I could probably do what you suggested however I would then need to know the uppercase equivalent of every possible extended character
in ISO-Latin-1 (which you are using by the looks of things), you can just subtract 32 from the asc() value of your lowercase "extended" chars to get the uppercase decimal value
I don't know how you'd achieve this with a regex. You'd need to get the asc() value of a backreference, subtract 32 and get the chr() equivalent of that, but backreferences are the last thing evaluated in the string to replace argument for REReplace() (i.e. \1 will be considered a string value "\1" until all other expressions in the string to replace argument have been evaluated).
Mark
-- ** Archive: http://www.mail-archive.com/dev%40lists.cfdeveloper.co.uk/
To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] For human help, e-mail: [EMAIL PROTECTED]
