[xsl] Copy the Document Type Definition of XML file

Hi!

I'm trying to copy resp. to modify an XML document via XSL. Here I use
the usual identity template:

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

Unfortunately the DTD definition of the source document is not copied at
this way.

Does anybody know how can I accomplish this?

The only idea I had was that you can use <xsl:output> to write a DTD
declaration. However, I have no idea how I can read out the
declaration from the source XML document. Is this possible with XSL? Or 
should I take e.g. Java and pass the info as a parameter to the stylesheet?

Thanks for any suggestions!

Rossen K.


--~------------------------------------------------------------------
XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/
or e-mail: <mailto:xsl-list-unsubscribe(_at_)lists(_dot_)mulberrytech(_dot_)com>
--~--