Question

我有2个表，我试图将数据从一个插入到另一个，并保持id之间的映射。我找到here有同样问题的人，但解决方案对我不好。

这是一个例子：

两个表

CREATE TABLE [source] (i INT identity PRIMARY KEY, some_value VARCHAR(30))
CREATE TABLE [destination] (i INT identity PRIMARY KEY, some_value VARCHAR(30))

CREATE TABLE [mapping] (i_old INT, i_new INT) -- i_old is source.i value, i_new is the inserted destination.i column

一些示例数据

INSERT INTO [source] (some_value)
SELECT TOP 30 name
FROM sysobjects

INSERT INTO [destination] (some_value)
SELECT TOP 30 name
FROM sysobjects

在这里，我想将所有内容从源传输到目标，但是能够在两个表上保持映射：我尝试使用OUTPUT子句，但我不能引用插入的列之外的列：

INSERT INTO [destination] (some_value)
--OUTPUT inserted.i, s.i INTO [mapping] (i_new, i_old) --s.i doesn't work
SELECT some_value
FROM [source] s

任何人都有解决方案吗？

Answer 1

如果您没有权限修改表格，请尝试以下sql：我们的想法是使用临时表作为目标表和映射表之间的桥梁。

SQL查询：

declare @source table (i INT identity PRIMARY KEY, some_value VARCHAR(30))
declare @destination table (i INT identity PRIMARY KEY, some_value VARCHAR(30))
declare @mapping table (i_old INT, i_new INT) -- i_old is source.i value, i_new is the inserted destination.i column


declare @tempSource table
(

 id_source INT identity , source_value VARCHAR(30)
 ,Id_New int,source_new VARCHAR(30)
)


insert into @source
output inserted.i, inserted.some_value into @tempSource(id_source,source_value)
SELECT TOP 10 name
FROM sysobjects


--select * from @tempsource


insert into @destination
OUTPUT inserted.i, inserted.some_value  INTO @tempSource (Id_New,source_new)
select source_value from @tempSource


insert into @mapping
select Id_source, Id_New from 
(
    select a.id_source, a.source_value
    from
     @tempSource a
    where id_source is not null and source_value is not null
) aa
inner join 
(
    select a.Id_New, a.source_new
    from
     @tempSource a
    where Id_New is not null and source_new is not null
) bb on aa.source_value = bb.source_new


select * from @mapping

映射表结果：

i_old       i_new
----------- -----------
1           1
2           2
3           3
4           4
5           5
6           6
7           7
8           8
9           9
10          10

Answer 2

不确定是否写入方式但是有效：D

MERGE [#destination] AS D
USING [#source] AS s
    ON s.i <> s.i
WHEN NOT MATCHED BY TARGET
THEN 
    INSERT (some_value) VALUES (some_value)
OUTPUT inserted.i, s.i INTO [#mapping] (i_new, i_old);

插入 - 选择保持身份映射

2 个答案: