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() < 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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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