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RE: DTD
- From: "Michael Kay" <michael dot h dot kay at ntlworld dot com>
- To: <xsl-list at lists dot mulberrytech dot com>
- Date: Wed, 1 May 2002 08:55:29 +0100
- Subject: RE: [xsl] DTD
- Reply-to: xsl-list at lists dot mulberrytech dot com
XSLT doesn't generate a DTD for the result tree. It would serve no useful
purpose.
When you do a transformation, you typically have in mind the structure of
the XML you want to generate. If you want to validate that the result
document actually conforms to this structure, you can add a DOCTYPE
declaration to the output file using <xsl:output>. The responsibility of
producing output that is actually valid according to the target DTD falls
entirely on the stylesheet author.
Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com
> -----Original Message-----
> From: owner-xsl-list@lists.mulberrytech.com
> [mailto:owner-xsl-list@lists.mulberrytech.com]On Behalf Of tinku
> Sent: 01 May 2002 07:40
> To: XSL-List@lists.mulberrytech.com
> Subject: [xsl] DTD
>
>
> Hi,
>
> XSLT is used to transform an input XML tree into a result XML
> tree. Input XML tree has a DTD on which it is validated. But when
> we transform an input XML tree, XSLT produces a new XML tree which
> is completely different from the input tree. I wonder, how XSLT
> generates a new DTD for that result tree?.
>
> Thanks.
> Tinku.
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