Sql 根据另一列的值减去行

我有两张桌子要买，我想减去购买日期。根据客户ID，有重复的客户ID，所以我想减去客户ID 105和105、108和108等的购买日期

我有以下代码，但它是从下一个购买日期减去每个购买日期

     SELECT DATEDIFF(DAY,P1.PURCHASEDATE,P2.PURCHASEDATE) AS "diff in days since last purchase"
    FROM Purchases P1
    JOIN Purchases P2
    ON P1.CustomerID= P2.CustomerID   

---

请尝试在

上的
中添加一个不相等的：
P1.PURCHASEID P2.PURCHASEID
，意思如下：

SELECT DATEDIFF(DAY,P1.PURCHASEDATE,P2.PURCHASEDATE) AS "diff in days"
FROM Purchases P1
JOIN Purchases P2
(ON P1.CustomerID= P2.CustomerID and P1.PURCHASEID <> P2.PURCHASEID  )

选择DATEDIFF（天，P1.PURCHASEDATE，P2.PURCHASEDATE）作为“以天为单位的差异”
从购买P1
加入购买P2
（在P1.CustomerID=P2.CustomerID和P1.PURCHASEID P2.PURCHASEID上）
尝试在
上的
中添加一个不相等的：
P1.PURCHASEID P2.PURCHASEID
，意思如下：

SELECT DATEDIFF(DAY,P1.PURCHASEDATE,P2.PURCHASEDATE) AS "diff in days"
FROM Purchases P1
JOIN Purchases P2
(ON P1.CustomerID= P2.CustomerID and P1.PURCHASEID <> P2.PURCHASEID  )

选择DATEDIFF（天，P1.PURCHASEDATE，P2.PURCHASEDATE）作为“以天为单位的差异”
从购买P1
加入购买P2
（在P1.CustomerID=P2.CustomerID和P1.PURCHASEID P2.PURCHASEID上）
您可以使用外部应用：

;WITH Purchases AS (
SELECT *
FROM (VALUES
(1,'2012-08-15',1,105,'a510'),
(2,'2012-08-15',2,102,'a510'),
(3,'2012-08-15',3,103,'a506'),
(4,'2012-08-16',1,105,'a510'),
(5,'2012-08-17',5,106,'a507'),
(6,'2012-08-17',5,107,'a509'),
(7,'2012-08-18',4,108,'a502'),
(8,'2012-08-19',2,108,'a510'),
(9,'2012-08-19',3,109,'a502'),
(10,'2012-08-20',3,110,'a503')
) as t(PurchaseID,PurchaseDate,Qty,CustomerID,ProductID)
)

SELECT  p1.*,
        DATEDIFF(DAY,P2.PurchaseDate,P1.PurchaseDate) as ddiff
FROM Purchases p1
OUTER APPLY (
    SELECT TOP 1 *
    FROM Purchases
    WHERE p1.CustomerID = CustomerID
        AND PurchaseDate < p1.PurchaseDate
    ORDER BY PurchaseDate DESC
) p2

您还可以使用（SQL Server 2012及更高版本）：

您可以使用外部应用程序：

;WITH Purchases AS (
SELECT *
FROM (VALUES
(1,'2012-08-15',1,105,'a510'),
(2,'2012-08-15',2,102,'a510'),
(3,'2012-08-15',3,103,'a506'),
(4,'2012-08-16',1,105,'a510'),
(5,'2012-08-17',5,106,'a507'),
(6,'2012-08-17',5,107,'a509'),
(7,'2012-08-18',4,108,'a502'),
(8,'2012-08-19',2,108,'a510'),
(9,'2012-08-19',3,109,'a502'),
(10,'2012-08-20',3,110,'a503')
) as t(PurchaseID,PurchaseDate,Qty,CustomerID,ProductID)
)

SELECT  p1.*,
        DATEDIFF(DAY,P2.PurchaseDate,P1.PurchaseDate) as ddiff
FROM Purchases p1
OUTER APPLY (
    SELECT TOP 1 *
    FROM Purchases
    WHERE p1.CustomerID = CustomerID
        AND PurchaseDate < p1.PurchaseDate
    ORDER BY PurchaseDate DESC
) p2

您还可以使用（SQL Server 2012及更高版本）：

因此，您的结果应仅包含表中至少有两个条目的客户。对如果一个客户有多个条目，您希望看到第一次购买和最后一次购买之间的差异吗？还是在最后一个和最后一个之间？或者还有什么？所以您的结果应该只包含表中至少有两个条目的客户。对如果一个客户有多个条目，您希望看到第一次购买和最后一次购买之间的差异吗？还是在最后一个和最后一个之间？或者别的什么？