Subject: [xsl] Node Position & Relationship! From: "Adam J Knight" <adam@xxxxxxxxxxxxxxxxx> Date: Sat, 12 Feb 2005 00:08:36 +1000 |
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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