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> From: Thomas B. Passin [mailto:email@example.com]
> 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  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. "
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...