Lua string.match() problem

2
votes

I want to match a few lines for a string and a few numbers. The lines can look like

"  Code                                             : 75.570 "

or

"  ..dll                                   :          13.559       1"

or

"  ..node                                    :  4.435    1.833    5461"

or

"  ..NavRegions                                     :  0.000         "

I want something like

local name, numberLeft, numberCenter, numberRight = line:match("regex");

But I'm very new to the string matching.

4
stringluamatch
One question: Do the strings for the name always have either zero or two periods at the front, with the rest of the string being letters? - Jeffrey Blake
There are the two space characters at the beginning of the line, the name follows and there are again space characters, the colon, and 1 or 2 or 3 numbers(unknown amount of space characters again in between and space characters at the end or not. It's really complex, does anybody has a better idea than string matching or maybe a workaround? - luac
Retagged because Lua doesn't contain the regex libraries. For more details on Lua pattern matching, see lua.org/pil/20.2.html - BMitch

4 Answers

5
votes

This pattern will work for every case:

%s*([%w%.]+)%s*:%s*([%d%.]+)%s*([%d%.]*)%s*([%d%.]*)

Short explanation: [] makes a set of characters (for example the decimals). The last to numbers use [set]* so an empty match is valid too. This way the number that haven't been found will effectively be assigned nil.

Note the difference between using + - * in patterns. More about patterns in the Lua reference.

This will match any combination of dots and decimals, so it might be useful to try and convert it to a number with tonumber() afterwards.

Some test code:

s={
"  Code                                             : 75.570 ",
"  ..dll                                   :          13.559       1",
"  ..node                                    :  4.435    1.833    5461",
"  ..NavRegions                                     :  0.000         "
}
for k,v in pairs(s) do
    print(v:match('%s*([%w%.]+)%s*:%s*([%d%.]+)%s*([%d%.]*)%s*([%d%.]*)'))
end
1
votes

Here is a starting point:

s="  ..dll                                   :          13.559       1"
for w in s:gmatch("%S+") do
        print(w)
end

You may save these words in a table instead of printing, of course. And skip the second word.

1
votes

@Ihf Thank you, I now have a working solution.

local moduleInfo, name = {};
for word in line:gmatch("%S+") do
    if (word~=":") then
        word = word:gsub(":", "");
        local number = tonumber(word);
        if (number) then
            moduleInfo[#moduleInfo+1] = number;
        else
            if (name) then
                name = name.." "..word:gsub("%$", "");
            else
                name = word:gsub("%$", "");
            end
        end
    end
end

@jpjacobs Really nice, thanks too. I'll rewrite my code for synthetic reasons ;-) I'll implement your regex of course.

0
votes

I have no understanding of the Lua language, so I won't help you there. But in Java this regex should match your input

"([a-z]*)\\s+:\\s+([\\.\\d]*)?\\s+([\\.\\d]*)?\\s+([\\.\\d]*)?"

You have to test each group to know if there is data left, center, right

Having a look at Lua, it could look like this. No guarantee, I did not see how to escape . (dot) which has a special meaning and also not if ? is usable in Lua.

"([a-z]*)%s+:%s+([%.%d]*)?%s+([%.%d]*)?%s+([%.%d]*)?"