Hi Dan,
Oleg's solution is neater anyway though, as long as it's unlikely
that there'll be hordes of children of a (because if there were
lots of children then doing count()s would involve a lot of node
visits which would take time).
See my earlier message. I'm not sure Oleg's solution actually
meets the specified problem.
True; it does with a minor change though:
b and count(b|c|d) = count(*)
Cheers,
Jeni
---
Jeni Tennison
http://www.jenitennison.com/
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list