I have been coding a program in Lua that automatically formats IRC logs from a roleplay. In the roleplay logs there is a specific guideline for "Out of character" conversation, which we use double parentheses for. For example: ((<Things unrelated to roleplay go here>))
. I have been trying to have my program remove text between double brackets (and including both brackets). The code is:
ofile = io.open("Output.txt", "w")
rfile = io.open("Input.txt", "r")
p = rfile:read("*all")
w = string.gsub(p, "%(%(.*?%)%)", "")
ofile:write(w)
The pattern here is > "%(%(.*?%)%)"
I've tried multiple variations of the pattern. All resulted in fruitless results:
1. %(%(.*?%)%) --Wouldn't do anything.
2. %(%(.*%)%) --Would remove *everything* after the first OOC message.
Then, my friend told me that prepending the brackets with percentages wouldn't work, and that I had to use backslashes to 'escape' the parentheses.
3. \(\(.*\)\) --resulted in the output file being completely empty.
4. (\(\(.*\)\)) --Same result as above.
5. (\(\(.*?\)\) --would for some reason, remove large parts of the text for no apparent reason.
6. \(\(.*?\)\) --would just remove all the text except for the last line.
The short, absolute question: What pattern would I need to use to remove all text between double parentheses, and remove the double parentheses themselves too?
You're friend is thinking of regular expressions. Lua patterns are similar, but different.
%
is the correct escape character.Your pattern should be
%(%(.-%)%)
. The-
is similar to*
in that it matches any number of the preceding sequence, but while*
tries to match as many characters as it can (it's greedy),-
matches the least amount of characters possible (it's non-greedy). It won't go overboard and match extra double-close-parenthesis.