1. Using Relational Databases and SQL
Chapter 6
Set Functions
Steven Emory
Department of Computer Science
California State University, Los Angeles

2. Topics for Today Set (Aggregate) Functions GROUP BY Clause
HAVING Clause

3. Set Functions Definition
A set function, or group aggregate function, is a function that operates on groups

4. Aggregate Functions Aggregate/Non-aggregate similarities
Both take some kind of input
Both perform operations on the input
Both have an single output.
Aggregate/Non-aggregate differences
Input to an aggregate function is a group of data
Input to a non-aggregate function is a single item

5. Examples Function Example: Set Function Example:
SELECT LEFT(Title, 1) FROM Movies;
Set Function Example:
SELECT MPAA, COUNT(MPAA) FROM Movies GROUP BY MPAA;

6. Aggregate Functions There are only 5 general aggregate functions
COUNT(*), COUNT(fieldname)‏
AVG(fieldname)‏
MIN(fieldname)
MAX(fieldname)‏
SUM(fieldname)‏

7. COUNT COUNT(*)‏ COUNT(fieldname)‏ Counts the number of rows in a table
Excludes NULLs (doesn't count them)‏
-- This query returns 6. SELECT COUNT(*) AS 'Number of Movies' FROM Movies;
COUNT(fieldname)‏
Same as above
-- This query also returns 6. SELECT COUNT(ArtistID) AS 'Number of Movies' FROM Movies;

8. AVG AVG(fieldname)‏ Averages all the data under fieldname
Excludes NULLs (doesn't count NULL as 0).
-- Averages all movie runtimes. SELECT AVG(Runtime) AS 'Average Runtime' FROM Movies;

9. MIN and MAX MIN(fieldname)‏ MAX(fieldname)‏
Returns the minimum value under fieldname
-- Returns the minimum movie runtime. SELECT MIN(Runtime) AS 'Shortest Runtime' FROM Movies;
MAX(fieldname)‏
Returns the maximum value under fieldname
-- Returns the maximum movie runtime. SELECT MAX(Runtime) AS 'Longest Runtime' FROM Movies;

10. SUM SUM(fieldname)‏ Sums all the data under fieldname
Excludes NULLs (doesn't count NULL as 0).
-- Sums all of the movie runtimes. SELECT SUM(Runtime) AS 'Total Runtime' FROM Movies;

11. Filtering Aggregate Calculations
To exclude items from being aggregated, you may use the WHERE clause.
Example: Count the number of PG-13 movies. SELECT COUNT(*) FROM Movies WHERE MPAA = 'PG-13';
Example: Count the number of rated R movies. SELECT COUNT(*) FROM Movies WHERE MPAA = 'R';

12. Mixing Field Types Can we calculate both with a single query?
| MPAA | COUNT(*) | | PG-13 | | | R | | rows in set (0.01 sec)
Well, we would need to mix non-aggregated fieldnames with aggregated ones
-- Example: What does this do? Does it work? No! SELECT MPAA, COUNT(MPAA) FROM Movies;

13. Grouping Tables Solution: You can divide the table into groups.
-- Groups the movies table by MPAA rating. SELECT MPAA FROM Movies GROUP BY MPAA;
-- Groups and counts movies by MPAA rating. SELECT MPAA, COUNT(MPAA) FROM Movies GROUP BY MPAA;

14. How GROUP BY Works
GROUP BY begins by sorting the table based on the grouping attribute (in our case, Gender)‏
If any aggregates are present, GROUP BY causes each aggregate to be applied per-group rather than per-table
GROUP BY then condenses the table so that each group only appears once in the table (if listed) and displays any aggregated group values along with it

15. GROUP BY Example

16. Grouping on Multiple Fields
GROUP BY can use multiple fieldnames (similar to how you can sort using multiple fieldnames)‏
-- Example: Report the number of movies by MPAA rating and year of release. SELECT MPAA, YEAR(ReleaseDate), COUNT(*) FROM Movies GROUP BY MPAA, YEAR(ReleaseDate);
In the SELECT clause that contains one or more aggregates, you should only list table attributes that are als

17. Filtering Based on Aggregates
Can we use aggregate functions in the WHERE clause?
-- List all genres that have an average movie runtime of over 2 hours. SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) WHERE AVG(Runtime) > 120 GROUP BY Genre;
The answer is no because WHERE filters during aggregation! We need something that filters after!

18. The HAVING Clause Solution is to use the HAVING clause Example:
-- List all genres that have an average movie runtime of over 2 hours. SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre HAVING AVG(Runtime) > 120;

19. How HAVING Works In previous example:
This is calculated first... SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre;
Then the result is filtered using the HAVING clause... SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre HAVING AVG(Runtime) > 120;

20. How HAVING Works So in other words: Since HAVING filters on groups:
WHERE filters per row (DURING aggregation)‏
HAVING filters per group (AFTER aggregation)‏
Since HAVING filters on groups:
You cannot use just any fieldname you want to in the SELECT or HAVING clause with an aggregate query; you can only the use ones you choose to group by
Example on next page...

21. Having Examples This works: This doesn't work:
SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre HAVING AVG(Runtime) > 120;
This doesn't work:
SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre HAVING AVG(Runtime) > Runtime;
HAVING only sees group attributes and aggregates.

22. Having Examples Why doesn't it work?
Because Runtime is an attribute of a movie, and not an attribute of a group. You can only use group attributes in a HAVING clause.
Now since Genre is a property of the aggregated group (since we are grouping by Genre), we can use it in the HAVING clause.
SELECT Genre, COUNT(*), AVG(Runtime) FROM Movies JOIN XRefGenresMovies USING(MovieID) GROUP BY Genre HAVING (AVG(Runtime) > AND Genre <> ‘Horror’);

23. HAVING Summary So in a HAVING clause: Anything else and...
You can use aggregate functions
You can use constant values
You can use grouping attributes
Anything else and...
Happy error time!
Usually “ERROR 1111 (HY000): Invalid use of group function” or “ERROR 1054 (42S22): Unknown column 'Runtime' in having clause” are the most common errors.

24. An Advanced HAVING Problem
List the country and average age of all actors born in that country, for only those countries that have an average (actor) age greater than 40. Remember that nobody every says “I'm years old!”

25. Solution
SELECT BirthCountry, TRUNCATE(AVG(TRUNCATE(DATEDIFF(C urDate(), BirthDate)/365, 0)), 0) AS 'Average Age' FROM People P JOIN XRefActorsMovies A ON P.PersonID = A.ActorID GROUP BY BirthCountry HAVING TRUNCATE(AVG(TRUNCATE(DATEDIFF(C urDate(), BirthDate)/365, 0)), 0) > 40;

26. Aggregating Distinct Values
A normal SELECT DISTINCT query filters out duplicates in a second pass
Aggregates are computed in the first pass, so if a field contains duplicate values, and you aggregate on that field, SELECT DISTINCT WILL NOT filter out duplicate values from being aggregated.
The solution is to use the DISTINCT keyword in the aggregate function:
SELECT COUNT(DISTINCT MPAA) FROM Movies;

27. Aggregating Distinct Values
Example:
-- Returns 6 since there are 6 movies. SELECT COUNT(MPAA) FROM Movies;
-- Returns 6 since there are 6 movies and 6 is unique. SELECT DISTINCT COUNT(MPAA) FROM Movies;
-- Returns 2 since only PG-13 and R rated movies are currently in the database. SELECT COUNT(DISTINCT MPAA) FROM Movies;