题解 | #统计每个用户的平均刷题数#

SELECT UP.university,QD.difficult_level,COUNT(QPD.question_id)/COUNT(DISTINCT QPD.device_id) AS avg_answer_cnt
FROM question_practice_detail AS QPD
INNER JOIN user_profile AS UP
ON UP.device_id=QPD.device_id
AND UP.university='山东大学'
INNER JOIN question_detail AS QD
ON QD.question_id=QPD.question_id
GROUP BY QD.difficult_level;