题解 | #统计每个学校各难度的用户平均刷题数#

select
    d.university,
    d.difficult_level,
    sum(d.queNums) / count(d.device_id) as avg_answer_cnt
from
    (
        select
            a.university,
            a.device_id,
            c.difficult_level,
            count(c.difficult_level) as queNums
        from
            user_profile a
            join question_practice_detail b on a.device_id = b.device_id
            join question_detail c on b.question_id = c.question_id
        group by
            a.university,
            a.device_id,
            c.difficult_level
    ) d
group by
    d.university,
    d.difficult_level

暂时这样写吧，第一思路