问题描述
我有一个表 Items (ItemID, Name, ...) 其中 ItemID 是自动生成的身份
I have a table Items (ItemID, Name, ...) where ItemID is auto-generated identity
我想在同一张表上的 FROM 选择中将行添加到该表中.并将 OriginalItemID 和 NewlyGeneratedID 之间的引用保存到表变量中.
I want to add rows into this table FROM select on this same table.AND save into table variable the references between OriginalItemID and NewlyGeneratedID.
所以我希望它看起来像下面这样:
So I want it to look like the following:
DECLARE @ID2ID TABLE (OldItemID INT, NewItemID INT);
INSERT INTO Items OUTPUT Items.ItemID, INSERTED.ItemID INTO @ID2ID
SELECT * FROM Items WHERE Name = 'Cat';
BUT Items.ItemID
显然在这里不起作用.是否有一种解决方法可以使 OUTPUT 从 SELECT 语句中获取原始 ItemID?
BUT Items.ItemID
obviously does not work here. Is there a workaround to make OUTPUT take original ItemID from the SELECT statement?
推荐答案
如果您使用的是 SQL Server 2008+,则可以使用 MERGE 用于获取当前 ID 和新 ID.this question 中描述了该技术.
If you are on SQL Server 2008+, you can use MERGE for getting both the current and the new ID. The technique is described in this question.
对于您的示例,语句可能如下所示:
For your example the statement might look like this:
MERGE INTO
Items AS t
USING
(
SELECT *
FROM Items
WHERE Name = 'Cat'
) AS s
ON
0 = 1
WHEN NOT MATCHED BY TARGET THEN
INSERT (target_column_list) VALUES (source_column_list)
OUTPUT
S.ItemID, INSERTED.ItemID INTO @ID2ID (OldItemID, NewItemID)
;
这篇关于T-SQL:在 OUTPUT 子句中插入原始值的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!