题解 | #统计每个学校各难度的用户平均刷题数#

SELECT a.university,c.difficult_level,COUNT(FORMAT(b.question_id,4))/COUNT(DISTINCT(b.device_id)) AS avg_answer_cnt
FROM question_practice_detail AS b 
LEFT JOIN user_profile AS a ON a.device_id=b.device_id
LEFT JOIN question_detail AS c ON b.question_id=c.question_id
GROUP BY a.university,c.difficult_level;