这题比较简单,就是在上一题的基础上,加上分组查询条件,限定山东大学即可,考察的是分组过滤having的使用 select up.university, qd.difficult_level, round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt from question_practice_detail as qpd join question_detail as qd on qpd.question_id = qd.question_id join user_profile as up on up.device_id=qpd.device_id group by up.university,qd.difficult_level having up.university='山东大学' order by up.university,qd.difficult_level;