题解 | #统计每个学校的答过题的用户的平均答题数#

select up.university,
    count(qpd.question_id) / count(distinct qpd.device_id) as "平均刷题数"
from question_practice_detail as qpd
left join user_profile as up 
on qpd.device_id=up.device_id 
group by university

记录自己出现的问题

1：审题没省清楚，每个学校就要想要到group by university

2：一开始university的别名用成了qpd

3：设计两列的平均值就不是用AVG了，而是count后相除