题解 | #统计每个学校各难度的用户平均刷题数#

select university, difficult_level, count(qpd.id)/count(distinct qpd.device_id) from user_profile up,
question_practice_detail qpd, question_detail qd
where up.device_id = qpd.device_id and qpd.question_id = qd.question_id
group by university, difficult_level

原来多表查询也可以不用Join 然后用户平均刷题数 所以是用户做分母