题解 | #获得积分最多的人(二)#

select u.id as id, u.name as id, ttt.total_sum as grade_num from user u join (select tt.user_id, tt.total_sum, dense_rank()over(order by tt.total_sum desc) as rn from (select distinct t.user_id,t.total_sum from (select user_id, grade_num, sum(grade_num)over(partition by user_id) as total_sum from grade_info) t order by t.total_sum desc) tt)ttt on u.id = ttt.user_id where ttt.rn = 1