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(_at_)brightidea(_dot_)com(_dot_)au> 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."
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