题解 | #统计每个用户的平均刷题数#

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

不太习惯对表重命名，大佬指导会不会影响查询速度，还是单纯的美观