How to get the length of a string capture?

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#1

Hello,
I’m making a function to simplify the pattern of #("string") into its actual number (6 in this case). This is what I have right now:

local function SimplifyStrings(str)
	return str:gsub(
		"#%(\"(.+)\"%)",
		string.len("%1")
	)
end

print(SimplifyStrings("#(\"does it work?\")"))

The problem here is that it pre-processes the length of %1 to a 2. How could I get around this issue?
(No, changing string.len to # doesn’t help)

1 Like
#2

You should use string.match instead. Which will return the string which matches our pattern, then we can just get the length of it using the # operator

local function SimplifyStrings(str)
   return #str:match("#%(\"(.+)\"%)$")
end

print(SimplifyStrings("#(\"does it work?\")")) -- 13
print(SimplifyStrings("#(\"string\")")) -- 6
#3

Yes, but I want it to replace more occurrences than 1. In my case when I use it there will probably be around 50 of these.

#4

I’m not sure if this is exactly what you wanted, and im also not the best at string patterns so there is probably a more elegant solution

But here is what I got…

local function SimplifyStrings(str)
	local sum = "";
	for w in str:gmatch("[^#%(\"(.+)\"%)]+") do
		sum ..= string.len(w) .. " "
	end
	return sum
end

print(SimplifyStrings("#(\"string\")#(\"does it work?\")")) -- 6 13
1 Like
#5

Yeah this works good, but I also want it to keep other things that don’t match the pattern.

#6

Update: I got it to work.

local function SimplifyStrings(String)
	for capture in string.gmatch(String, "#%(\"([^\"]+)") do
		String = String:gsub(
			"#%(\"([^\"]+)\"%)",
			#capture,
			1
		)
	end
	
	return String
end

print(SimplifyStrings(
	"#(\"Did it work?\") some strings in between #(\"Yes it did!\")"
))

Output:
12 some strings in between 11