题解 | #统计每个用户的平均刷题数#

select up.university,qd.difficult_level,
    round(count(qpd.device_id)/count(distinct qpd.device_id),4) as avg_answer_cnt
from user_profile as up
left join question_practice_detail as qpd
on up.device_id	=qpd.device_id
left join question_detail as qd
on qpd.question_id = qd.question_id
where up.university = '山东大学'
group by up.university,qd.difficult_level