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) 
from user_profile as up 
inner join question_detail as qd 
inner join question_practice_detail as qpd 
on up.device_id = qpd.device_id 
and qd.question_id = qpd.question_id 
and up.university like "%山东%"
group by qd.difficult_level;  

其实就是考多表链接

有几个要注意的点：

平均答题数目就是答题的总数量/总人数（人数可能会有一个人答多题的现象，所有要用distinct去重）

因为是指定山东大学所以用了and up.university like "%山东%"，这里制定了山东大学其实直接用and up.university = "山东大学"更对。

最后group by up.university,qd.difficult_level; 是按题目对题目难度进行分组