• From: Wayne Steele <xmlmaster@h...>
• To: xmldude@b..., xml-dev@l...
• Date: Fri, 13 Apr 2001 10:53:59 -0700

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

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@b...>
>Reply-To: Kevin Burges <xmldude@b...>
>To: xml-dev <xml-dev@l...>
>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
>ID/IDREF)
>
>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@b...
>
>+++++++++++++ Cool music - http://mp3.com/marshan
>++ Attitude Rock Webzine - http://burieddreams.com/attitude

_________________________________________________________________
Get your FREE download of MSN Explorer at http://explorer.msn.com

• Prev by Date: Re: "Binary XML" proposals
• Next by Date: Re: Linkbases, Topic Maps, and RDF Knowledge Bases -- help me
• Previous by thread: Inline DTD in stylesheet
• Next by thread: New Technology Area announced for XML Europe 2001
• Index(es):
  • Date
  • Thread