题解 | #统计每个学校各难度的用户平均刷题数#

没啥说的跟上一道题差不多我看的视频还没到join所以直接where筛选就完事了

SELECT 
    university,
    difficult_level,
    COUNT(q_p.question_id) / COUNT(DISTINCT q_p.device_id) avg_answer_cnt
FROM user_profile u,
     question_detail q,
     question_practice_detail q_p
WHERE
    u.device_id = q_p.device_id
    and q_p.question_id = q.question_id
GROUP BY
    university,difficult_level;