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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