I think union (|) operator is the answer..

//tree_node[@id=9]/ancestor::tree_node |
//tree_node[@id=9]/preceding-sibling::* |
//tree_node[@id=9] |
//tree_node[@id=9]/following-sibling::*

Probably, this should also work (seems more efficient
also) -
//tree_node[@id=9]/(ancestor::tree_node |
preceding-sibling::* | self::* | following-sibling::*)

Regards,
Mukul

--- Adam J Knight <adam@xxxxxxxxxxxxxxxxx> wrote:

> Hi guys,
> 
> Another chapter in this tragedy.
> 
> Question: 
> 
> The expression
> '//tree_node[@id=9]/ancestor::tree_node' returns a
> nodeset
> containing node with id(7). I have verified this
> with xpath visualizer.
> 
> How do I return the ALL siblings of this node also.
> 
> The result being
> Node(id=7)
> Node(id=8) : sibling of Node(id=7)
> Node(id=9) : sibling of Node(id=7)
> Node(id=14) : sibling of Node(id=7)
> 
> HELP APPRECIATED!!!!
> 
> ALL TO0 FAMILIAR XML STRUCTURE:
> <tree>
>   <tree_node id="7" value="Test Level One A">
>     <tree_node id="8" value="Test Level Two A"/>
>     <tree_node id="9" value="Test Level Two B">
>       <tree_node id="11" value="Test Level Three
> B"/>
>       <tree_node id="10" value="Test Level Three A">
>         <tree_node id="12" value="Test Level Four
> A"/>
>         <tree_node id="13" value="Test Level Four
> B"/>
>       </tree_node>
>     </tree_node>
>     <tree_node id="14" value="Test Level Two C"/>
>   </tree_node>
> </tree>




		
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