题解 | #统计每个用户的平均刷题数#

select
    university,
    difficult_level,
    count(a.device_id) / count(distinct a.device_id)
from
    user_profile as a,
    question_practice_detail as b,
    question_detail as c
where
    a.device_id = b.device_id
    and b.question_id = c.question_id
group by
    university,
    difficult_level
having
    university = '山东大学'