题解 | #统计每个学校各难度的用户平均刷题数#

select university, detail.difficult_level, count(detail.question_id)/count(distinct detail.device_id)
from 
(select qpd.device_id ,qd.question_id, difficult_level 
 from 
 question_detail qd, question_practice_detail qpd 
 where qd.question_id = qpd.question_id) as detail,
user_profile
where detail.device_id = user_profile.device_id
group by university, difficult_level;