题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    university,
    difficult_level,
    count( q.question_id )/ count( DISTINCT ( q.device_id )) avg_answer_cnt 
FROM
    user_profile u,
    question_practice_detail q,
    question_detail qd 
WHERE
    u.device_id = q.device_id 
    AND q.question_id = qd.question_id 
GROUP BY
    university,
    difficult_level;