如何計算每行的剩餘金額？

Question

我有一張表，我想爲每個行ID找到剩餘總量。但是，金額的順序是按升序排列的。

id amount 
1 3 
2 2 
3 1 
4 5 

結果應該是這樣的：

id remainder 
1 10 
2 8 
3 5 
4 0 

如何做到這一點有什麼想法？我猜測，過分的條款是要走的路，但我不能把它拼在一起。謝謝。

Answer 1

SQL Server 2008中的回答，我不能提供一個SQL小提琴，現在看來，這條的begin關鍵字，從而提高對語法錯誤。不過，我覺得這個測試我的機器上：

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; 

爲了達到所需輸出：

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; 

理由把守運行總計：http://www.ienablemuch.com/2012/05/recursive-cte-is-evil-and-cursor-is.html

選擇兩害取其輕;-)

Answer 2

既然你沒有指定你的RDBMS，我只是假設它是PostgreSQL的;-)

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

輸出：

| 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

它工作在SQL Server 2012太：http://www.sqlfiddle.com/#!6/c446a/1

思考s解決方案SQL Server 2008 ...

---

順便說一句，你的ID只是一個行號？如果是這樣，只是這樣做：

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; 

輸出：

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

但如果你真的需要的ID完整和移動在頂部的最小量，這樣做：

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; 

輸出：

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

查看查詢進程裂變這裏：http://www.sqlfiddle.com/#!6/c446a/11

Answer 3

我只是想提供一個更簡單的方式按降序爲了做到這一點：

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

這將在甲骨文，SQL Server 2012中，和Postgres工作。

的通解requres自連接：

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