题解 | #统计每个用户的平均刷题数#

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