题解 | #获得积分最多的人(一)#

1.分析
现根据对user_id相同的求和，再用左连接。

2.代码

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