Try  "count(parent::para/preceding-sibling::para)+1"
  This will give your parent para's position with respect to the other para
  nodes in the sibling list.

> -------Original Message-------
>
> From: Philip Vallone
> Date: 17/01/2009 13:42:43
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject:  Get position of parent
>
> Hello List,
>
> What is the best way to get the position of a parent node? In the below xml,
> assume my context node is para:
>
> /table/tgroup/tbody/row/entry[1]/para
>
> If my context node is para, how do I get the position of its parent entry?
>
> <table frame="all" align="center" id="C-TABLE3" width="90%">
>   <title>Title</title>
>   <tgroup cols="3">
>    <colspec colnum="1" colname="spycolgen1" colwidth="*"/>
>    <colspec colnum="2" colname="spycolgen2" colwidth="*"/>
>    <colspec colnum="3" colname="spycolgen3" colwidth="*"/>
>    <tbody>
>     <row>
>      <entry>
>       <!--get position of parent::entry-->
>       <para id="table3-para 1">context
> node</para>
>      </entry>
>      <entry>
>       <para>test</para>
>      </entry>
>      <entry>
>       <para>test</para>
>      </entry>
>     </row>
>    </tbody>
>   </tgroup>
> </table>
>
>
>
>
> Thanks
> Phil
>



-- 
Vasu Chakkera
Numerical Algorithms Group Ltd.
Oxford
www.vasucv.com