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Transforming XML to XML

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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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