题解 | #统计每个用户的平均刷题数#

SELECT
    u.university,
    d.difficult_level,
    COUNT(d.question_id) / COUNT(DISTINCT q.device_id) AS avg_answer_cnt
FROM
    question_practice_detail q
    LEFT JOIN user_profile u ON q.device_id = u.device_id
    LEFT JOIN question_detail d ON q.question_id = d.question_id
WHERE
    u.university = '山东大学'
GROUP BY
    d.difficult_level