Checking for an escaped quote

blabla bla "I need this part \" and this part" blablal

I need this part " and this part

theoriginalbit, on 22 September 2014 - 03:54 AM said:

What have you got so far? I assume it is thinking that the escaped quote is the end of the string?

'hello \" world'

local myData = "blabla bla \"I need this part \\\" and this part\" blablal"

local foundInQuotes, inQuotes, i = "", false, 1

while i <= #myData do
	if myData:sub(i,i) == "\\" then
		i = i + 1
		
		if inQuotes then
			if myData:sub(i,i) == "n" then
				foundInQuotes = foundInQuotes .. "\n"
			else foundInQuotes = foundInQuotes .. myData:sub(i,i) end
		end
		
	elseif myData:sub(i,i) == "\"" then
		inQuotes = not inQuotes

	elseif inQuotes then
		foundInQuotes = foundInQuotes .. myData:sub(i,i)
	end
	
	i = i + 1
end

print(foundInQuotes)

theoriginalbit, on 22 September 2014 - 08:14 AM said:

well if you are doing it letter by letter, you could simply have a boolean flag for when you discover a \ and then just ignore the next character, or alternatively, validate the next character is a valid character, since this is also a valid string

'hello \" world'

you don't have to only have them terminated within a double quoted string.

function readString( self )
	local char = self.reader:peek()

	local doubleQuote = char == "\""
	local singleQuote = char == "'"
	if doubleQuote or singleQuote then
		self.reader:pop()
		
		local s = ""
		local backslash = false
		while doubleQuote or singleQuote do
			if backslash then
				backslash = false
				s = s .. self.reader:pop()
			else
				local char = self.reader:peek()
				if char == ( doubleQuote and "\"" or "'" ) then
					doubleQuote = false
					singleQuote = false
					self.reader:pop()
					return s
				elseif char == "\\" then
					backslash = true
					s = s .. "\\"
				else
					s = s .. self.reader:pop()
				end
			end
			
		end
	end
end

local t = {}
t.reader = StringReader.new( '"bla bla bla \" blabla"' )
print(readString(t))

"bla bla bla "

'"bla bla bla \" blabla"'

'"bla bla bla " blabla"'

"bla bla \\\" bla"

"bla bla \" bla"

MKlegoman357, on 22 September 2014 - 03:48 PM said:

Just an idea, but I think it's not completely your fault. The way you define this string:

'"bla bla bla \" blabla"'

Well, the Lua parser converts it to this, if I'm not mistaken:

'"bla bla bla " blabla"'

That is probably why your other attempts failed when they tried to find the backslash. To produce a string with a backslashed double quote, inside double quotes you should use this:

"bla bla \\\" bla"

…and then the Lua parser will convert it to this:

"bla bla \" bla"

local s = "\"This is \"a string\" with escaped\" quotes\""
print(s:gsub( "\"(.*)\"", "%1"))
--> This is "a string" with escaped" quotes

MKlegoman357, on 23 September 2014 - 05:40 PM said:

Lol, that was exactly what I mentioned to you in my first post :D/>

Engineer, on 23 September 2014 - 03:00 PM said:

I just realized something which makes this unnecessary to even try. If one escapes a quote it is for the compiler/parser, not the string itself.

theoriginalbit, on 24 September 2014 - 12:36 AM said:

I wish I'd've come back and read this thread sooner. I assumed you already knew this, hence my suggestion of the boolean flag.

Spoiler