题解 | 统计每个用户的平均刷题数

select
    up.university,
    qd.difficult_level,
    round(count(qpd.id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
from
    user_profile up
    join question_practice_detail qpd on qpd.device_id = up.device_id
    join question_detail qd on qpd.question_id = qd.question_id
where
    university in ('山东大学')
group by
    up.university,
    difficult_level
ORDER BY
    qd.difficult_level;