题解 | #获得积分最多的人(二)#

窗口函数+嵌套子查询

select id,name,grade_sum 
from
(select b.id as id, b.name as name, sum(a.grade_num) as grade_sum, 
dense_rank() over(order by sum(grade_num) desc) as rk 
from grade_info a left join user b on a.user_id = b.id group by user_id)
where rk = 1
order by id;