Question

我有一个表格，我希望为每一行ID找到剩余金额。但是，金额的顺序是按升序排列的。

id   amount
1    3
2    2
3    1
4    5

结果应如下所示：

id   remainder
1    10
2    8
3    5
4    0

有关如何实现这一目标的任何想法？我猜测over子句是要走的路，但是我不能把它拼凑起来。谢谢。

Answer 1

由于你没有指定你的RDBMS，我只假设它是Postgresql; - ）

select  *, sum(amount) over() - sum(amount) over(order by amount) as remainder
from tbl;

输出：

| ID | AMOUNT | REMAINDER |
---------------------------
|  3 |      1 |        10 |
|  2 |      2 |         8 |
|  1 |      3 |         5 |
|  4 |      5 |         0 |

工作原理：http://www.sqlfiddle.com/#!1/c446a/5

它也适用于SQL Server 2012：http://www.sqlfiddle.com/#!6/c446a/1

考虑SQL Server 2008的解决方案......

顺便说一下，你的身份证只是一个行号吗？如果是，请执行以下操作：

select 
  row_number() over(order by amount) as rn
  , sum(amount) over() - sum(amount) over(order by amount) as remainder
from tbl
order by rn;

输出：

| RN | REMAINDER |
------------------
|  1 |        10 |
|  2 |         8 |
|  3 |         5 |
|  4 |         0 |

但如果您确实需要完整的身份证并将最小的金额移到最上面，请执行以下操作：

with a as
(
  select  *, sum(amount) over() - sum(amount) over(order by amount) as remainder,
      row_number() over(order by id) as id_sort,
      row_number() over(order by amount) as amount_sort
  from tbl
)
select a.id, sort.remainder
from a
join a sort on sort.amount_sort = a.id_sort
order by a.id_sort;

输出：

| ID | REMAINDER |
------------------
|  1 |        10 |
|  2 |         8 |
|  3 |         5 |
|  4 |         0 |

请在此处查看查询进度：http://www.sqlfiddle.com/#!6/c446a/11

Answer 2

我只想提供一种更简单的方法来按降序执行此操作：

select id, sum(amount) over (order by id desc) as Remainder
from t

这适用于Oracle，SQL Server 2012和Postgres。

一般解决方案需要自我加入：

select t.id, coalesce(sum(tafter.amount), 0) as Remainder
from t left outer join
     t tafter
     on t.id < tafter.id
group by t.id

Answer 3

SQL Server 2008回答，我无法提供SQL小提琴，它似乎剥离了begin关键字，导致语法错误。我在我的机器上测试过这个：

create function RunningTotalGuarded()
returns @ReturnTable table(
    Id int, 
    Amount int not null, 
    RunningTotal int not null, 
    RN int identity(1,1) not null primary key clustered
)

as
begin

  insert into @ReturnTable(id, amount, RunningTotal) 
  select id, amount, 0 from tbl order by amount;

  declare @RunningTotal numeric(16,4) = 0;
  declare @rn_check int = 0;

  update @ReturnTable
    set 
        @rn_check = @rn_check + 1
        ,@RunningTotal = 
        case when rn = @rn_check then
            @RunningTotal + Amount
        else
            1 / 0
        end
        ,RunningTotal = @RunningTotal;     
  return;    
end;

要获得所需的输出：

with a as
(
    select *, sum(amount) over() - RunningTotal as remainder
        , row_number() over(order by id) as id_order
    from RunningTotalGuarded()
)
select a.id, amount_order.remainder
from a
inner join a amount_order on amount_order.rn = a.id_order;

守卫跑步总数的基本原理：http://www.ienablemuch.com/2012/05/recursive-cte-is-evil-and-cursor-is.html

选择较小的邪恶; - ）

如何计算每行的剩余量？

3 个答案: