题解 | #获得积分最多的人(二)#

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