题解 | #统计每个学校各难度的用户平均刷题数#

#0基础学习SQL

select
    university,
    difficult_level,
    count(q.question_id)/count(distinct u.device_id) #本题的关键在于这个答题数量除以用户
from
    user_profile u
inner join    question_practice_detail q 
on q.device_id = u.device_id

inner join   question_detail qd
on q.question_id = qd.question_id

group by university,difficult_level  #不同学校，不同难度，进行分组