Subject: Re: Position() of parent node
From: David Carlisle <davidc@xxxxxxxxx>
Date: Tue, 6 Feb 2001 18:56:25 GMT
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> 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
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