[xsl] Get parent's node position - Urgent

Subject: [xsl] Get parent's node position - Urgent
From: "Paulo Henrique S. Bermejo" <bermejo@xxxxxxxxxxx>
Date: Tue, 25 Sep 2001 12:03:48 -0300
Hi all,

I would like take the "position" of parent's node.
I try:
<xsl:value-of select="position(parent::*)"/>
But didnt't work because position don't accept params.

To get the name of parent node I know:
<xsl:value-of select="name(parent::*)"/>
But I need the position.

Thanks all,


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