RE: [xsl] Node Position & Relationship!

Subject: RE: [xsl] Node Position & Relationship!
From: "Andrew Welch" <ajwelch@xxxxxxxxxxxxxxx>
Date: Fri, 11 Feb 2005 14:41:41 -0000
> 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.

You need to read about the "xpath axis'" (plural)

To test if a node is the child of a tree node:

  <xsl:if test="parent::tree">

To test if a tree node is the top-most tree node (dependant on

  <xsl:if test="not(parent::tree)">

Or, (again depending on the structure of your xml):

  <xsl:if test="not(ancestor::tree)">

To test if a node has siblings:

  <xsl:if test="preceding-sibling::tree or following-sibling::tree">


  <xsl:if test="../tree">

There are many, many ways.  If you are certain of the structure of your
xml you could even do it at the template match level:

  <xsl:template match="/tree">
    (level 0 tree)

  <xsl:template match="/tree/tree">
    (level 1 tree)

...and so on.


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