题解 | #获得积分最多的人(三)#

SELECT
	u.id,
	u.NAME,
	c.m AS gn 
FROM
	user AS u LEFT JOIN(
 
select user_id,m,dense_rank() over (order by m desc) as t_rank from ( 
select  user_id,sum(ng) as m from (
select user_id,
case when type='add' then grade_num else -(grade_num) end  as ng from grade_info) a
group by user_id) b )c on u.id=c.user_id where c.t_rank=1 ;