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Transforming XML to XML
- To: xsl-list at mulberrytech dot com
- Subject: Transforming XML to XML
- From: Matt Gushee <mgushee at havenrock dot com>
- Date: Mon, 6 Nov 2000 03:13:10 -0700 (MST)
- References: <3A12B6D8.43411332@brokat.com>
- Reply-To: xsl-list at mulberrytech dot com
Zeljko Rajic writes:
> I'm a newbie to XML/XSL so maybe my question may look trivial:
Well, the solution is simple, but not at all obvious.
> How do I transform a XML document into another XML document? My problem actually
> is to write a XSL transformation rule that creates the XML docuement
> description: <?xml version="1.0" encoding="UTF-8"?>
>
> I've tried to following XSL syle sheet:
>
> <?xml version="1.0" encoding="UTF-8"?>
> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
> xmlns:fo="http://www.w3.org/1999/XSL/Format">
>
> <xsl:template match="/">
> <xsl:processing-instruction name="xml">version="1.0"
> encoding="UTF-8"</xsl:processing-instruction>
Ah, an easy mistake to make! The XML declaration looks and smells like
a processing instruction, but it is not a processing instruction.
> </xsl:template>
>
> </xsl:stylesheet>
Instead, try this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fo="http://www.w3.org/1999/XSL/Format">
==> <xsl:output method="xml" encoding="UTF-8"?/> <==
<xsl:template match="/">
....
Don't forget to put something in that first template (like
'<xsl:apply-templates/>') to kick off further processing.
Best of luck.
Matt Gushee
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