Subject: AW: [xsl] how to get position of node in node-set From: "Kloeck, Erwin" <Erwin.Kloeck@xxxxxx> Date: Thu, 11 Sep 2003 18:55:11 +0200 |
Thanks again David, I went for the xsl solution and it works beautifully. Erwin -----Ursprüngliche Nachricht----- Von: David Carlisle [mailto:davidc@xxxxxxxxx] Gesendet: Donnerstag, 11. September 2003 18:33 An: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Betreff: Re: [xsl] how to get position of node in node-set Node sets are sets and so don't have an intrinsic order. So nodes don't really have a position in a set (you can't for example sort the nodes and save the sorted list in a set) You can ask what is the position if the nodes in the set were sorted into document order, although actually it's not so easy. If the nodes are siblings and you are on <entry text="ccc"/> then you want count(preceding-sibling::entry)+1 however you only want to count nodes that are in $legend so that would be count(preceding-sibling::entry[@text=$legend])+1 if your text attribute uniquely determines the node, or something like count(preceding-sibling::entry[count(.|$legend)=count($legend)]+1 otherwise. They are all pure xpath solutions, or you could use xsl, as in <xsl:for-each select="$legend"> <xsl:if test="@text='ccc'"><xsl:value-of select="position()"/></xsl:if> David ________________________________________________________________________ This e-mail has been scanned for all viruses by Star Internet. The service is powered by MessageLabs. For more information on a proactive anti-virus service working around the clock, around the globe, visit: http://www.star.net.uk ________________________________________________________________________ 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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