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Re: [xsl] XSL from XSL
Subject: Re: [xsl] XSL from XSL From: JBryant@xxxxxxxxx Date: Fri, 18 Nov 2005 09:59:08 -0600 |
I've been around that block a few times, and I have found that the key thing to remember is that XSL is XML. An XSL stylesheet is just another XML document and can be treated the same as any other XML document by an XSL transform. As for copying the whole thing, an identity transform will do. If you use the following stylesheet as all three arguments (in, xsl, and out) for an XSLT processor, you'll get a copy of thing as output. <?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"/> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> </xsl:stylesheet> Jay Bryant Bryant Communication Services (presently consulting at Synergistic Solution Technologies)
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