题解 | #统计每个学校各难度的用户平均刷题数#

select 
up.university,
qd.difficult_level,
round(count(qpd.question_id)/count(distinct up.device_id),4) as avg_answer_cnt
from user_profile as up join question_practice_detail as qpd
on up.device_id = qpd.device_id
join question_detail as qd
on qpd.question_id = qd.question_id
group by up.university,qd.difficult_level

select 后面的列要表明出自哪个表

一个人可能会回答多种难度的question ，所以在计算不同难度的回复情况时，应该给device_id去重

由题目给出的输出案列，可以判断出用内连接 inner join 的的方式连接3个表并提取出的需要的信息