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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Transforming XML to XML
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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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