SQL23 统计每个学校各难度的用户平均刷题数

SELECT
u.university,
qd.difficult_level,
count(q.question_id)/count(distinct(q.device_id)) AS avg_answer_cnt
FROM question_practice_detail AS q
LEFT JOIN user_profile AS u
ON u.device_id=q.device_id
LEFT JOIN question_detail AS qd
ON q.question_id=qd.question_id
GROUP BY u.university, qd.difficult_level;