题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    uq.university,
    qd.difficult_level,
    COUNT(uq.question_id) / COUNT(DISTINCT uq.device_id)
FROM
    question_detail qd
JOIN (
        SELECT
            u.university university,
            u.device_id device_id,
            q.question_id question_id
        FROM
            user_profile u
        JOIN question_practice_detail q ON u.device_id = q.device_id
    ) uq 
ON qd.question_id = uq.question_id
GROUP BY
    uq.university,
    qd.difficult_level;