[David Carlisle]
For-each does operate on each node in a node-set, but they
are operated
on separately, as if they were single, unrelated nodes.
With for-each,
for example, you can only get a position() of 1 on any one
iteration,
Not at all. see the example in the spec that uses position() in a
for-each to not put a comma after the last item in a list.
The rare time I do not test what I say or check the Rec before posting,
I get it messed up! Now I am scratching my head trying to remember just
what were the situations I could not get what I wanted with position()
within a for-each block. I do remember times like that, but not
details. Must have been a long time ago.
But it is clear in the Rec, just what David says (of course!) -
"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."
Sorry to be promulgating mis-information :-(
Cheers,
Tom P
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