题解 | #统计每个用户的平均刷题数#

SELECT
    up.university,
    qd.difficult_level,
-- qpd.question_id / up.device_id 的列信息需要加表前缀，因为是内连接，加不同表前缀(前提是该表有相应的列字段)结果是相同的
    ROUND(COUNT(qpd.question_id) / COUNT(DISTINCT up.device_id), 4) AS avg_answer_cnt
FROM
    question_practice_detail AS qpd
    INNER JOIN user_profile AS up ON qpd.device_id = up.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