select university,difficult_level,round(count(qpd.question_id)/count(distinct qpd.device_id),4) avg_answer_cnt 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 发现多表连接查询的没有on条件也...