Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community!
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.Start your free trial
Ken Stone29,703 Points
Alias the count
I found that aliasing the count can make it a bit more legible.
select count(*) as num_classes, teacher_id, teachers.first_name, teachers.last_name from teachers join classes on classes.teacher_id = teachers.id group by teacher_id having num_classes = 7 order by num_classes desc;
Is there a way to find out the max(num_classes)? besides order by and looking at the results? Then aliasing max instead of having to hard code the number 7 in the query?
Ben DeitchTreehouse Teacher
WITH NUM_CLASSES_COUNT AS ( select count(*) as num_classes, teacher_id, teachers.first_name, teachers.last_name from teachers join classes on classes.teacher_id = teachers.id group by teacher_id order by num_classes desc ) SELECT * FROM NUM_CLASSES_COUNT WHERE NUM_CLASSES = ( SELECT MAX(NUM_CLASSES) FROM NUM_CLASSES_COUNT )
Does this code work as well?
SELECT FIRST_NAME, LAST_NAME FROM TEACHERS JOIN CLASSES ON CLASSES.TEACHER_ID = TEACHERS.ID WHERE PERIOD_ID IN (1, 2, 3, 4, 5, 6, 7) GROUP BY TEACHERS.ID;
Guilherme Mergulhao4,002 Points
This is the way I managed to do:
SELECT T.FIRST_NAME || ' ' || T.LAST_NAME AS FULL_NAME, COUNT(C.PERIOD_ID) AS CLASSES FROM CLASSES AS C INNER JOIN TEACHERS AS T ON T.ID = C.TEACHER_ID GROUP BY T.FIRST_NAME HAVING CLASSES = 7 ORDER BY FULL_NAME ASC