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(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] 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
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Vasu Chakkera
Numerical Algorithms Group Ltd.
Oxford
www.vasucv.com
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