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Re: [xsl] Reading the XML declaration using XSL

Subject: Re: [xsl] Reading the XML declaration using XSL
From: Dimitre Novatchev <dnovatchev@xxxxxxxxx>
Date: Thu, 20 Jan 2005 18:58:09 +1100

The xml declaration is not represented in the XML Infoset --
therefore, it is not accessible to XPath and XSLT. Similarly,
important parts of the DTD are also lost (such as the content model of
elements), the exact lexical representation that was used for any
empty element, the order of attributes in the source xml document,
whether a namespace was declared on an element or was inherited.

Did I miss something?

So, in brief, this is something belonging to the category
"Irrelevent/Cannot be done in XSLT due to nonexistent information".

Dimitre Novatchev.

On Wed, 19 Jan 2005 22:59:05 -0600, Kenneth Stephen
<marvin.the.cynical.robot@xxxxxxxxx> wrote:
> Hi,
>    I'd like to change the way an XML file gets rendered in a browser.
> The effect that I want to achieve is what Mozilla does when one does a
> "View source" - i.e. displays the XML as a text document with syntax
> highlighting. I can write an XSL program to read XML and produce
> syntax highlighted output, and associate the XSL to the XML via the
> xml-stylesheet PI. However, if the XML file that I'm processing has an
> XML declaration at the top, then I'm at a loss as to how to process
> that. I've been unable to figure out a way to use XSL to read the XML
> declaration and process that. Any ideas / suggestions?
> Thanks,
> Kenneth

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