Re: [xsl] Node Position & Relationship!

[António Mota <amsmota@xxxxxxxxx>]

I'm doing it like this:

<xsl:variable name="level" select="count(ancestor::Menu)"/>
(...)
<xsl:choose>
	<xsl:when test="$level=0">
		<xsl:text>1</xsl:text>   ----> this node is visible
	</xsl:when>

where Menu is your tree_node.


On Sat, 12 Feb 2005 00:08:36 +1000, Adam J Knight
<adam@xxxxxxxxxxxxxxxxx> wrote:
> Given the following xml structure, I want to create an xsl if element that
> tests the current node to see if it is a level 0 tree_node element.
> 
> <?xml version="1.0"?>
> <tree>
>   <tree_node id="7" value="Test">
>       <tree_node id="8" value="Test Sub"/>
>       <tree_node id="9" value="Test Sub One">
>           <tree_node id="10" value="Test Sub Two"/>
>     </tree_node>
>   </tree_node>
> </tree>
> 
> Here is my attached, pretty sad!!!!!
> 
> <xsl:if test="{count(self::*)=1}">
>           <xsl:apply-templates select="tree_node"/>
> </xsl:if
> 
> Can someone give me the correct way to achieve this.
> Also how can I found out if a particular node is a child, ancestor or peer
> Of any given node.
> 
> Thanks to anyone who response, muchly appreciated.
> 
> Cheers,
> Adam
>  
> NB: "Pray as if everything depended upon God and work as if everything
> depended upon man."