题解 | #统计每个学校的答过题的用户的平均答题数#

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

两个JOIN的条件要分开列出