Re: [xsl] Get position of parent

[Mukul Gandhi <gandhi.mukul@xxxxxxxxx>]

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