题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    user_profile.university,
    question_detail.difficult_level,
    ROUND(
        COUNT(question_practice_detail.question_id) / COUNT(DISTINCT question_practice_detail.device_id),
        4
    ) as avg_answer_cnt
FROM
    user_profile,
    question_practice_detail,
    question_detail
WHERE
    user_profile.device_id = question_practice_detail.device_id
    AND
    question_practice_detail.question_id = question_detail.question_id
GROUP BY
    university,
    difficult_level
;

难题记录