Question

Does anyone know how to disable the current link in a breadcrumb trail.

I am working on a bspoke CMS and i have been asked to remove the current link in the breadcrumb trail.

All that i know in XSLT has failed and i have quite some time on, XSLT is no my strength and i need some helpful input.
Please help

here is the XSLT code:

[code]
    <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
    <xsl:output method="html" omit-xml-declaration="yes" />

    <!--
    Variable to determine if the current open page is displayed as a link or plain text
    select = 1 - displays plain text link
    select = 0 - displays link (default value)
      -->
    <xsl:variable name="DisableCurrentPageLink" select="0" />

    <xsl:template match="BreadCrumbTrail">
    <xsl:apply-templates select="Page" />
    </xsl:template>

    <xsl:template match="Page">
    <span class="BCTDelimiter">»</span>
    <xsl:choose>
    <xsl:when test="$DisableCurrentPageLink = 1">
    <xsl:choose>
    <xsl:when test="@IsOpenPage = 1">
    <span class="breadcrumb-link">
    <xsl:value-of select="@Title"/>
    </span>
    </xsl:when>
    <xsl:otherwise>
    <a href="{@URL}" class="breadcrumb-link" title="{@Title}">
    <xsl:value-of select="@Title" />
    </a>
    </xsl:otherwise>
    </xsl:choose>
    </xsl:when>
    <xsl:otherwise>
    <a href="{@URL}" class="breadcrumb-link" title="{@Title}">
    <xsl:value-of select="@Title" />
    </a>
    </xsl:otherwise>
    </xsl:choose>
    </xsl:template>
    </xsl:stylesheet>


[/code]

Answer 1

It’s hard without seeing the input form. That code will make an unlinked span for pages that have <page IsOpenPage="1"> as the markup, but does you CMS add that attribute for the relevant pages?

If the currently open page is always last in the trail you could change

 <xsl:when test="@IsOpenPage = 1">

to

 <xsl:when test="not(following-sibling::Page)">