Lua: Pattern match after a string?

Assuming that, by "word" you mean "a string of characters without spaces", you can do this:

for line in ioFile:lines() do
  local variable, value = line:match("(%S+)%s+(.+)")
  ... --Do stuff with variable and value
end

One alternative with string.match was shown by Nicol Bolas, here is another alternative:

function splitOnFirstSpace(input)
    local space = input:find(' ') or (#input + 1)
    return input:sub(1, space-1), input:sub(space+1)
end

Usage:

local command, param = splitOnFirstSpace(line)

If no argument is given (splitOnFirstSpace('no-param-here')), then param is the empty string.

I do not believe Lua is packaged with a split() function like Ruby or Perl.

I found that this guy built a lua version of Perl's split function: http://lua-users.org/lists/lua-l/2011-02/msg01145.html

If you can guarantee that the argument will only have 1 word before it, with that word not containing any spaces, you can read in that line, run the split function on it, and use the return array's 1 index value as what you want.

You could error check that too and make sure you get a 'C:\' within your expected directory, or check to make sure the string is == to 'On' or 'Off'. Because of using the hardcoded index value I really advocate you error check your expected value. Nothing is worse than having an assumed value be wrong.

If an error is detected make sure to log or print it to the screen so you know about it.

This could catch bugs where maybe the string that was input is improper.

Some simple code that models what I suggest you do:

line = "directory C:\Program Files\abc\def/";

contents = line.split(" "); --Split using a space
directory = contents[2]; --Here is your directory

if(errorCheckDir(directory))
--Use directory
end

EDIT: In response to comments below Lua indeed begins indexing at 1, not 0. Also, in the case that the directory contains spaces (which is probable) instead of simply using contents[2], I would loop through all of contents except index 1, and piece back together the directory making sure to add the required space between each index that you attach.

So in the case above, contents[2] and contents[3] would have to be stitched back together with a space in between to recover the proper directory.

directory = contents[2].." "..contents[3]

This can be easily automated using a function which has a loop in it and returns back the proper directory:

function recoverDir(contents)
    directory = "";
    --Recover the directory
    for i=2, table.getn(contents) do
        directory = directory..contents[i].." ";
    end
    --strip extra space on the end
    dirEnd = string.len(directory);
    directory = string.sub(directory,1,dirEnd-1);
    return directory; --proper directory
end