联接后在SQL中减去两列

我有一张叫“订单”的桌子

PurchaseID VARCHAR
purchaseDate DATETIME
purchasePrice FLOAT

我想找出一天与另一天的购买价格之间的差异——puyrchaseID将是相同的。（当然，这只是一个示例表）

那真的有效吗

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=?

在占位符中填入要比较的ID和日期

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查阅数据库管理系统文档，查看您是否掌握了数学知识。

您可能还想添加

和a.purchaseDate b.purchaseDate

您是否尝试过类似的方法

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

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如果愿意，也可以在查询中执行减法。您可能还希望确保不会在同一天购买两件物品：

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

选择a.purchasePrice作为purchasePriceDay1，b.purchasePrice作为purchasePriceDay2，（purchasePriceDay2-purchasePriceDay1）作为purchasePriceDelta
从a命令到b命令
其中a.PurchaseID=b.PurchaseID
a.采购日期

为什么？如果我想知道价格在同一天没有变化，那该怎么办：PIt与您的解决方案相同，只是使用子选择而不是联接。无论哪种方式，如果您使用索引进行搜索，它一点也不重要。@Benoit我们如何修改此查询以仅包括“ABS（a.PurchaseID-b.PurchaseID）>10”行

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; 

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