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(question_id) / COUNT(distinct up.device_id), 4) 
    AS avg_answer_cnt 
    # ROUND(COUNT(question_id)/COUNT(distinct qpd.device_id), 4)
    # AS avg_answer_cnt
	#两个都可以，这里体现了对内连接的理解，
FROM user_profile AS up
	#这里也可以写成 FROM qpd 下面连接的表换成up，
INNER JOIN 
    question_practice_detail AS qpd
    ON up.device_id = qpd.device_id

GROUP BY university

ORDER BY university ASC