Re: [xsl] Node Position & Relationship!

[Mukul Gandhi <mukul_gandhi@xxxxxxxxx>]

Hi Adam,
  I'll try to answer your question in parts..

q1) to create an xsl if element that tests the current
node to see if it is a level 0 tree_node element.
Ans -
<xsl:if test="self::tree_node and
not(parent::tree_node)">

</xsl:if>
If you are at a given node, this if condition
evaluates true if the node is at level 0(i.e., if it
is a tree_node element, and it has no tree_node
parent)

In the XML you gave, this condition will be true for
<tree_node id="7" value="Test"> node..

q2) how can I find out if a particular node is a
child, ancestor or peer of any given node.
Ans - Suppose node X is the context node, and there is
another node Y ;
X is child of Y, if this condition is true
<xsl:if test="parent::Y">

</xsl:if>

Y is an ancestor of X, if this condition is true
<xsl:if test="ancestor::Y">

</xsl:if>

Y is peer of X(following or preceding sibling), if
this condition is true
<xsl:if test="(following-sibling::Y) or
(preceding-sibling::Y)">

</xsl:if>

Regards,
Mukul
  
--- 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."



		
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