RE: [xsl] xsl:copy-of

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


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