题解 | #统计每个学校各难度的用户平均刷题数#

三表连接，注意的是，最后是 question_id 总数量/device_id总数量，也就是用户数,而不是answer_cnt总数量/总用户数

select c.university
      ,d.difficult_level
      ,count(c.question_id)/COUNT(DISTINCT c.device_id)
from 
     (select a.*
            ,b.university
            ,b.answer_cnt
      FROM question_practice_detail as a
      inner join user_profile as b
      on a.device_id=b.device_id) c
inner join question_detail as d
on c.question_id=d.question_id
group by c.university,d.difficult_level