select a.university,c.difficult_level,count(b.question_id)/count(distinct b.device_id)from user_profile as a inner join question_practice_detail as b inner join question_detail as c on a.device_id = b.device_id and b.question_id=c.question_id group by a.university,c.difficult_level 先确定要筛选的条件,大学,不同难...