Subject: [xsl] Re: Template for node-set parents From: Darren Oh <darren@xxxxxxx> Date: Mon, 5 Nov 2012 11:07:33 -0500 |
Thanks for all the suggestions. I succeeded by using an EXSL extension function. I am writing code for PHP, so I have to use what PHP supports. Here is my toy code: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:exsl="http://exslt.org/common" extension-element-prefixes="exsl" version="1.0"> <xsl:output method="xml"/> <xsl:strip-space elements="*"/> <xsl:variable name="sorted"> <xsl:for-each select="/response/data/result"> <xsl:sort select="COMPANY_SORT" data-type="text" order="descending"/> <xsl:copy-of select="."/> </xsl:for-each> </xsl:variable> <xsl:template match="*"> <xsl:copy> <xsl:apply-templates/> </xsl:copy> </xsl:template> <xsl:template match="/response/data/result"> <xsl:variable name="position" select="position()"/> <xsl:copy-of select="exsl:node-set($sorted)/*[$position]"/> </xsl:template> </xsl:stylesheet> On Oct 19, 2012, at 1:03 PM, Darren Oh wrote: > I am trying to generate a stylesheet that copies an XML source document. The only change should be that nodes selected by an XPath expression are sorted. I want this to work for any XML source document. The only information available to generate the stylesheet is the XPath expression and the sort criteria. I think this requires creating a template for the parents of the nodes selected by the XPath expression. How can I do this?
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