This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
Re: Position() of parent node
- To: xsl-list at lists dot mulberrytech dot com
- Subject: Re: [xsl] Position() of parent node
- From: David Carlisle <davidc at nag dot co dot uk>
- Date: Tue, 6 Feb 2001 18:56:25 GMT
- References: <83329AAEE258D411849400C00D0039D80AC946@ils>
- Reply-To: xsl-list at lists dot mulberrytech dot com
> Can anyone provide me with the syntax for getting the position() value of
> the current nodes' parent node
A node does not, of itself, have a position() it only has a position in
a given node list (and the same node may be in many node lists).
so for example if you go
<xsl:apply-templates select="parent::xxx"/>
then the position() of the node will be 1.
Probably what you want is the sibling number of the parent node, which
is
<xsl:value-of select="count(1parent::*/preceding-sibling::*)"/>
David
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list