Question

基本上我有两个名为Purchase和PurchaseRefund的表。

根据Purchase，您可以拥有一个PurchaseRefund但多个PurchaseId行。

我要做的是编写一个带有参数@purchase_id和@refund_id

的查询

假设我们有一个购买行，其TotalAmount为10英镑，SubTotal为10英镑

然后我们有两个PurchaseRefund行......

216 - Has a refundAmount of £2.50   
217 - Has a refundAmount of £2.25

因此，当查询以@refund_id作为 216 运行时，它应显示

SubTotal: £10  
Refund: £2.5  
Total: £7.50

当查询以@refund_id作为 217 运行时，它应显示

SubTotal: £7.50  
Refund: £2.25  
Total: £5.25

我希望这是有道理的

Answer 1

您需要从Purchase加入PurchaseRefund并获取符合您条件的所有行 - 例如：

DECLARE @Purchase TABLE (PurchaseID INT, TotalAmount DECIMAL(10,2))

INSERT INTO @Purchase VALUES(42, 10.0)

DECLARE @PurchaseRefund TABLE (RefundID INT, PurchaseID INT, RefundAmount DECIMAL(10,2))

INSERT INTO @PurchaseRefund VALUES (216, 42, 2.50)
INSERT INTO @PurchaseRefund VALUES (217, 42, 2.25)

DECLARE @MaxRefundID INT = 217

SELECT 
   p.PurchaseID,
   SUM(r.RefundAmount) AS 'Refund', 
   MAX(p.TotalAmount) - SUM(r.RefundAmount)  AS 'Total'
FROM @Purchase p
INNER JOIN @PurchaseRefund r ON p.PurchaseID = r.PurchaseID
WHERE r.RefundId <= @MaxRefundID
GROUP BY p.PurchaseID

这项工作大部分都可以 - 我得到@MaxRefundID = 216的值：

PurchaseID  Refund  Total
   42        2.50   7.50

我得到@MaxRefundID = 217的值：

PurchaseID  Refund  Total
  42         4.75   5.25

Answer 2

如果这是用于显示/报告，我会在其中执行运行总计，使用sql执行此操作是一个庞大的PIA。

SELECT 
p.Amount,
r.Refund,  
Sum(rPrevious.Refunds) As RunningTotal,
FROM Purchases p
Inner Join Refunds r On r.PurchaseId = p.PurchaseID And r.RefundID = @RefundID
Inner Join Refunds rPrevious on r.PurchaseID = p.PurchaseID AND rPrevious.RefundID < @RefundID
WHERE p.PurchaseID = @PurchaseID
Group By p.Amount,r.Refund

会给你216 10,2.50,0 和 217这样的10,2.25,2.5

Select (Amount - RunningTotal) as Subtotal,
Refund, 
(Amount - Refund - RunningTotal) as Total
From
( SELECT 
  p.Amount,
  r.Refund  
  Sum(rPrevious.Refunds) As RunningTotal,
  FROM Purchases p
  Inner Join Refunds r On r.PurchaseId = p.PurchaseID And r.RefundID = @RefundID
  Inner Join Refunds rPrevious on rPrevious on r.PurchaseID = p.PurchaseID 
                                  AND rPrevious.RefundID < @RefundID
  WHERE p.PurchaseID = @PurchaseID
  Group By p.Amount,r.Refund
) RunningTotals

应该完成这项工作，但没有经过检查，我可能会对你的专栏名称采取一些许可。

理论虽然你想要的行是合理的购买金额 - 退款总额＆lt; @RefundID 退款金额你的总数只有一个 - 另一个。或购买金额 - 退款总额＆lt; = @RefundID

内部查询技巧只是为了获得退款总额而节省了很多麻烦？两次，一次用于小计，一次用于总计。

缺少括号...

Answer 3

这建立在marc_s的解决方案之上，以获得您想要的确切数字。

DECLARE @Purchase TABLE (PurchaseID INT, TotalAmount DECIMAL(10,2))

INSERT INTO @Purchase VALUES(42, 10.0)

DECLARE @PurchaseRefund TABLE (RefundID INT, PurchaseID INT, RefundAmount DECIMAL(10,2))

INSERT INTO @PurchaseRefund VALUES (216, 42, 2.50)
INSERT INTO @PurchaseRefund VALUES (217, 42, 2.25)

DECLARE @RefundID int = 217

SELECT 
   p.TotalAmount - (SELECT SUM(RefundAmount) FROM @PurchaseRefund WHERE RefundID < r.RefundID) as SubTotal,
   r.RefundAmount AS 'Refund', 
   p.TotalAmount - (SELECT SUM(RefundAmount) FROM @PurchaseRefund WHERE RefundID <= r.RefundID) as Total
FROM @Purchase p
INNER JOIN @PurchaseRefund r ON p.PurchaseID = r.PurchaseID
where r.RefundID = @RefundID
GROUP BY r.RefundID, p.TotalAmount, r.RefundAmount

Answer 4

管理它可能不是最好但它有效....

DECLARE     @subtotal                                       DECIMAL(18,2)


SELECT      @subtotal                                       = COALESCE  (SUM(PurchaseRefund.RefundAmount), 0) 
                                                            FROM        PurchaseRefund 
                                                            WHERE       PurchaseId      = @PurchaseId 
                                                            AND         id              < @PurchaseRefundId


SELECT          pure.Id, 
                pure.ConsumerId, 
                pure.ConsumerCode, 
                pure.RetailerStoreId,
                pure.RefundAmount, 
                pure.TimestampReceived, 
                pure.TimestampPayPalRequest, 
                pure.TimestampPayPalResponse, 
                pure.TimestampResponed          AS TimestampResponded, 
                pure.RefundKey, 
                pure.ResponseCode, 
                pure.RetailerId,
                reco.Headline                   AS OfferUsed,
                retr.Name                       AS RetailerName,
                rest.Name                       AS RetailerStoreName,
                purc.CurrencyCode,
                reco.DiscountType,
                reco.DiscountValue,
                purc.PayKey,
                purc.TransactionId,
                2                                           AS transactionType,
                purc.isInStore,
                purc.Total - @subtotal                      AS SubTotal, 
                purc.Total - @subtotal - pure.RefundAmount  AS Total, 
                purc.TrackingId, 
                purc.ResponseCode, 
                pure.TerminalId,
                pure.Id                     AS PurchaseRefundId,
                purc.Id                     AS PurchaseId       

FROM            PurchaseRefund              pure

INNER JOIN      Purchase                    purc
ON              pure.PurchaseId             = purc.Id

INNER JOIN      Retailer                    retr
ON              purc.RetailerId             = retr.Id

LEFT OUTER JOIN RetailerCoupon              reco
ON              purc.RetailerCouponId       = reco.Id

LEFT OUTER JOIN RetailerStore               rest
ON              purc.RetailerStoreId        = rest.id

WHERE           purc.Id                     = @PurchaseId
AND             pure.Id                     = @PurchaseRefundId

Answer 5

SELECT purchase.id, purchase.amount, sum(purchaseRefund.amount), (purchase.amount - sum(purchaseRefund.amount)) as total
FROM Purchase as purchase
LEFT JOIN PurchaseRefund as purchaseRefund on purchaseRefund.purchaseId = purchase.id
WHERE purchase.id = 216

SQL - 查询请求

5 个答案: