题解 | #统计每个学校的答过题的用户的平均答题数#

select t2.university, round(avg(cnt), 4) as avg_answer_cnt from ( # 先统计每个学生的答题数量 select device_id, count(question_id) as cnt from question_practice_detail group by device_id ) t1 left join user_profile t2 on t1.device_id = t2.device_id group by t2.university order by t2.university;