SELECT u.university,d.difficult_level,COUNT(qpd.question_id) / COUNT(DISTINCT u.device_id) FROM user_profile u JOIN question_practice_detail qpd ON u.device_id=qpd.device_id JOIN question_detail d ON qpd.question_id=d.question_id WHERE u.university='山东大学' GROUP BY u.university,d.difficult_level; 这一个和上一个一样，只不过添加了一个过滤条件而已