题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    up.university as university,
    qd.difficult_level as difficult_level,
    COUNT(qpd.question_id)/COUNT(distinct qpd.device_id) as avg_answer_cnt
FROM
    question_practice_detail as qpd
INNER JOIN
     user_profile as up
ON
    up.device_id=qpd.device_id
INNER JOIN
    question_detail as qd
ON
    qd.question_id=qpd.question_id
GROUP BY
    up.university,
    qd.difficult_level;