Question

I am looking for solution for a scenario which is troubling me a lot.

I am working on mule 3.3.

I have some incoming XML and a second XMLcoming from the enricher.

Now the xml from the enricher is to be added into my input XML.

My flow looks like below (abstract)

<flow name="main" >
    <file:inbound  ....> 
    <enricher target="#[variable:myProperty]">
        <vm:outbound .... />
    </enricher>

    <xslt transformer .... />
    .......
    .......
    <file:outbound ..>
</flow>

My Mule Flow part and the XSL as given below

    <mulexml:xslt-transformer   maxIdleTransformers="2" maxActiveTransformers="5"   xsl-file="C:\NSBTransformerXSL.xsl" outputEncoding="UTF-8" doc:name="XSLT">


        <mulexml:context-property key="RefXML"  value="#[header:INVOCATION:RefXML]" />
    </mulexml:xslt-transformer>

My XSL is given below

<xsl:param name="RefXML"></xsl:param>

<xsl:template match="@*|node()">        
    <xsl:copy   >             
        <xsl:apply-templates select="@*|node()"/>                  
    </xsl:copy>    
</xsl:template>

<xsl:template match="TXRequest">
    <xsl:copy copy-namespaces="no" >
        <xsl:apply-templates select="@* | node()"/>         
        <xsl:copy-of select="$RefXML"/>                         
    </xsl:copy>
</xsl:template>

Thanks..

Answer 1

Use an XSL-T transformer and pass the XML fragment you got from the enricher to your XSL as a named parameter.

That way you can easily combine two XMLs.

The correct way of doing this would be to parse RefXML as a DOM element, then pass it as an XSL parameter, but a bug in Mule prevents this 🙁

So the only option is a verbatim copy of the string value of RefXML:

<xsl:value-of select="$RefXML" disable-output-escaping="yes" />

Not super satisfying but it works.