Subject: RE: [xsl] xsl:copy-of From: "Michael Kay" <mhkay@xxxxxxxxxxxx> Date: Mon, 21 May 2001 11:33:26 +0100 |
> This does the job fine. Now I want to make my XSLT more > generic so I don't > have to specifically say that the "<BODY>" is a CDATA > section. Basically I > want an exact copy of the original document whatever the > original document > may be. If you're saying, you want CDATA sections in the input to result in CDATA sections in the output, you can't: the XML parser throws away the information as to whether the input was CDATA before the XSLT transformer gets to see it. This is because in the XPath data model <BODY><![CDATA[SomeData]]></BODY> and <BODY>SomeData</BODY> are defined to be precisely equivalent: they are simply different ways of writing the same thing. Mike Kay XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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