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Re: Transforming XML to XML
Matt,
thanks for your hint. It worked! :)
Doing some tests I found out that Xalan also takes the first child element of
the root element to find out about the document type. So for example if the
first child element is a <html> it automatically omits the "<?xml ... ?>" XML
declaration. This is what confused me a little !
Greetings
Zeljko Rajic
Matt Gushee wrote:
>
> 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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list