题解 | #统计每个用户的平均刷题数#

select
    user_p.university,
    qd.difficult_level,
    count(qd.question_id)/count(distinct qpd.device_id) avg_answer_cnt
from
    (
        select
            up.device_id,
            up.university
        from
            user_profile up
        where
            up.university = '山东大学'
    ) user_p
    left join question_practice_detail qpd on user_p.device_id = qpd.device_id
    inner join question_detail qd on qpd.question_id = qd.question_id
group by
    qd.difficult_level