Re: how to get new position() of a node in a sorted re

Good luck,
Wendell

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,

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Wendell Piez                            mailto:wapiez@xxxxxxxxxxxxxxxx
Mulberry Technologies, Inc.                http://www.mulberrytech.com
17 West Jefferson Street                    Direct Phone: 301/315-9635
Suite 207                                          Phone: 301/315-9631
Rockville, MD  20850                                 Fax: 301/315-8285
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