Thanks Bob, That is exactly what I needed, just didn't know how to go about it... :)
Laurie. From: Bob Stayton [mailto:[email protected]] Sent: 2009-08-05 00:51 To: Laurie Burley; [email protected] Subject: Re: [docbook-apps] breadcrumbs root node You could change your select statement to require a parent on each element: <xsl:for-each select="$this.node/ancestor::*[parent::*]"> Since the root node does not have a parent, it would not be selected. I have not tested it, but I think it would work. Bob Stayton Sagehill Enterprises [email protected]<mailto:[email protected]> ----- Original Message ----- From: Laurie Burley<mailto:[email protected]> To: [email protected]<mailto:[email protected]> Sent: Monday, August 03, 2009 1:21 AM Subject: [docbook-apps] breadcrumbs root node Hi, Has anyone figured out a way to remove the root node from breadcrumbs in their html output? I am using the following code to generate my breadcrumbs: <xsl:template name="breadcrumbs"> <xsl:param name="this.node" select="."/> <table class="headertable"> <tr> <td width="95%"> <div class="breadcrumbs"> <xsl:for-each select="$this.node/ancestor::*"> <span class="breadcrumb-link"> <a> <xsl:attribute name="href"> <xsl:call-template name="href.target"> <xsl:with-param name="object" select="."/> <xsl:with-param name="context" select="$this.node"/> </xsl:call-template> </xsl:attribute> <xsl:apply-templates select="." mode="title.markup"/> </a> </span> <xsl:text> > </xsl:text> </xsl:for-each> <!-- And display the current node, but not as a link --> <span class="breadcrumb-node"> <xsl:apply-templates select="$this.node" mode="title.markup"/> </span> </div></td> <!--add a logo to the top right corner--> <xsl:call-template name="logo"/> </tr> </table> </xsl:template> Any help would be greatly appreciated, Kind Regards, Laurie Burley
