Subject: Re: Get position of parent
From: Mukul Gandhi <gandhi.mukul@xxxxxxxxx>
Date: Sat, 17 Jan 2009 20:06:23 +0530
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I think, it could be something like,
count(../preceding-sibling::entry) + 1
On Sat, Jan 17, 2009 at 7:12 PM, Philip Vallone
<philip.vallone@xxxxxxxxxxx> wrote:
> 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
--
Regards,
Mukul Gandhi
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