题解 | #获得积分最多的人(二)#

第一步通过group by 和sum对不用用户的累计积分分组求和，即
 ( select user_id,
          sum(grade_num) as t_num
           from
           grade_info
           group by user_id ) as a
第二步，通过rank()over()函数对第一步的累计积分进行排序，即
select user_id,t_num,
            rank() over (order by t_num desc) as r
        from
            (第一步 ) as b
注：因题目提示有多个积分第一的情况所以，用rank
第三步，将第二步生成的b表和user用join连接，并select 出需要显示的列，即id,  name, b.t_num as grade_num，最后排序，即
  select
    id,
    name,
    b.t_num as grade_num
from
    (
        select
            user_id,
            t_num,
            rank() over (
                order by
                    t_num desc
            ) as r
        from
            (
                select
                    user_id,
                    sum(grade_num) as t_num
                from
                    grade_info
                group by
                    user_id
            ) as a
    ) as b
    join user as u on u.id = b.user_id
where
    b.r = 1
    order by id