1.type='add'积分为正，否则积分为负，按照user_id计算每个用户的积分和

select user_id,
sum(case when type='add' then grade_num 
    else -1*grade_num end) sumgrade 
from grade_info g 
group by user_id

2.使用窗口函数rank()对上表按照积分倒序排序，这样积分最高的用户排序都为1

为方便观看，将步骤1的表记为'user_sum'

select us.user_id,us.sumgrade,rank() over (order by sumgrade) as ranking 
from user_id us

3.将步骤2的表记为rank，接下来就是找出rank表中ranking=1的信息并链接user表中的相应信息即可

select u.id,u.name,r.sumgrade as grade_sum  
from user u join rank r on u.id=r.user_id 
where r.ranking=1 
order by u.id

将上述三个步骤整合即为：

select u.id,u.name,rank.sumgrade 
from user u 
join 
(select user_id,sumgrade,rank() over (order by sumgrade desc) as ranking 
 from 
(select user_id,sum(case when type='add' then grade_num else -1*grade_num end) sumgrade 
from grade_info g group by user_id)) 
rank 
on u.id=rank.user_id 
where rank.ranking=1 order by u.id