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[Manos Batsis] > From: Thomas B. Passin [mailto:tpassin@c...] > Not with xsl:for-each, though. Position() would always be 1. Nope, sorry. Although the position() function returns the index of the current node *in the current node list*, the spec clearly says [1] that "The expression must evaluate to a node-set. The template is instantiated with the selected node as the current node, and with a list of all of the selected nodes as the current node list. " [Tom P] Whoops, you're right. I know I was thinking of something specific that would always give position()=1, but now I can't remember what... Tom P
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