[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Identify last node in nested nodeset with same nam
Mat,
Luckily, there's an easier and more "natural" way to do this -- don't write "tags", as you do, but create full-fledged elements. Using this, with recursion, yields a simple approach. With this input: <data> <menu name="link1"/> <menu name="link2"> <menu name="link2a"/> <menu name="link2b"/> </menu> </data> This transform, using call-template: <?xml version="1.0" encoding="iso8859-1"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes"/> <xsl:strip-space elements="*"/> <xsl:template match="/data"> <xsl:call-template name="write-menu"> <xsl:with-param name="items" select="menu"/> </xsl:call-template> </xsl:template> <xsl:template name="write-menu"> <xsl:param name="items" select="/.."/> <ul> <xsl:for-each select="$items"> <li> <xsl:value-of select="@name"/> <xsl:if test="menu"> <xsl:call-template name="write-menu"> <xsl:with-param name="items" select="menu"/> </xsl:call-template> </xsl:if> </li> </xsl:for-each> </ul> </xsl:template> </xsl:stylesheet> Produces: <?xml version="1.0" encoding="UTF-8"?> <ul> <li>link1</li> <li>link2<ul> <li>link2a</li> <li>link2b</li> </ul> </li> </ul> As an aside, if iterating over a set of nodes, last() in a test can identify the last node. Regards, --A From: Mat Bergman <matbergman@xxxxxxxxx> Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Identify last node in nested nodeset with same name Date: Thu, 23 Jun 2005 19:26:19 -0700 (PDT) _________________________________________________________________ Dont just search. Find. Check out the new MSN Search! http://search.msn.click-url.com/go/onm00200636ave/direct/01/
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