Subject: Re: [xsl] flattening an xml hierarchy From: "Vasu Chakkera" <vasucv@xxxxxxxxx> Date: Thu, 8 Jan 2009 16:18:49 +0000 |
USe identity transformation. <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="node()|@*"/> </xsl:copy> </xsl:template> <xsl:template match = "Dontwant1"> <xsl:apply-templates select="node()|@*"/> </xsl:template> <xsl:template match = "dontwant2"/> </xsl:stylesheet> This will do your job. Although.. You have 2 definitions of what you want to Remove. 1. Node and its children 2. Node . I am not sure what you meant by "remove" But anyway, in either case, if you want to remove node and all its children, you just define an empty template if you want to ignore the node alone, then you define the template for that node and not bothor about copying the current node, and copy all its children. HTH 2009/1/8 Tim <timlhunt@xxxxxxxxxxxxxx>: > Hi, > I'd like to remove some elements from an xml hierachy using xslt: > > Original xml example: > <Dontwant1> > <dontwant2>foo</dontwant2> > <x3> > <x4>stuff</x4> > <y5>more stuff</y5> > </x3> > </Dontwant1> > > Like to have transform output: > <x3> > <x4>stuff</x4> > <y5>more stuff</y5> > </x3> > > Thanks for your insghts. > > Tim > > -- Vasu Chakkera Numerical Algorithms Group Ltd. Oxford www.vasucv.com
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] flattening an xml hierarc, Vasu Chakkera | Thread | Re: [xsl] flattening an xml hierarc, Evan Lenz |
Re: [xsl] flattening an xml hierarc, Martin Honnen | Date | Re: [xsl] flattening an xml hierarc, Tim |
Month |