[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] 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
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