XSLT: how to deep copy (including DOCTYPE, encoding, etc.)

Question

I've written the following stylesheet to copy an existing XML. during testing, i noticed that the DOCTYPE was not copied at all and the XML encoding defaultet to UTF-8 (though the original XML has ISO-8859-1). to fix this, i have to set the encoding and doctype-system hard coded:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <!-- adjust doctype-system to generate a different DOCTYPE SYSTEM value: -->
    <xsl:output method="xml" encoding="ISO-8859-1" doctype-system="mine.dtd"/>
    <!-- copy nodes: -->
    <xsl:template match="node()|@*" >
       <xsl:copy>
            <xsl:apply-templates select="node()|@*" />
       </xsl:copy>
    </xsl:template>
</xsl:stylesheet>but this results in a specialized copy-XSL. is there a generalized way to copy a XML with all content (including <! and <? tags, etc.)? (of course,

Answer 1

The encoding of an XML document is not a significant feature. If you look at the XML specification you'll see that it defines an XML document as a sequence of Unicode characters. How you encode that in bytes is not part of the specification.
That means that if you take a document and make one version of it encoded in UTF-8 and another version encoded in ISO-8859-1, those are both the same XML document.
You'll notice that XSLT doesn't give you a way to find the encoding of any of its input documents, because it is irrelevant to XML. So if you want to treat the encoding as a significant feature, then that's nothing to do with XML. You'll have to find some other way to deal with it.