Subject: Re: [xsl] how to get new position() of a node in a sorted result tree? From: Wendell Piez <wapiez@xxxxxxxxxxxxxxxx> Date: Wed, 01 Aug 2001 18:08:52 -0400 |
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() < 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,
====================================================================== 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 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ======================================================================
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