[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Get parent's node position - Urgent
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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