Subject: RE: [xsl] Get position of parent From: "Philip Vallone" <philip.vallone@xxxxxxxxxxx> Date: Sat, 17 Jan 2009 14:31:16 -0500 |
Thank you. Regards, Phil -----Original Message----- From: Michael Kay [mailto:mike@xxxxxxxxxxxx] Sent: Saturday, January 17, 2009 11:53 AM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: RE: [xsl] Get position of parent It rather depends what you mean by "position", but if you mean the number of preceding siblings plus one, you can use count(../preceding-sibling::*)+1 or if you prefer <xsl:number select=".." count="*"/> Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Philip Vallone [mailto:philip.vallone@xxxxxxxxxxx] > Sent: 17 January 2009 13:42 > 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
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] Get position of parent, Michael Kay | Thread | [xsl] Re: [XSL] Two "Philisophical", Alain |
RE: [xsl] Get position of parent, Michael Kay | Date | [xsl] Re: [XSL] Two "Philisophical", Alain |
Month |