题解 | #统计每个学校各难度的用户平均刷题数#

SELECT m.university, m.difficult_level, count(m.question_id)/count(distinct m.device_id) as avg_answer_cnt
FROM(SELECT u.university, q2.difficult_level, q1.question_id,u.device_id
FROM user_profile u,question_practice_detail q1, question_detail q2
WHERE u.device_id=q1.device_id and q1.question_id=q2.question_id
) AS m
group by m.university, m.difficult_level
##这是使用了子查询，因为个人觉得比较好盘逻辑