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Re: Where can I find the XSLT DTD?
- To: Yann Desnoues <Yann dot Desnoues at alcatel dot fr>
- Subject: Re: Where can I find the XSLT DTD?
- From: "Scott Boag/CAM/Lotus" <Scott_Boag at lotus dot com>
- Date: Thu, 3 Feb 2000 14:02:02 -0500
- Cc: xsl-list at mulberrytech dot com
- Reply-To: xsl-list at mulberrytech dot com
If you look in the Xalan stuff, you'll see a xsl-html40s.dtd which loosely
validates XSLT that is producing HTML. It's not perfect, but it's a help.
-scott
Yann Desnoues
<Yann.Desnoues@alcate To: xsl-list@mulberrytech.com
l.fr> cc: (bcc: Scott Boag/CAM/Lotus)
Sent by: Subject: Re: Where can I find the XSLT DTD?
owner-xsl-list@mulber
rytech.com
02/03/00 01:16 PM
Please respond to
xsl-list
"John E. Simpson" wrote:
> At 05:00 PM 2/3/2000 +0100, Yann Desnoues wrote:
> >Linda van den Brink wrote:
> >
> > > There's an appendix to the XSLT spec "DTD Fragment for XSLT
Stylesheets
> > > (Non-Normative)" at http://www.w3.org/TR/xslt#dtd
> > >
> > > I hope it's any help to you...
> >
> >Thank you but it doesn't help as it is not a complete DTD.
> >The entity result-elements is not defined.
> >Hence the DTD is not valid.
>
> Of course result-elements is not defined. It *can't* be.
>
> If you're transforming to HTML, then there's one "valid XSLT DTD" with
one
> definition of result-elements. If you're transforming to MathML, there's
a
> completely different result-elements. In fact, there are as many
> definitions of result-elements as there are possible XML vocabularies in
> the universe. Hence there can *be* no result-elements, and that's why the
> appendix to the XSLT Rec is both a fragment and non-normative.
I'll never be able to validate ANY of my XSL doc?
But may the good question is what to use as an XSL editor?
>
>
> =============================================================
> John E. Simpson
> simpson@polaris.net
> -------------------------------------------------------------
> I put contact lenses in my dog's eyes. They had little
> pictures of cats on them. Then I took one out and he ran
> around in circles. (Stephen Wright)
>
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