Thanks for the answer.
Does XSLT processor modifies the DTD's present in the original
input file?
How does the XSLT processor restores the default values present in
the DTD in the input file?
-------
Tinku
On Wed, 01 May 2002 Michael Kay wrote :
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@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx
> -----Original Message-----
> From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of
tinku
> Sent: 01 May 2002 07:40
> To: XSL-List@xxxxxxxxxxxxxxxxxxxxxx
> Subject: 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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| Current Thread |
- DTD
- tinku - Wed, 1 May 2002 02:42:47 -0400 (EDT)
- <Possible follow-ups>
- tinku - Wed, 1 May 2002 04:17:36 -0400 (EDT) <=
- tinku - Wed, 1 May 2002 05:31:39 -0400 (EDT)
- Dion Houston - Wed, 1 May 2002 05:53:39 -0400 (EDT)
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