题解 | #获得积分最多的人(二)#

with f as (select u.id
                , u.name
                , sum(grade_num) as `grade_num`
           from grade_info gi
                    inner join user u on gi.user_id = u.id
           group by u.id, u.name
           order by grade_num desc)
select id, name, grade_num
from (select *
           , dense_rank() over (order by grade_num desc ) as dr
      from f) as g
where dr = 1;