读懂题目的隐含意,分组包含了学校分组和难度分组,最后使用having分组 select university, difficult_level, count(qpd.question_id) / count(distinct qpd.device_id) as avg_answer_cnt from question_practice_detail as qpd left join user_profile as up on up.device_id=qpd.device_id left join question_detail as qd on qd....