题解 | #统计每个用户的平均刷题数#

select university, difficult_level, count(qpd.question_id)/count(distinct up.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 qpd.question_id=qd.question_id
group by university, difficult_level
having university='山东大学'

这题好像这样也行

where university='山东大学'

group by university, difficult_level