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/
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