题解 | #统计每个学校的答过题的用户的平均答题数#

select u.university  as university,(count(q.question_id)/count(distinct(q.device_id)) ) as avg_answer_cnt
from user_profile as u 
join question_practice_detail as  q
on u.device_id=q.device_id
group by university;

select u.university  as university,(count(q.question_id)/count(distinct(q.device_id)) ) as avg_answer_cnt

from user_profile as u

join question_practice_detail as  q

on u.device_id=q.device_id

group by university

order by avg_answer_cnt;

为啥加上 order by avg_answer_cnt;就不行呢？