[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] how to get new position() of a node in a sorted result
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
|
PURCHASE STYLUS STUDIO ONLINE TODAY!Purchasing Stylus Studio from our online shop is Easy, Secure and Value Priced! Download The World's Best XML IDE!Accelerate XML development with our award-winning XML IDE - Download a free trial today! Subscribe in XML format
|