题解 | #获得积分最多的人(一)#

select
    user.name name,
    t.total_number as grade_num
from
    user
    left join (
        select
            user_id,
            sum(grade_num) total_number
        from
            grade_info
        where
            type = "add"
        group by
            user_id
        order by
            total_number desc
    ) t on user.id = t.user_id
order by t.total_number desc limit 1