each state. If we didn't use the DISTINCT keyword we would count all authors who lived in a city.

13 NULLs and GROUPS u NULLs never equal another NULL u BUT... GROUP BY will create a separate group for the NULLs u Think of it as a Group of Unknowns

14 Example u The Type column contains NULLs SELECT type, COUNT(*) FROM titles GROUP BY type; u Returns count of 1, if we used a COUNT(type) instead of Count(*) we'd get back a zero instead. Why?

15 Discussion u Count(*) counts whole rows and there is 1 row of a NULL type group u Count(type) counts the non-NULL type columns in the NULL type group and there are zero non-NULL values in the NULL group.

16 More NULLs u More than one NULL in a column? SELECT advance, COUNT(*) FROM titles GROUP BY advance; u Two books have a NULL advance and they are grouped. [Note: zero is different group]

17 GROUP BY with WHERE u The WHERE clause allows grouping of a subset of rows. The WHERE clause acts first to find the rows you want u Then the GROUP BY clause divides the rows into groups SELECT type, AVG(price) FROM titles WHERE advance > 5000 GROUP BY type;

18 No WHERE u Same statement, no WHERE SELECT type, AVG(price) FROM titles GROUP BY type; u NULL group returned [In the previous example, the WHERE clause eliminated the NULLs]

19 ORDER the GROUPS u GROUP BY puts rows into sets, but doesn't put them in order. SELECT type, AVG(price) FROM titles WHERE advance > 5000 GROUP BY type ORDER BY 2;

20 Exercise u Show the average position that an author appears on a book if the author has a royalty share less than 100%. Also, show the number of books written by the author. List the author using his social security number and sort by social security number within number of books order. Show the authors with the most number of books first.

21 Discussion u SELECT au_id, AVG(au_ord), COUNT(title_id) FROM titleauthors WHERE royaltyshare < 1.0 GROUP BY au_id ORDER BY 3 DESC, au_id;

22 HAVING Clause u HAVING is like a WHERE clause for a GROUP u WHERE limits rows u HAVING limits GROUPs

23 HAVING Syntax u SELECT select_list FROM table_list [WHERE conditions] GROUP BY group_list [HAVING conditions];

24 HAVING Aggregates u The WHERE conditions apply before Aggregates are calculated u Then the HAVING conditions apply after Aggregates are calculated

25 HAVING vs. WHERE u WHERE comes after the FROM u HAVING comes after the GROUP BY u WHERE conditions cannot include Aggregates u HAVING conditions almost always include Aggregates

26 Example u SELECT type, count(*) FROM titles GROUP BY type HAVING COUNT(*) > 1; u NOTE: Cannot use WHERE instead of HAVING since WHERE does not allow Aggregates

27 HAVING without Aggregates u Applies to grouping columns SELECT type FROM titles GROUP BY type HAVING type LIKE 'p%'; You could of used the WHERE clause to find types that began with 'p', as well

28 Exercise u List the editor positions that have at least three editors

29 Answer u SELECT ed_pos, count(*) FROM editors GROUP BY ed_pos HAVING count(*) >= 3;

30 HAVING Conditions u You may use more than one condition on a HAVING clause SELECT pub_id, SUM(advance), AVG(price) FROM titles GROUP BY pub_id HAVING SUM(advance) > AND AVG(price) '0800';

31 Exercise u List the publisher id and the average advance for each book that the publisher sells and the total number of books they sell, but only if the total cost of all the books they sell (that are priced more than $10.00) is more than eighty dollars and they sell more than one book. Sort by pub_id and book count.

32 Discussion u SELECT pub_id, AVG(advance), COUNT(*) FROM titles WHERE price > 10 GROUP BY pub_id HAVING SUM(price) > 80 AND Count(*) > 1 ORDER BY 1, 3;

33 Discussion u The WHERE clause first eliminates all books that do not cost more than $10 u Then the GROUP BY forms the pub_id groups u Then the HAVING clause eliminates any groups whose total cost ( sum(price) ) is not greater than $80 and any pub_id that has not sold more than one book.

34 Section 5 - Last Slide u Please complete Assignment 4