Re: [xsl] Generating a list of items NOT present in source XML

Subject: Re: [xsl] Generating a list of items NOT present in source XML
From: Greg Faron <gfaron@xxxxxxxxxxxxxxxxxx>
Date: Fri, 17 May 2002 16:27:49 -0600
At 11:57 AM 5/17/2002, you wrote:
<xsl:with-param name="count" select="count + 1" />

[ Just for those following this thread (and later in the archives), the above line (and others like it) should probably read as <xsl:with-param name="count" select="$count + 1" /> Note the dollar-sign before 'count'. Without it, you will have an infinite recursion. ]

Jeni,

I modified your code to the following stylesheet. The first template matches / and calls the other template with a sorted node-set. I have a question at the end of the sheet.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"; xmlns:msxsl="urn:schemas-microsoft-com:xslt" extension-element-prefixes="msxsl">
<xsl:output method="html" indent="yes"/>
<xsl:template match="/">
<missing>
<xsl:variable name="sorted-obj-rtf">
<xsl:for-each select="root/obj">
<xsl:sort data-type="number" select="childnode"/>
<xsl:copy-of select="."/>
</xsl:for-each>
</xsl:variable>
<xsl:call-template name="show-missing">
<xsl:with-param name="nodes" select="msxsl:node-set($sorted-obj-rtf)/obj"/>
<xsl:with-param name="start-number" select="1"/>
<xsl:with-param name="end-number" select="9"/>
<xsl:with-param name="index" select="'./childnode'"/>
</xsl:call-template>
</missing>
</xsl:template>
<!--


Named Template "show-missing"

  -->
  <xsl:template name="show-missing">
    <xsl:param name="nodes" select="/.."/>
    <xsl:param name="start-number" select="1"/>
    <xsl:param name="end-number" select="1"/>
    <xsl:param name="index"/>
    <xsl:if test="$nodes and $index and ($end-number >= $start-number)">
      <xsl:choose>
        <!-- For each of these tests, instead of comparing against
             $nodes[1], I would like to compare against
             $nodes[1]/($index converted to XPath path)/. -->
        <xsl:when test="$start-number > $nodes[1]">
          <xsl:call-template name="show-missing">
            <xsl:with-param name="nodes" select="$nodes[position() > 1]"/>
            <xsl:with-param name="start-number" select="$start-number"/>
            <xsl:with-param name="end-number" select="$end-number"/>
            <xsl:with-param name="index" select="$index"/>
          </xsl:call-template>
        </xsl:when>
        <xsl:when test="$start-number = $nodes[1]">
          <xsl:call-template name="show-missing">
            <xsl:with-param name="nodes" select="$nodes[position() > 1]"/>
            <xsl:with-param name="start-number" select="$start-number + 1"/>
            <xsl:with-param name="end-number" select="$end-number"/>
            <xsl:with-param name="index" select="$index"/>
          </xsl:call-template>
        </xsl:when>
        <xsl:otherwise>
          <obj>
            <childnode>
              <xsl:value-of select="$start-number"/>
            </childnode>
          </obj>
          <xsl:call-template name="show-missing">
            <xsl:with-param name="nodes" select="$nodes"/>
            <xsl:with-param name="start-number" select="$start-number + 1"/>
            <xsl:with-param name="end-number" select="$end-number"/>
            <xsl:with-param name="index" select="$index"/>
          </xsl:call-template>
        </xsl:otherwise>
      </xsl:choose>
    </xsl:if>
  </xsl:template>
</xsl:stylesheet>

The above works for the simple XML document given in the previous post because the string() of the obj node is the same as the string() of the only contained obj/childnode node. What is the best way (or is there a way) to allow $index to be converted from a string to the XPath that the string represents?


Greg Faron Integre Technical Publishing Co.



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