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, Paulo. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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