题解 | #统计每个用户的平均刷题数#

select up.university , qd.difficult_level ,count(qpd.question_id)/count(distinct qpd.device_id) as avg_answer_cnt 

from user_profile as up join  question_practice_detail as qpd on up.device_id = qpd.device_id 

left join question_detail as qd on qpd.question_id = qd.question_id 
 
 where up.university = '山东大学'
 group by  qd.difficult_level

select 。。（字段运算）。 from 表

join表 on 连接条件 。。

where 筛选条件

group by 分组的字段

having 聚合的运算