题解 | #统计每个学校各难度的用户平均刷题数#

SELECT u.university, qd.difficult_level,
    COUNT(qp.question_id)/COUNT(DISTINCT qp.device_id) AS avg_answer_cnt
FROM user_profile u 
JOIN question_practice_detail qp 
    ON u.device_id=qp.device_id
JOIN question_detail qd 
    ON qp.question_id=qd.question_id
GROUP BY u.university, qd.difficult_level