题解 | #获得积分最多的人(二)#

#找到了id跟总分
select
* from (
    select a.user_id,b.name,a.add0
    from
            (
                #add
                select user_id,sum(grade_num) add0
                from grade_info 
                where type='add'
                group by user_id
            ) a left  join user b
            on a.user_id=b.id
        order by a.add0 desc
        ) t1

where t1.add0=
(
    select max(a.add0)
    from
            (
                #add
                select user_id,sum(grade_num) add0
                from grade_info 
                where type='add'
                group by user_id
            ) a left  join user b
            on a.user_id=b.id
        order by a.add0 desc
)     
        ;