题解 | #统计每个用户的平均刷题数#

SELECT 
    up.university,
    qd.difficult_level,
    COUNT(qpd.question_id)/ COUNT(DISTINCT qpd.device_id) AS avg_answer_cnt
    FROM
        question_practice_detail as qpd
    LEFT JOIN
        user_profile as up on up.device_id = qpd.device_id 
    INNER JOIN
        question_detail AS qd ON qpd.question_id = qd.question_id 
    WHERE
        up.university = "山东大学"
    GROUP BY
        difficult_level
        ;