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 qpd.device_id), 4) as avg_answer_cnt
from
    question_detail as qd
    inner join question_practice_detail as qpd on qpd.question_id = qd.question_id
    inner join user_profile as up on qpd.device_id = up.device_id
group by
    up.university,
    qd.difficult_level
order by
    up.university,
    qd.difficult_level;
# >>>>>>>>>>虽然多表有点晕,但是画一下各个表的联系图,就可以知道链接的时候用什么做链接条件了,剩下的交给别名,正常查就行.
# 需要注意表的链接顺序和链接条件,不能把还没连上的表用来当链接条件