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I created related link and display by use XSLT like:
<ul class="footer-menu1" > <xsl:for-each select="$currentPage/ancestor-or-self::* [@isDoc][@level=1]/* [name() = 'navigationRelatedlink' and not(@isDoc)]/links/link"> <li> <xsl:element name="a"> <xsl:if test="./@newwindow = '1'"> <xsl:attribute name="target">_blank</xsl:attribute> </xsl:if> <xsl:choose> <xsl:when test="./@type = 'external'"> <xsl:attribute name="href"> <xsl:value-of select="./@link"/> </xsl:attribute> </xsl:when> <xsl:when test="./@type = 'media'"> <xsl:attribute name="href"> <xsl:value-of select="umbraco.library:GetMedia(./@link, 0)/umbracoFile" /> </xsl:attribute> </xsl:when> <xsl:otherwise> <xsl:attribute name="href"> <xsl:value-of select="umbraco.library:NiceUrl(./@link)"/> </xsl:attribute> </xsl:otherwise> </xsl:choose> <xsl:value-of select="./@title"/> </xsl:element> <!-- Child node of relate link -->
<!-- Child node of relate link --> </li> </xsl:for-each> </ul>
Is there anyway to show that child node of each related link?
Please help me .
Hi,
I'm assuming you'll only show child nodes if the related links is of type "internal" - correct? In that case you can use GetXmlNodeById to get the selected node, and then loop through it's children:
<xsl:if test="@type = 'internal'"> <xsl:variable name="selectedNodeChildren" select="umbraco.library:GetXmlNodeById(@link)/* [@isDoc]" /> <xsl:if test="$selectedNodeChildren"> <ul> <xsl:for-each select="$selectedNodeChildren"> <li> <a href="{umbraco.library:NiceUrl(@id)}"> <xsl:value-of select="@nodeName"/> </a> </li> </xsl:for-each> </ul> </xsl:if></xsl:if>
-Tom
wonderfullllll!!! I can display a child node.
I stuck in this problem for 3 days .
Many thanks Tom :)
is working on a reply...
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Secondary node of Related Link
I created related link and display by use XSLT like:
<ul class="footer-menu1" >
<xsl:for-each select="$currentPage/ancestor-or-self::* [@isDoc][@level=1]/* [name() = 'navigationRelatedlink' and not(@isDoc)]/links/link">
<li>
<xsl:element name="a">
<xsl:if test="./@newwindow = '1'">
<xsl:attribute name="target">_blank</xsl:attribute>
</xsl:if>
<xsl:choose>
<xsl:when test="./@type = 'external'">
<xsl:attribute name="href">
<xsl:value-of select="./@link"/>
</xsl:attribute>
</xsl:when>
<xsl:when test="./@type = 'media'">
<xsl:attribute name="href">
<xsl:value-of select="umbraco.library:GetMedia(./@link, 0)/umbracoFile" />
</xsl:attribute>
</xsl:when>
<xsl:otherwise>
<xsl:attribute name="href">
<xsl:value-of select="umbraco.library:NiceUrl(./@link)"/>
</xsl:attribute>
</xsl:otherwise>
</xsl:choose>
<xsl:value-of select="./@title"/>
</xsl:element>
<!-- Child node of relate link -->
#####I want to display all child nodes of each related link ######
<!-- Child node of relate link -->
</li>
</xsl:for-each>
</ul>
Is there anyway to show that child node of each related link?
Please help me .
Hi,
I'm assuming you'll only show child nodes if the related links is of type "internal" - correct? In that case you can use GetXmlNodeById to get the selected node, and then loop through it's children:
-Tom
wonderfullllll!!! I can display a child node.
I stuck in this problem for 3 days .
Many thanks Tom :)
is working on a reply...