Subject: RE: [xsl] Finding position of a node relative to the root instead of the parent node From: bryan.s.schnabel@xxxxxxxxxxxxxx Date: Mon, 22 Dec 2003 15:00:05 -0800 |
Perhaps a more straight forward way to get the position relative to the nr element might be to use the descendant axis. Just changing your code a bit. . . So this: descendant::movie[position() <= $thisMany] instead of this: //movie[position() <=number($thisMany)] and this: descendant::movie[position() >number($thisMany)] instead of this: //*[name()='movie'][position() >number($thisMany)] gets you something like this: <tr> <td> <br> <b><i>Charlie's Angles: Full Throttle</i></b> <br> <b><i>28 Days Later</i></b> <br> <b><i>The Santa Clause 2</i></b> </td> <td> <br> <b><i>Northfork</i></b> <br> <b><i>Rudy: The Rudy Giuliani Story</i></b> <br> <b><i>Russian Ark</i></b> </td> </tr> -----Original Message----- From: Cynthia DeLaria [mailto:cdelaria@xxxxxxxxx] Sent: Monday, December 22, 2003 1:12 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Cc: btusdin@xxxxxxxxxxxxxxxx Subject: [xsl] 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
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