This could just be my own misunderstanding of something, or a feature that doesn't exist.
I frequently use the writer's search to locate or replace items in my documents, but I recently tried to use a wildcard search and replace, and it failed miserably. I want to find all instances of the end of a sentence that is followed by a single space and replace the single space with a double space. The search term, with "regular expressions" selected, is <[.!?]["']* [^ ]> (ignore the <> - they are my separators for this because it contains most of the usual quoting characters; a period, exclamation mark or question mark, optionally followed by one or more quote marks, followed by a space, followed by anything that is not a space. This works just fine. What is the replacement expression? I have looked at the help sections, but there is no information on replacement expressions. As far as I can tell, it is not available in Word, either. It is, however, a fairly simple expression in vi, sed and all related Linux/Unix text editor programs (I'm guessing emacs can also do this, but I've never used it). The vi command is :g/\([.!?]["']* \)\([^ ]\)/s//\1 \2/g (globally search for the expression, grouping the front and back separately, replace with part one, space, part two, globally on each line as well). Is this possible in OOo at all? (And, no, to convert to vi, do this and convert back is not practical - they are already conversions from text files, and this problem wasn't present in them.) Thanks. mhr
