Subject: Re: [xsl] Adding Missing Elements From: "Joris Gillis" <roac@xxxxxxxxxx> Date: Tue, 13 Sep 2005 21:48:36 +0200 |
"The best way to do this" does not exist, it's amatter of personal personal style.Given these possible variations and based on some processing elsewhere in the document, I would like to make sure that <node> always has <item>B</item> in its list...possibly adding <list> if it is not present.
Can someone give me some suggestions on the best way to do this?
<xsl:stylesheet version="1.0" xmlns:var="myvar" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="xml" indent="yes"/>
<var:root> <list> <item>B</item> </list> </var:root>
<xsl:template match="node()|@*"> <xsl:copy> <xsl:apply-templates select="node()|@*" /> </xsl:copy> </xsl:template>
<xsl:template match="node|list|item"> <xsl:element name="{local-name()}"> <xsl:apply-templates select="@*" /> <xsl:apply-templates select=" $var[current()/self::node][not(current()/list)]| $var[current()/self::list]/item[not(.=current()/item)]| node()" /> </xsl:element> </xsl:template>
regards, -- Joris Gillis (http://users.telenet.be/root-jg/me.html) B+Et ipsa scientia potestas estB; - Francis Bacon , Meditationes sacrae
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