题解 | #统计每个学校各难度的用户平均刷题数#

count(qpd.question_id)/ count(distinct qpd.device_id) as avg_answer_cnt
from question_practice_detail as qpd // 这个表里的字段分别出现在了下面两个表里，所以以这个表作为主表来join另外两个表
inner join user_profile as u on u.device_id = qpd.device_id//可以连续join两个表
inner join question_detail as qd on qpd.question_id = qd.question_id
group by u.university, qd.difficult_level//group by 的字段需要select里也有