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My solution assumed XSLT 2.0. I think that these days, if you want to use XSLT 1.0, you should say so - it's been superseded for nearly 6 years now.
I'm afraid I don't do 1.0 solutions for people - my memories of how to get round its restrictions are fading rapidly. But there are probably other people on the list who can help.
On 01/11/2012 21:14, Darren Oh wrote:
Re: [xsl] Template for node-set parents
Subject: Re: [xsl] Template for node-set parents From: Michael Kay <mike@xxxxxxxxxxxx> Date: Thu, 01 Nov 2012 21:29:44 +0000 |
My solution assumed XSLT 2.0. I think that these days, if you want to use XSLT 1.0, you should say so - it's been superseded for nearly 6 years now.
I'm afraid I don't do 1.0 solutions for people - my memories of how to get round its restrictions are fading rapidly. But there are probably other people on the list who can help.
Michael Kay Saxonica
On 01/11/2012 21:14, Darren Oh wrote:
Thanks for the suggestion. I got stuck when trying to produce a sorted node-set. Instead of a node-set, I got a result tree fragment, for which no node-set operations are possible. Here is a simplified example to illustrate the problem. Whereas it is possible to select a node from $result, it is not possible to select a node from $sorted. Any ideas?
<?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:output method="xml"/> <xsl:template match="*"> <xsl:copy> <xsl:apply-templates/> </xsl:copy> </xsl:template> <xsl:variable name="result" select="/response/data/result"/> <xsl:variable name="sorted"> <xsl:apply-templates select="/response/data/result"> <xsl:sort select="COMPANY_SORT" data-type="text" order="descending"/> </xsl:apply-templates> </xsl:variable> <xsl:template match="/"> <xsl:copy-of select="$result[1]"/> <xsl:copy-of select="$sorted[1]"/> </xsl:template> </xsl:stylesheet>
On Oct 19, 2012, at 1:27 PM, Michael Kay wrote:
Try:
1. define a global variable $v1 that selects the result of the path expression in document order.
2. define another global variable $v2 that selects the sorted result of the path expression
3. Use a base template rule that's the identity copy
4. Add a template rule that matches nodes in $v1 (match="node()[. intersect $v1]). In this rule, determine the index position of this node in $v1 (count ($v1[. << $this]) + 1), and output the corresponding node from $v2 (copy-of select="$v2[$n]").
Michael Kay Saxonica
On 19/10/2012 18:03, 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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