使用rank()over()和sum()内嵌定位并列第一的用户

select u.id,u.name,t.grade_sum
from (select user_id,
             sum(grade_num) as grade_sum,
             rank()over(order by sum(grade_num) desc) as a
      from grade_info
      group by user_id) as t
join user u
on u.id = t.user_id
where t.a = 1