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I have searched a lot for a clear answer. But I think I have to ask: How do I view 8 random nodes from a variable parent?
Thank you...
Okay, now I got my XSLT to view the nodes. What I need now is so view only 8 of them and in random order.
<xsl:for-each select="umbraco.library:GetXmlNodeById(@parentID)/*"> <p> <a href="/skiferie?stay={@nodeName}"><xsl:value-of select="@nodeName" /></a></p></xsl:for-each>
This is getting wired...
<xsl:for-each select="umbraco.library:GetXmlNodeById(@parentID)/*"> <xsl:if test="position() < 8"> <p> <a href="/skiferie?stay={@nodeName}"><xsl:value-of select="stayName" /></a></p> </xsl:if> </xsl:for-each>
That doesn't return enything... If I remove the if-test, I get all the content I need (and more).
Any one who has an idea?
is working on a reply...
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Get specific number of random nodes from a variable parent
I have searched a lot for a clear answer. But I think I have to ask: How do I view 8 random nodes from a variable parent?
Thank you...
Okay, now I got my XSLT to view the nodes. What I need now is so view only 8 of them and in random order.
<xsl:for-each select="umbraco.library:GetXmlNodeById(@parentID)/*">
<p>
<a href="/skiferie?stay={@nodeName}"><xsl:value-of select="@nodeName" /></a></p>
</xsl:for-each>
This is getting wired...
<xsl:for-each select="umbraco.library:GetXmlNodeById(@parentID)/*">
<xsl:if test="position() < 8">
<p>
<a href="/skiferie?stay={@nodeName}"><xsl:value-of select="stayName" /></a></p>
</xsl:if>
</xsl:for-each>
That doesn't return enything... If I remove the if-test, I get all the content I need (and more).
Any one who has an idea?
is working on a reply...
This forum is in read-only mode while we transition to the new forum.
You can continue this topic on the new forum by tapping the "Continue discussion" link below.