Re: [xsl] node-set processing

Subject: Re: [xsl] node-set processing
From: Kamal Bhatt <kbhatt@xxxxxxxxx>
Date: Mon, 05 Dec 2005 14:49:02 +1100
G. Ken Holman wrote:

At 2005-12-05 10:20 +1100, Kamal Bhatt wrote:

For example, does a nodeset keep the position of the original source

Not even .. position() is only relative to the "current node set" as a result of the last push (<xsl:apply-templates>) or pull (<xsl:for-each>) ... not a node set variable and not the source node tree.

or will position give me the position in the node set?


Can you even run position on a node-set?


I tell my students in such a case when using XSLT 1.0 to just index into the node set using another variable that is a number, using a recursive template. Very much programming and quite unlike templating.

I hope the example below helps.

. . . . . . Ken

T:\ftemp>type kamal.xml
         <Cell>Cell 1</Cell>
         <Cell>Cell 2</Cell>
          <Cell>This is the longest cell</Cell>
          <Cell>Cell 3</Cell>

T:\ftemp>type kamal.xsl
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet xmlns:xsl="";

<xsl:output method="text"/>

<xsl:template match="/Table">
  <xsl:call-template name="find-longest">
    <xsl:with-param name="set" select=".//Cell"/>

<xsl:template name="find-longest">
<xsl:param name="set"/>
<xsl:param name="index" select="1"/>
<xsl:param name="longest" select="0"/>
<xsl:when test="$index > count($set)">
<!--finished looking-->
<xsl:value-of select="$longest"/>
<xsl:when test="string-length($set[$index])>$longest">
<!--found new longest-->
<xsl:call-template name="find-longest">
<xsl:with-param name="set" select="$set"/>
<xsl:with-param name="index" select="$index + 1"/>
<xsl:with-param name="longest" select="string-length($set[$index])"/>
<!--this isn't any longer-->
<xsl:call-template name="find-longest">
<xsl:with-param name="set" select="$set"/>
<xsl:with-param name="index" select="$index + 1"/>
<xsl:with-param name="longest" select="$longest"/>

T:\ftemp>xslt kamal.xml kamal.xsl con

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