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Hi Mark,
If by "DTD reference" you mean a DOCTYPE declaration's SYSTEM and/or PUBLIC identifiers, the short answer is (sadly) "no".
You can specify a string value for these identifiers in the doctype-system and doctype-public attributes on xsl:output, but they have to be string literals, they can't be copied from the input. (The parser doesn't preserve these values for the processor in any case, so there's nowhere to copy them from.)
Workarounds include using literal values (which must be known to the stylesheet in advance) and post-processing.
At 01:14 PM 8/13/2003, you wrote:
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Re: [xsl] How to "copy" a DTD reference?
Subject: Re: [xsl] How to "copy" a DTD reference? From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> Date: Wed, 13 Aug 2003 14:11:59 -0400 |
Hi Mark,
If by "DTD reference" you mean a DOCTYPE declaration's SYSTEM and/or PUBLIC identifiers, the short answer is (sadly) "no".
You can specify a string value for these identifiers in the doctype-system and doctype-public attributes on xsl:output, but they have to be string literals, they can't be copied from the input. (The parser doesn't preserve these values for the processor in any case, so there's nowhere to copy them from.)
Workarounds include using literal values (which must be known to the stylesheet in advance) and post-processing.
Cheers, Wendell
At 01:14 PM 8/13/2003, you wrote:
Is there a way to copy a DTD reference in the original source document using XSL?
====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ======================================================================
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