Subject: Re: [xsl] Get position of parent From: Vasu Chakkera <vasucv@xxxxxxxxx> Date: Sat, 17 Jan 2009 15:19:55 +0000 |
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: [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 > -- Vasu Chakkera Numerical Algorithms Group Ltd. Oxford www.vasucv.com
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