Subject: RE: [xsl] Generating namespace declarations in output documents: How? From: Jarno.Elovirta@xxxxxxxxx Date: Thu, 6 Sep 2001 06:10:01 +0300 |
> So I've got this document more or less of the form > > <foo> > ...some elements... > </foo> > > and I need to transform it into > > <bar xmlns:xsi="http://www.w3.org/2000/10/XMLSchema-instance" > xsi:noNamespaceSchemaLocation="[somewhere]"> > ...some elements... > </bar> > > (yes, the transformation from foo to bar is pretty trivial at > the moment, but it's expected to get nastier later). > > Of course the first thing I tried (since this is the first time > I've used XSL) is exactly what they tell you not to do in > section 7.1.3 of the XSLT spec -- > > <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > version="1.0"> > <xsl:template match="foo"> > <xsl:element name="bar"> > <xsl:attribute name="xmlns:xsi"> > http://www.w3.org/2000/10/XMLSchema-instance > </xsl:attribute> > <xsl:attribute name="xsi:noNamespaceSchemaLocation"> > [somewhere] > </xsl:attribute> > <xsl:apply-templates/> > </xsl:template> > ...some templates... > </xsl:transform> > > -- you know, the part where it says "Thus, while it is not an error > to do [what I just did], it will not result in a namespace declaration > being output." > > So what *will* result in a namespace declaration being output? Let the XSLT engine worry about outputing the namespace declarations - just do <xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xsi="http://www.w3.org/2000/10/XMLSchema-instance" version="1.0"> <xsl:template match="foo"> <xsl:element name="bar"> <xsl:attribute name="xsi:noNamespaceSchemaLocation"> [somewhere] </xsl:attribute> </xsl:element> <xsl:apply-templates/> </xsl:template> </xsl:transform> Jarno XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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