题解 | 统计每个学校各难度的用户平均刷题数

SELECT
    B.university,
    C.difficult_level,
    COUNT(A.question_id) /COUNT(DISTINCT A.device_id) as avg_answer_cnt
FROM
    question_practice_detail A
    INNER JOIN user_profile B ON A.device_id = B.device_id
    LEFT JOIN question_detail C
    ON A.question_id = C.question_id
GROUP BY
    B.university,
    C.difficult_level;