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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Finding position of a node relative to the root in
 Everything returns "1 down from the root" with this... Is there something missing? -----Original Message----- From: cknell@xxxxxxxxxx [mailto:cknell@xxxxxxxxxx] Sent: Monday, December 22, 2003 2:33 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: RE: Finding position of a node relative to the root instead of the parent node <?xml version="1.0" encoding="UTF-8" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html" indent="yes" encoding="UTF-8" /> <xsl:template match="/"> <html> <body> <xsl:apply-templates /> </body> </html> </xsl:template> <xsl:template match="nr/featured/movie"> <div><xsl:value-of select="@title" /> is <xsl:value-of select="count(ancestor::*) -1" /> down from the root.</div> </xsl:template> <xsl:template match="nr/also_new/movie"> <div><xsl:value-of select="@title" /> is <xsl:value-of select="count(ancestor::*) -1" /> down from the root.</div> </xsl:template> </xsl:stylesheet> -- Charles Knell cknell@xxxxxxxxxx - email -----Original Message----- From: Cynthia DeLaria <cdelaria@xxxxxxxxx> Sent: Mon, 22 Dec 2003 14:12:27 -0700 To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Cc: <btusdin@xxxxxxxxxxxxxxxx> Subject: Finding position of a node relative to the root instead of the parent node Good Day, I have searched the xsl list unsuccessfully for an answer to this question, although I'm sure something like this has been addressed. I think I'm just not sure what to search on to find it. I have the following xml snippet: <nr> <featured> <movie title="Charlie's Angles: Full Throttle">Description</movie> <movie title="28 Days Later">Description</movie> <movie title="The Santa Clause 2">Description</movie> </featured> <also_new> <movie title="Northfork" /> <movie title="Rudy: The Rudy Giuliani Story" /> <movie title="Russian Ark" /> </also_new> </nr> Basically, in the fully-flushed out version of the XML, the <featured> movies have images and full descriptions, while the <also_new> movies have only a title and rating. What I need to do is find a way to find the position of each <movie> node relative to the root, as I need to create a "print the new releases" page that lists all new releases in two columns. This is what I tried, but it gives me the position based on the parent (i.e. <also_new> or <featured>) not relative to the root. <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html" encoding="ISO-8859-1" /> <xsl:template match="/nr"> <html> <head> <title>New Releases E-Newsletter</title> <head> <body bgcolor="#ffffff" text="#000000" link="#023f7e" alink="#ff0000" vlink="#023f7e"> Just print out this list.<br /> <table width="100%"> <xsl:variable name="thisMany"><xsl:value-of select="(count(//*[name()='movie']) div 2)" /></xsl:variable> <tr> <td valign="top" width="50%" class="body2"> <br /> <xsl:for-each select="//movie[position() < number($thisMany)]"> <b><i><xsl:value-of select="./@title" /></i></b><br /> </xsl:for-each> <br /> </td> <td valign="top" width="50%" class="body2"> <br /> <xsl:for-each select="//*[name()='movie'][position() > number($thisMany)]"> <b><i><xsl:value-of select="./@title" /></i></b><br /> </xsl:for-each> <br /> </td> </tr> </table> </body> </html> </xsl:template> </xsl:stylesheet> Is it possible to get the position relative to the root node? It seems like this should be very simple, but all of the things I have tried have not worked to produce the intended outcome. Thank you! Cynthia XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list 
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