题解 | #统计每个学校各难度的用户平均刷题数#

现做后面两个表的内连接，在与第一个表做内连接，这个avg_answer_cnt感觉怪怪的

SELECT university, difficult_level, count(*)/count(distinct A.device_id) as avg_answer_cnt FROM
user_profile A inner join (select device_id, difficult_level from question_practice_detail B inner join question_detail C on B.question_id = C.question_id) D on A.device_id = D.device_id
group by difficult_level, university
order by university;