题解 | #获得积分最多的人(一)#

简单的求最大值的相关信息，被我绕啊绕的，记录一下

我的方法：

with tmp as(
select user_id,name,sum(grade_num)sg from grade_info g join user on g.user_id=user.id
group by user_id,name
)
select name,sg from tmp where sg=(select max(sg)from tmp)

改进：不用group by，用sum窗口函数，取最大值用降序+limit 1

一个select搞定

select name,sum(grade_num)over(partition by user_id)sg from grade_info g join user on g.user_id=user.id
order by sg desc limit 1