> It will not generate the mixed-case version that is, it will 
> give either ROBART J(when upper-case() is used) or robart 
> j(when lower-case() is used). In any case we won't get Robart 
> J which we are getting with the first solution.

Yes, that's true if you use current-grouping-key(). However, you can always
re-evaluate the group-by expression. Of course, by definition, it might give
you different answers for different members of the group, for example ROBART
J on one member and Robart J on another.

Michael Kay
http://www.saxonica.com/