Question

我有下表：

+ ---------+------------+--------------+-------------+-------------+----------------+
| RecordId | Client     | Subscription | Status      | Date        | Payment Method |
+ ---------+------------+--------------+-------------+-------------+----------------+
| ABC1234  | John Doe   | NetFlex      | Renewed     | 13 Nov 2018 | Credit Card    |
| ABC1233  | Jane Smith | SplitiFy     | Not Renewed | 01 Sep 2018 | NULL           |
| ABC1232  | John Doe   | NetFlex      | Renewed     | 13 Nov 2017 | PayPart        |
| ABC1231  | Jane Smith | SplitiFy     | Renewed     | 01 Sep 2017 | Cash           |
| ABC1230  | John Doe   | NetFlex      | Renewed     | 13 Nov 2016 | Cash           |
+ ---------+------------+--------------+-------------+-------------+----------------+

我希望有一个新列，可以在其中为特定客户提供以前的付款方式，如果没有以前的付款，则将其设置为NULL。换句话说，我想获得以下信息

+ ---------+------------+--------------+-------------+-------------+----------------+-------------------------+
| RecordId | Client     | Subscription | Status      | Date        | Payment Method | Previous Payment Method |
+ ---------+------------+--------------+-------------+-------------+----------------+-------------------------+
| ABC1234  | John Doe   | NetFlex      | Renewed     | 13 Nov 2018 | Credit Card    | PayPart                 |
| ABC1233  | Jane Smith | SplitiFy     | Not Renewed | 01 Sep 2018 | None           | Cash                    |
| ABC1232  | John Doe   | NetFlex      | Renewed     | 13 Nov 2017 | PayPart        | Cash                    |
| ABC1231  | Jane Smith | SplitiFy     | Renewed     | 01 Sep 2017 | Cash           | NULL                    |
| ABC1230  | John Doe   | NetFlex      | Renewed     | 13 Nov 2016 | Cash           | NULL                    |
+ ---------+------------+--------------+-------------+-------------+----------------+-------------------------+

有人知道如何创建后一个表吗？

Answer 1

如果您使用的是SQL Server 2012或更高版本，则可以使用LAG，它将在不使用自联接的情况下返回同一结果集中前一行的数据

因此，在这里我们可以结合使用LAG和PARTITION BY以获得预期结果

SELECT RecordId,Client,Subscription,Status,Date,PaymentMethod 
       ,LAG (PaymentMethod, 1, 0) OVER (PARTITION BY Client,Subscription ORDER BY Date) 
         AS PreviousPaymentMethod 
 FROM Table

如果条件为真，则获取记录的最后一个上一个值

1 个答案: