Subject: Re: [xsl] empty element question From: "G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx> Date: Fri, 16 Oct 2009 16:15:20 -0400 |
How do I transform this input document:
<div> <p>string 1<lb/>string 2</p> </div>
to this output document?
<div> <p>string 1</p> <p>string 2</p> </div>
T:\ftemp>type morgan.xml <div> <p>string 1<lb/>string 2</p> </div>
T:\ftemp>xslt2 morgan.xml morgan.xsl <?xml version="1.0" encoding="UTF-8"?><div> <p>string 1</p><p>string 2</p> </div> T:\ftemp>type morgan.xsl <?xml version="1.0" encoding="US-ASCII"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:template match="*[lb]"> <!--remember where we are--> <xsl:variable name="this" select="."/> <!--determine all of the groups of nodes--> <xsl:for-each-group select="node()" group-starting-with="lb"> <!--return to the node being copied--> <xsl:for-each select="$this"> <xsl:copy> <xsl:apply-templates select="@*"/> <!--but only copy that content not including the break--> <xsl:apply-templates select="current-group()[not(self::lb)]"/> </xsl:copy> </xsl:for-each> </xsl:for-each-group> </xsl:template>
<xsl:template match="@*|node()"><!--identity for all other nodes--> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template>
-- Upcoming: hands-on code list, UBL, XSLT, XQuery and XSL-FO classes in Copenhagen Denmark and Washington DC USA, October/November 2009 Interested in other classes? http://www.CraneSoftwrights.com/s/i/ Crane Softwrights Ltd. http://www.CraneSoftwrights.com/s/ Training tools: Comprehensive interactive XSLT/XPath 1.0/2.0 video Video lesson: http://www.youtube.com/watch?v=PrNjJCh7Ppg&fmt=18 Video overview: http://www.youtube.com/watch?v=VTiodiij6gE&fmt=18 G. Ken Holman mailto:gkholman@xxxxxxxxxxxxxxxxxxxx Male Cancer Awareness Nov'07 http://www.CraneSoftwrights.com/s/bc Legal business disclaimers: http://www.CraneSoftwrights.com/legal
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