题解 | #统计每个学校各难度的用户平均刷题数#

SELECT
    t1.university,
    t3.difficult_level,
    COUNT(t2.question_id) / COUNT(distinct (t2.device_id)) as avg_answer_cnt
FROM
    user_profile t1,
    question_practice_detail t2,
    question_detail t3
WHERE
    t1.device_id = t2.device_id
    and t2.question_id = t3.question_id
GROUP BY
    t1.university,
    t3.difficult_level;