题解 | #统计每个学校的答过题的用户的平均答题数#

SELECT
	university,
	ROUND( count(*)/ count( DISTINCT u.device_id ), 4 ) AS avg_answer_cnt 
FROM
	user_profile u
	INNER JOIN question_practice_detail q ON u.device_id = q.device_id 
GROUP BY
	university 
ORDER BY
	university ASC