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Get parent's node position - Urgent
- To: <xsl-list at lists dot mulberrytech dot com>
- Subject: [xsl] Get parent's node position - Urgent
- From: "Paulo Henrique S. Bermejo" <bermejo at eps dot ufsc dot br>
- Date: Tue, 25 Sep 2001 12:03:48 -0300
- Reply-To: xsl-list at lists dot mulberrytech dot com
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,
Paulo.
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