Thank you.
Regards,
Phil
-----Original Message-----
From: Michael Kay [mailto:mike(_at_)saxonica(_dot_)com]
Sent: Saturday, January 17, 2009 11:53 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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(_dot_)vallone(_at_)verizon(_dot_)net]
Sent: 17 January 2009 13:42
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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