string.match throwing error: attempt to index field '?' (a string value)
The error is in the line data[i] = line This line makes data[i] a string variable which cannot have other strings indexed to it. Change that line to: data[i] = {} and everything works fine....
Lua pattern similar to regex positive lookahead?
str:gsub( '§\\n(.)', '%1' ) should do what you want. This deletes the delimiter given that it is followed by another character, putting this character back into to string. Test code local str = { 'abc§\\ndef§\\nghi\\n', 'abc§\\ndef§\\nghi§\\n', 'abc§\\ndef§\\nghi§§\\n', } for i = 1, #str do print( ( str[ i ]:gsub( '§\\n(.)',...
Lua XML extract from pattern
string,lua,pattern-matching,lua-patterns
In general, it's not a good idea to use pattern matching (either Lua pattern or regex) to extract XML. Use a XML parser. For this problem, you don't need to escape \ or <(even if you do, Lua pattern uses % to escape magic characters). And use brackets to get...
How to match single digits and digits separated by a hyphen with string.match?
Easiest is to do it separately: function roll(input) local min,high = string.match(input, '(%d+)-(%d+)') if min == nil then min = string.match(input, '(%d+)') end return min, high end print(roll '10') print(roll '10-100') ...
Lua string.gsub inside string.gmatch?
Well for one thing, you miss-spelled strawberry here: fruits = {apple = "green", orange = "orange", stawberry = "red"} You can also work with lua tables as sets, which means that nested loop searching for duplicates is unnecessary. It can be simplified to something like: fruits = {apple = "green",...
Store Lua string.match output to array
arrays,string,lua,lua-patterns
You can store the result into a table: local t = {string.match(array[i], pattern)} if #t ~= 0 then words = t end end The return value of parse_string is a table now: local result = (parse_strings (a, 'alarm dentist (%w+) (%w+)')) for k, v in ipairs(result) do print(k, v) end...
Lua: RegEx to Pattern
string,lua,string-matching,gsub,lua-patterns
You could try like this. > f = "foo [\"12\"] bar" > x = string.gsub(f, "%[\"(%d+)\"%]", "[%1]") > print(f) foo ["12"] bar > print(x) foo [12] bar \d which matches any digits would be represented as %d in lua....
Check for valid domain
string,lua,string-matching,lua-patterns,lua-5.1
I'm not sure the requirement you described are correct, but this logic follows your requirements: -- Only characters A-Z and numbers 0-9 and the - so long as it's not at the start or end of string. -- Be at least 3 characters excluding the extension. -- No spaces allowed...
lua pattern, exclusion doesn't work with end of string
string,lua,pattern-matching,lua-patterns
Given: local s = [[5427 root 21620 S mpd /etc/mpd2.conf 5437 root 25660 S mpd]] The following pattern string.match(s,"(%d+)%s+root%s+%d+%s+S%s+mpd[%s]-$") returns: 5437 root 25660 S mpd whereas this: string.match(s,"(%d+%s+root%s+%d+%s+S%s+mpd[%s]%p?[%w%p]+)") returns: 5427 root 21620 S mpd /etc/mpd2.conf...
Lua string.find can't find the last word in a line
In this line: local s,e=string.find(line,"%w+ ",pos) -- ^ There is a whitespace in the pattern "%w+ ", so it matches a word followed by a whitespace. When you input, e.g, word1 word2 word3 and press Enter, word3 has no whitespace following it. There is no whitespace in the example of...
Lua pattern help (Double parentheses)
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...
Replacing String in String Corona SDK
string,sdk,lua,corona,lua-patterns
You can use string.gsub with lua string-patterns: string.gsub("this is a !a joke !/a! haha", "!a %a+ !/a!", "cheer") If need be you can even capture the string that's between !a: local str = "this is a !a joke !/a! haha" str:gsub ("!a (%a+) !/a!", "sad %1") Which after substitution gives...
Have Lua pathological patterns with exponential running time?
lua,time-complexity,lua-patterns
Yes, lua patterns can take exponential time. Try running: string.find('aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa', 'a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?' .. 'a?a?a?a?a?a?a?a?a?a?a?a?a?a?a?aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa') They can still run reasonably fast if you keep the patterns simple though so I would try testing some real examples on your own data....
How to split a string before a dot in Lua?
Since regular matching is greedy, you just need to match anything until you see . (don't forget to escape it): print(string.match("That.Awkward.Moment.2014.720p.BluRay.x264.YIFY.srt", '(.+)%.(.+)')) will print That.Awkward.Moment.2014.720p.BluRay.x264.YIFY srt ...
tonumber and regex
regex,lua,numbers,lua-patterns
The whole thing can be shortened to this one liner: local hh, mm, ss = dt:match "(%d%d?):(%d%d?):(%d%d?)" As for the error occuring in your tonumber, it is because gsub returns 2 values after its operation. First being the substituted sring and the second being a number. tonumber assumes the second...
finding a url in a string lua pattern
url,lua,pattern-matching,string-matching,lua-patterns
-- all characters allowed to be inside URL according to RFC 3986 but without -- comma, semicolon, apostrophe, equal, brackets and parentheses -- (as they are used frequently as URL separators) local string_with_URLs = [[ <a href="http://www.lua.org:80/manual/5.2/contents.html">L.ua 5.2</a> [url=127.0.0.1:8080]forum link[/url] [markdown link](https://74.125.143.101/search?q=Who+are+the+Lua+People%3F) auth link: ftp://user:[email protected]/path - not recognized yet ]]...
Combine multiple string.match functions into a single line
To match on one of a set of characters you need to use the [set] notation1. So the following should do what you want and provide a slightly better indication of the failure: local p = string.find(text, '[&"#[email protected]%%!^*]'); if p then print("Invalid character found at "..p) end You could even...
How to split string using a whole word as separator?
local str = "importanttext1 has gathered importanttext2 from stackoverflow." local s1 = str:match("(.+)has") local s2 = str:match("gathered%s+(%S+)") print(s1) print(s2) Note that "%s" matches a whitespace character while "%S" matches a non-whitespace character....
Lua: split string into words unless quoted
There may be ways to do this with clever parsing, but an alternative way may be to keep track of a simple state and merge fragments based on detection of quoted fragments. Something like this may work: local text = [[I "am" 'the text' and "some more text with '"...
Unexpected Regular expression result using dot symbol
You need to escape . with %..
Find string between tags not working
Try: print(string.match(str,"%[COLOR r;255|g;255|b;0%](.-)%[/COLOR%]")) Note that [,] are magic characters, so they need to be escaped....
Lua detect a character/string repeated in all text
Use the pattern "[?%s]*", which means zero or more of ? or whitespace characters. if text:gsub("[?%s]*", "") == "" then -- do something end ...