select university,difficult_level,count(q.question_id) /count(distinct q.device_id) as 'avg_answer_cnt' from question_practice_detail as q join user_profile as u on q.device_id=u.device_id JOIN question_detail qd ON q.question_id = qd.question_id GROUP BY university,difficult_level; 解题思路:感觉直接看问题比较好理...