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