题解 | #统计每个学校的答过题的用户的平均答题数#

select university, sum(per_cqs) / sum(1) as avg_answer_cnt
from (
select sum(1) as per_cqs, university
from user_profile, question_practice_detail
where user_profile.device_id = question_practice_detail.device_id
group by user_profile.device_id, university
) as t
group by university
order by university asc