In xquery you do
for $i at $pos in ../s return
if (...) then $pos else ()
However thats exactly the same as:
for $i in ../s return
if (...) then 1 + count($i/preceding-sibling::s) else ()
It's the same in this case, but
(a) it relies on the "in" expression ../s returning a sequence of
siblings, and
(b) the second solution is likely to be O(n^2) while the first is likely
to be O(n).
Michael Kay
Saxonica
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