XSLT Help

Hello experts,

Do you know what XSLT would maintain my message structure with a complete deep copy yet convert a special node of escaped xml string into normal XML format?

I have a scenario where SAP is sending an IDOC with some extra XML in a segment. This XML can vary in structure so I cannot define it inside the PI Data Type. However, PI always escapes this text. In this example file I want "myxmlnode" to be without the &lt ; and instead have <:

I want my XSLT mapping to run after GUI mapping to turn:

<?xml version="1.0" encoding="UTF-8"?>

<mynode1 xmlns="http://asdf"

date="2014-10-28T07:14:40">

  <mynode2 blahblah="asdf2">

  </mynode2>

<myxmlnode>&lt;foo xmlns:xsi=&quot;http://www.w3.org/2001/xmlschema-instance" xmlns:xsd=&quot;http://www.w3.org/2001/xmlschema"><foo2 comment=&quot;blah blah blah&quot; &gt;&lt;/foo&gt;</myxmlnode>

</mynode1>

Into:

<?xml version="1.0" encoding="UTF-8"?>

<mynode1 xmlns="http://asdf"

date="2014-10-28T07:14:40">

  <mynode2 blahblah="asdf2">

  </mynode2>

  <myxmlnode><foo xmlns:xsi="http://www.w3.org/2001/xmlschema-instance" xmlns:xsd="http://www.w3.org/2001/xmlschema"><foo2 comment="blah blah blah" ></foo></myxmlnode>

</mynode1>

I'm trying with the disable-output-escaping option but something is not working right:

<?xml version="1.0" encoding="utf-8"?>

    <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" >

      <xsl:output method="xml" indent="no" omit-xml-declaration="yes"/>

      <xsl:strip-space elements="*"/>

      <xsl:template match="@* | node()">

        <xsl:copy>

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

        </xsl:copy>

      </xsl:template>

      <xsl:template match="text()">

        <xsl:copy>

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

        </xsl:copy>

      </xsl:template>

    </xsl:stylesheet>

Many thanks for your advice,

Aaron