Re: DTD question

[Wendell Piez <wapiez at mulberrytech dot com>]

Chris,

If you want your serialized output to have a form something like the
following, which provides a validating parser with the hook to get the DTD
(and can do that for your human reader also):

<?xml version="1.0"?>
<!DOCTYPE B SYSTEM "b.dtd" >
<B>
...B or not B, either way...
</B>

Then you want your XSL to contain an <xsl:output> element at the top level,
like so:

<xsl:output method="xml" doctype-system="b.dtd"/>

This is described in the XSLT spec, section 16.

But XSLT does not provide a way to put the DTD in line, into an internal
subset of the DOCTYPE declaration, if that's what you were thinking of. If
you have something else in mind entirely, please clarify.

Good luck!
Wendell

At 06:57 PM 8/21/00 -0400, you wrote:
>I am transforming an XML file to another XML File (call it file B) in
order with a DTD that my company came up with.
>
>My question is this:  How do I put a link of the DTD in file B so an
outside part can see the structure?  Is there a way to do an include in the
XSL file that is doing the transformation?


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Wendell Piez                            mailto:wapiez@mulberrytech.com
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