题解 | #统计每个学校各难度的用户平均刷题数#

//做这种题时，首先也是最终要的是 单词别打错

select
    a.university,
    c.difficult_level,
    round(count(b.device_id)/count(distinct(a.device_id)),4)  avg_answer_cnt
from 
    question_practice_detail b
inner join
    user_profile a
on
    a.device_id = b.device_id
inner join
    question_detail c
on
    b.question_id = c.question_id
group by 
    a.university,
    c.difficult_level;