Question

我的SQL SERVER数据库中的各种表生成了以下XML

<XMLData>
...
<Type>1</Type>
...
</XMLData>

和

<XMLData>
...
<Type>2</Type>
...
</XMLData>

和

<XMLData>
...
<Type>3</Type>
...
</XMLData>

我需要的最终输出是单个组合如下：

<AllMyData>
    <XMLData>
        ...
        <Type>1</Type>
        ...
    </XMLData>
    <XMLData>
        ...
        <Type>2</Type>
        ...
    </XMLData>
    <XMLData>
        ...
        <Type>3</Type>
        ...
    </XMLData>
<AllMyData>

注意 - 我合并的所有独立元素都具有相同的标记名称。

提前感谢您查看此内容。

Answer 1

我在SQL中的各种表中生成了以下XML SERVER数据库

取决于你如何拥有它，但如果它在XML变量中，你可以这样做。

declare @XML1 xml
declare @XML2 xml
declare @XML3 xml

set @XML1 = '<XMLData><Type>1</Type></XMLData>'
set @XML2 = '<XMLData><Type>2</Type></XMLData>'
set @XML3 = '<XMLData><Type>3</Type></XMLData>'

select @XML1, @XML2, @XML3 
for xml path('AllMyData')

Answer 2

我无法评论，但可以回答，即使我认为评论更合适，我会扩展上面的rainabba回答以增加更多控制。我的.Net代码需要知道返回的列名，所以我不能依赖自动生成的名称，但需要上面提供的非常尖端的rainabba。

这样，xml可以有效地连接成一行，并将结果列命名为。您可以使用相同的方法将结果分配给XML变量，并从PROC返回该结果。

SELECT (
 SELECT XmlData as [*]
 FROM
     (
     SELECT
         xmlResult AS [*]
     FROM
         @XmlRes
     WHERE
         xmlResult IS NOT NULL
     FOR XML PATH(''), TYPE
     ) as DATA(XmlData)
 FOR XML PATH('')
) as [someColumnName]

Answer 3

如果使用for xml type，则可以组合XML列而不进行强制转换。例如：

select  *
from    (
        select  (
                select  1 as Type
                for xml path(''), type
                )
        union all
        select  (
                select  2 as Type
                for xml path(''), type
                )
        union all
        select  (
                select  3 as Type
                for xml path(''), type
                )
        ) as Data(XmlData)
for xml path(''), root('AllMyData'), type

打印：

<AllMyData>
    <XmlData>
        <Type>1</Type>
    </XmlData>
    <XmlData>
        <Type>2</Type>
    </XmlData>
    <XmlData>
        <Type>3</Type>
    </XmlData>
</AllMyData>

Answer 4

作为Mikael Eriksson答案的附录 - 如果您有一个流程需要不断添加节点，然后想要将其分组到一个节点下，这是一种方法：

declare @XML1 XML
declare @XML2 XML
declare @XML3 XML
declare @XMLSummary XML

set @XML1 = '<XMLData><Type>1</Type></XMLData>'
set @XMLSummary = (SELECT @XMLSummary, @XML1 FOR XML PATH(''))

set @XML2 = '<XMLData><Type>2</Type></XMLData>'
set @XMLSummary = (SELECT @XMLSummary, @XML2 FOR XML PATH(''))

set @XML3 = '<XMLData><Type>3</Type></XMLData>'
set @XMLSummary = (SELECT @XMLSummary, @XML3 FOR XML PATH(''))


SELECT @XMLSummary FOR XML PATH('AllMyData')

Answer 5

我需要做同样的事情，但不知道有多少行/变量，没有添加额外的模式，所以这是我的解决方案。遵循这种模式，我可以生成尽可能多的片段，组合它们，在PROCS之间传递它们甚至从procs返回它们，并且在任何时候，将它们包装在容器中，而无需修改数据或强制将XML结构添加到我的数据。我将此方法与HTTP端点一起使用以提供XML Web服务，并使用另一种将XML转换为JSON的技巧，以提供JSON Web服务。

    -- SETUP A type (or use this design for a Table Variable) to temporarily store snippets into. The pattern can be repeated to pass/store snippets to build
    -- larger elements and those can be further combined following the pattern.
    CREATE TYPE [dbo].[XMLRes] AS TABLE(
        [xmlResult] [xml] NULL
    )
    GO


    -- Call the following as much as you like to build up all the elements you want included in the larger element
    INSERT INTO @XMLRes ( xmlResult )
        SELECT
            (    
                SELECT
                    'foo' '@bar'
                FOR XML
                    PATH('SomeTopLevelElement')
            )

    -- This is the key to "concatenating" many snippets into a larger element. At the end of this, add " ,ROOT('DocumentRoot') " to wrapp them up in another element even
    -- The outer select is a time from user2503764 that controls the output column name

   SELECT (
    SELECT XmlData as [*]
    FROM
        (
        SELECT
            xmlResult AS [*]
        FROM
            @XmlRes
        WHERE
            xmlResult IS NOT NULL
        FOR XML PATH(''), TYPE
        ) as DATA(XmlData)
    FOR XML PATH('')
   ) as [someColumnName]

Answer 6

ALTER PROCEDURE usp_fillHDDT @Code  int

AS
BEGIN

 DECLARE @HD XML,@DT XML;  

    SET NOCOUNT ON;
    select invhdcode, invInvoiceNO,invDate,invCusCode,InvAmount into #HD
    from dbo.trnInvoiceHD where invhdcode=@Code

    select invdtSlNo No,invdtitemcode ItemCode,invdtitemcode ItemName,
    invDtRate Rate,invDtQty Qty,invDtAmount Amount ,'Kg' Unit into #DT from
     dbo.trnInvoiceDt  where invDtTrncode=@Code 

    set @HD = (select * from #HD HD  FOR XML AUTO,ELEMENTS XSINIL);
    set @DT = (select* from #DT DT FOR XML AUTO,ELEMENTS XSINIL);

    SELECT CAST ('<OUTPUT>'+ CAST (ISNULL(@HD,'') AS VARCHAR(MAX))+ CAST ( ISNULL(@DT,'') AS VARCHAR(MAX))+ '</OUTPUT>'   AS XML)

END

Answer 7

public String ReplaceSpecialChar(String inStr)
{
    inStr = inStr.Replace("&", "&amp;");
    inStr = inStr.Replace("<", "&lt;");
    inStr = inStr.Replace(">", "&gt;");
    inStr = inStr.Replace("'", "&#39;");
    inStr = inStr.Replace("\"", "&quot;");
    return inStr;
}

连接XML而不将类型转换为字符串

7 个答案: