题解 | 90#获得积分最多的人(二)#

select 
id
,name
,sum(grade_num) 
from user u 
join grade_info g on u.id=g.user_id
group by id,name
having sum(grade_num) >= all
(
    select sum(grade_num) over(partition by id )
    from user u 
    join grade_info g on u.id=g.user_id
)

思路：利用聚合函数求出每个人的总积分，然后利用 >=all 来得到结果。