correction: how to get new position() of a sorted resu

Hi,
    I apologize for the typo in my previous question. The original XSLT code
is:
        <xsl:template match="/">
                <xsl:apply-templates select="//article">
                        <xsl:sort data-type="number" order="descending" select="@date" />
                </xsl:apply-templates>
        </xsl:template>

        <xsl:template match="article">
                <xsl:if test="position() &lt; 100">
                    ...... <!-- do processing here -->
                </xsl:if>
        </xsl:template>

The original XML source file is:

<?xml version="1.0" encoding="ISO-8859-1"?>
<index section="Headlines">
   <article filename="file1.xml">
      <title>News 1 Title Text</title>
      <date>20001020</date>
      <ctprovider>Efe</ctprovider>
   </article>
   <article filename="file2.xml">
      <title>News 2 Title Text</title>
      <date>20001113</date>
      <ctprovider>Eastside Journal</ctprovider>
   </article>
   <article filename="file3.xml">
      <title>News 3 Title Text</title>
      <date>20001113</date>
      <ctprovider>Newsbytes News Network</ctprovider>
   </article>
    ............ <!-- more article elements here -->
</index>

    The question is how to get the position() from a sorted result tree. In
the above code, calling position() returns the ID of the original pre-sorted
tree.
    Thanks,


David


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