SELECT u.university,qe.difficult_level ,COUNT(qe.question_id) / COUNT(DISTINCT q.device_id) AS avg_answer_cnt FROM user_profile u JOIN question_practice_detail q ON u.device_id=q.device_id JOIN question_detail qe ON q.question_id =qe.question_id GROUP BY u.university,qe.difficult_level ; 这个做的时候一直没有分清查询那个平均数，搞了好大一会。做的总共问题的id除以做题人