题解 | #统计每个学校的答过题的用户的平均答题数#

select t.university,t. count /t. d_count as avg
from
(select a.university as university,count(distinct b.device_id)as d_count,count(b.question_id)as count
from user_profile as a join question_practice_detail as b on
a.device_id=b.device_id
group by university) t
group by t.university
order by t.university