题解 | #统计每个用户的平均刷题数#

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