题解 | #统计每个用户的平均刷题数#

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

加油💪