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Re: Inline DTD in stylesheet

Sadly, DTDs seem to be second class citizens for most XML specs. XSLT is no 

I don't know of an easy way to do it, but perhaps a kludge could be found.

Are you trying to:
   1. Always output the same Internal DTD subset; or
   2. Create a custom Internal DTD subset through XSLT

Also, is there a reason you can't use an External DTD subset?

-Wayne Steele

>From: Kevin Burges <xmldude@burieddreams.com>
>Reply-To: Kevin Burges <xmldude@burieddreams.com>
>To: xml-dev <xml-dev@lists.xml.org>
>Subject: Inline DTD in stylesheet
>Date: Thu, 12 Apr 2001 12:36:47 +0100
>I'm trying to create a simple stylesheet which extracts a list
>of field names and field references from an XML document. I want the
>XSL output to have an inline DTD so I can validate the output (using
>How can I get the stylesheet to generate the DTD in the output file?
>I've tried just including the DTD within the root stylesheet template,
>but no joy.
>Surely this must be a trivial problem?
>May the flares be with you,
>  Kevin                   mailto:xmldude@burieddreams.com
>+++++++++++++ Cool music - http://mp3.com/marshan
>++ Attitude Rock Webzine - http://burieddreams.com/attitude

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