题解 | #统计每个用户的平均刷题数#

select
    university,
    difficult_level,
    count(qpd.device_id) / count(DISTINCT qpd.question_id) as avg_answer_cnt
from
    user_profile up,
    question_practice_detail qpd,
    question_detail qd
where
    up.university = '山东大学' 
and 
    up.device_id = qpd.device_id
and
    qpd.question_id = qd.question_id
GROUP BY
    qd.difficult_level;