题解 | #统计每个学校各难度的用户平均刷题数#

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

先将三个表直接关联起来，按照大学和题的难易程度进行分类。

计算平均刷题数，用回答题目的次数 除以 回答的人数 （人数要去重）