减去两列SQL加入后

我有一台名为订单减去两列SQL加入后

PurchaseID VARCHAR 
purchaseDate DATETIME 
purchasePrice FLOAT

我想找到一个一天与另一天的购买价格之间的差额 - 该puyrchaseID将是相同的（这是只是过程的示例表）

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

那会竟在placeho工作

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2010-06-21 Jack Kada

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=?

填写带有ID和日期的邮件进行比较。

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

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2010-06-21 19:46:19

当然，请停止使用隐式连接！ – HLGEM 2010-06-21 19:56:04

我如何加入有什么问题？ – 2010-06-21 19:57:49

@Benoit我们如何修改这个查询，只包含行“ABS（a.PurchaseID - b.PurchaseID）> 10” – bluejamesbond 2015-10-22 05:30:26

你可能也想添加AND a.purchaseDate <> b.purchaseDate

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2010-06-21 19:46:36 PatrikAkerstrand

为什么？如果我想知道价格在同一天没有变化怎么办：P – 2010-06-21 19:49:51

你尝试过类似

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

来源

2010-06-21 19:46:37

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;

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2010-06-21 19:46:59

这只是毛重。 – 2010-06-21 19:51:25

这与您的解决方案相同，只是使用子选择而不是连接。无论哪种方式，如果您使用索引进行搜索，则完全不是严重问题。 – 2010-06-22 08:17:33

可以执行减去权查询，以及如果你想。您可能还想确保您不会在同一天内购买两件商品：

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

来源

2010-06-21 19:47:17 VeeArr

减去两列SQL加入后

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