Question

我有一个名为Orders的表

PurchaseID VARCHAR
purchaseDate DATETIME
purchasePrice FLOAT

我想找到一天购买价格与另一天购买价格之间的差异 - puyrchaseID将是相同的。（这只是一个示例表格）

SELECT a.purchasePrice AS purchasePriceDay1 ,  b.purchasePrice AS  purchasePriceDay1
from Orders a , Orders b
where a.PurchaseID = b.PurchaseID

这实际上是否有用

Answer 1

SELECT 
    a.PurchaseID, 
    ABS(a.PurchaseID - b.PurchaseID) AS diff
FROM 
    PurchaseID a INNER JOIN PurchaseID b ON a.PurchaseID=b.PurchaseID
WHERE a.PurchaseID=? 
    AND a.purchaseDate=? 
    AND b.purchaseDate=?

使用要比较的ID和日期填写占位符。

请参阅您的DBMS文档，看看您是否有数学absolute function。

Answer 2

您可能还想添加AND a.purchaseDate <> b.purchaseDate

Answer 3

你试过像

这样的事吗？

SELECT a.purchasePrice AS p1,  b.purchasePrice AS  p2, a.purchasePrice - b.purchasePrice as difference
from Orders a inner join Orders b on a.PurchaseID = b.PurchaseID

Answer 4

SELECT ((SELECT purchasePrice FROM Orders WHERE PurchaseID = 1 AND purchaseDate = '2010-06-21') - (SELECT purchasePrice FROM Orders WHERE PurchaseID = 1 AND purchaseDate = '2010-06-20')) AS diff;

Answer 5

如果您愿意，也可以在查询中执行减法。您可能还想确保在同一天没有购买两件商品：

SELECT a.purchasePrice AS purchasePriceDay1 ,  b.purchasePrice AS  purchasePriceDay2, (purchasePriceDay2-purchasePriceDay1) AS purchasePriceDelta
from Orders a , Orders b
where a.PurchaseID = b.PurchaseID
and a.PurchaseDate < b.PurchaseDate

在加入后在SQL中减去两列

5 个答案: