[xsl] how to get new position() of a node in a sorted result tree?

Subject: [xsl] how to get new position() of a node in a sorted result tree?
From: David Li <davidli@xxxxxxxxxxxx>
Date: Wed, 01 Aug 2001 17:33:40 -0400
Hi,
    Anyone knows how to get the new node position() of a sorted result tree?
    Specifically, I have the following XSLT code:
        <xsl:template match="/">
                <xsl:apply-templates select="//item">
                        <xsl:sort data-type="number" order="descending" select="@date" />
                </xsl:apply-templates>
                </font>
        </xsl:template>

        <xsl:template match="article">
                <!-- process only the first 100 -->
                <xsl:if test="position() &lt; 100">
                ......
                </xsl:if>
        </xsl:template>

    When I call the position() function, it returns the position ID in the original
tree not the position ID in the new sorted tree. How to get the new position in a
sorted tree?
    Thanks,

David Li


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