题解 | #统计每个学校各难度的用户平均刷题数#

没啥意思就跟上一个差不多就是多了一个学校的筛选条件，没得说

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.university = '山东大学'
    AND u.device_id = q_p.device_id
    AND q_p.question_id = q.question_id
GROUP BY
    q.difficult_level;