题解 | #获得积分最多的人(二)#

select
            t3.id
            ,t3.name
            ,t3.grade_num
from(
    select
                    t1.id
                    ,t1.name
                    ,sum(grade_num)  as  `grade_num`
                    ,rank() over (order by sum(grade_num) desc) as rank_
    from
                    user    t1
        inner join  grade_info  t2  on  t1.id = t2.user_id
    group by        
                    1,2
    ) t3
where
            t3.rank_ = 1
order by
                t3.id