Subject: RE: [xsl] Ordering nodes by number of children From: "Andreas L. Delmelle" <a_l.delmelle@xxxxxxxxxx> Date: Sun, 22 Feb 2004 15:20:24 +0100 |
> -----Original Message----- > From: Smilen Dimitrov > > I'm a newbie with XSL, and after long searching, and mainly > thanks to the information in the list, I finally managed to solve > the problem. Consider the following structure: > Hi, This bit: > > <xsl:for-each select="./*"> > > <xsl:sort select="count(child::*)" data-type="number" > order="descending"/> > > <xsl:call-template name="ArrangeDirectory"> > <xsl:with-param name="CurrentNode" select="."/> > </xsl:call-template> > > </xsl:for-each> > would, with the given input, actually have exactly the same output as: <xsl:apply-templates> <xsl:sort select="count(*)" data-type="number" order="descending" /> </xsl:apply-templates> So the complete solution revised: <?xml version="1.0" encoding="UTF-8" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" version="1.0" indent="yes" /> <xsl:template match="/"> <xsl:apply-templates /> </xsl:template> <xsl:template match="node"> <xsl:copy> <xsl:copy-of select="@*" /> <xsl:apply-templates> <xsl:sort select="count(*)" data-type="number" order="descending" /> </xsl:apply-templates> </xsl:copy> </xsl:template> </xsl:stylesheet> Cheers, Andreas XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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