Question

嗨，我写了一封SQL给我剩余的发票付款 REST =（发票总额 - 付款）

我有什么：

Invoice_ID|Invoice Total   | Payment   | REST     
----------|----------------|-----------|---------
01        |93              | 10        | 69     
01        |93              | 9         | 69         
01        |93              | 5         | 69

这就是我得到的，它是正确的，但我真正想要的是递归REST

Invoice_ID|Invoice Total   | Payment   | REST     
----------|----------------|-----------|---------
01        |93              | 10        | 83     
01        |93              | 9         | 74         
01        |93              | 5         | 69

Answer 1

尝试如下。在“付款”列上运行总计，并使用您的发票金额减去该值。

SELECT Invoice_ID,InvoiceTotal   ,Payment,
       InvoiceTotal- Sum(payment)
                   OVER (
                     partition BY Invoice_ID
                     ORDER BY rno) rest
FROM   (SELECT *,
               Row_number()
                 OVER (
                   partition BY Invoice_ID
                   ORDER BY Invoice_ID) rno
        FROM   table_name)a

Answer 2

CREATE TABLE #Table(Id INT IDENTITY(1,1),Invoice_ID INT,InvoiceTotal 
INT,Payment INT,REST INT)
INSERT INTO #Table(Invoice_ID ,InvoiceTotal ,Payment ,REST)
SELECT 01,93,10,69 UNION ALL     
SELECT 01,93,9,69   UNION ALL       
SELECT 01,93,5,69 

;WITH CTE (_Id ,_InvoiceID ,_InvoiceTotal ,_Payment ,_REST )AS
(
 SELECT Id ,Invoice_ID ,InvoiceTotal ,Payment ,(InvoiceTotal - Payment)
 FROM #Table WHERE Id = 1
 UNION ALL
 SELECT Id ,Invoice_ID ,InvoiceTotal ,Payment ,_REST - Payment
 FROM #Table
 JOIN CTE ON CTE._InvoiceID = #Table.Invoice_ID 
 WHERE  _Id = Id - 1 
 )
SELECT * FROM CTE

两列的递归减法？

2 个答案: