I have been trying to find all possible strings in between 2 strings
This is my input: "print/// to be able to put any amount of strings here endprint///"
The goal is to print every string in between print/// and endprint///
You can use Lua's string patterns to achieve that.
local text = "print/// to be able to put any amount of strings here endprint///"
print(text:match("print///(.*)endprint///"))
The pattern "print///(.*)endprint///" captures any character that is between "print///" and "endprint///"
Lua string patterns here
In this kind of problem, you don't use the greedy quantifiers * or +, instead, you use the lazy quantifier -. This is because * matches until the last occurrence of the sub-pattern after it, while - matches until the first occurence of the sub-pattern after it. So, you should use this pattern:
print///(.-)endprint///
And to match it in Lua, you do this:
local text = "print/// to be able to put any amount of strings here endprint///"
local match = text:match("print///(.-)endprint///")
-- `match` should now be the text in-between.
print(match) -- "to be able to put any amount of strings here "
Related
I need to encapsulate in some way pattern in lua pattern matching to find whole sequence of this pattern in string. What do I mean by that.
For example we have string like that:
"word1,word2,word3,,word4,word5,word6, word7,"
I need to match first sequence of words followed by coma (word1,word2,word3,)
In python I would use this pattern "(\w+,)+", but similar pattern in lua (like (%w+,)+), will return just nil, because brackets in lua patterns means completely different thing.
I hope now you see my problem.
Is there a way to do repeating patterns in lua?
Your example wasn't too clear in terms of what should happen to the word4,word5,word6 and word7,
This would give you any seqence of comma separated words without white space or empty positions.
local text = "word1,word2,word3,,word4,word5,word6, word7,"
-- replace any comma followed by any white space or comma
--- by a comma and a single white space
text = text:gsub(",[%s,]+", ", ")
-- then match any sequence of >=1 non-whitespace characters
for sequence in text:gmatch("%S+,") do
print(sequence)
end
Prints
word1,word2,word3,
word4,word5,word6,
word7,
You could do this easily using LPeg if that's available to you:
local lpeg = require "lpeg"
local str = "word1,word2,word3,,word4,word5,word6, word7,"
local word = (lpeg.R"az"+lpeg.R"AZ"+lpeg.R"09") ^ 1
local sequence = lpeg.C((word * ",") ^1)
print(sequence:match(str))
I'm trying to match any strings that come in that follow the format Word 100.00% ~(45.56, 34.76) in LUA. As such, I'm looking to do a regex close (in theory) to this:
%D%s[%d%.%d]%%(%d.%d, %d.%d)
But I'm having no luck so far. LUA's patterns are weird.
What am I missing?
Your pattern is close you neglected to allow for multiple instances of a digit you can do this by using a + at like %d+.
You also did not use [,( and . correctly in the pattern.
[s in a pattern will create a set of chars that you are trying to match such as [abc] means you are looking to match any as bs or c at that position.
( are used to define a capture so the specific values you want returned rather then the whole string in the event of a match, in order to use it as a char you for the match you need to escape it with a %.
. will match any character rather then specifically a . you will need to add a % to escape if you want to match a . specifically.
local str = "Word 100.00% ~(45.56, 34.76)"
local pattern = "%w+%s%d+%.%d+%%%s~%(%d+%.%d+, %d+%.%d+%)"
print(string.match(str, pattern))
Here you will see the input string print if it matches the pattern otherwise you will see nil.
Suggested resource: Understanding Lua Patterns
I am trying to split this statement in Lua
sendex,000D6F0011BA2D60,fb,btn,1,on,100,null
i need output like this way:
Mac:000D6F0011BA2D60
Value:1
command:on
value:100
how to split and get the values?
local input = "sendex,000D6F0011BA2D60,fb,btn,1,on,100,null"
local buffer = {}
for word in input:gmatch('[^,]+') do
table.insert(buffer, word)
--print(word) -- uncomment this to see the words as they are being matched ;)
end
print("Mac:"..buffer[2])
print("Value:"..buffer[5])
...
For a complete explanation of what string.gmatch does, see the Lua reference. To summarize, it iterates over a string and searches for a pattern, in this case [^,]+, meaning all groups of 1 or more characters that aren't a comma. Every time it finds said pattern, it does something with it and continues searching.
If your input is exactly like you have described, the code below works:
s="sendex,000D6F0011BA2D60,fb,btn,1,on,100,null"
Mac,Value,command,value = s:match(".-,(.-),.-,.-,(.-),(.-),(.-),")
print(Mac,Value,command,value)
It uses the non-greedy pattern .- to split the input into fields. It also captures the relevant fields.
Currently I have code that looks like this:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=)[^&]+", "%1hello", 1)
StringToPrint will look like this:
param=hello&456
I have replaced all of the characters before the & with the string "hello". This is where my question becomes a little strange and specific.
I want my string to appear as: param=helloZ&456. In other words, I want to preserve the character right before the & when replacing the string valueZ with hello to make it helloZ instead. How can this be done?
I suggest:
somestring:gsub("param=[^&]*([^&])", "param=hello%1", 1)
See the Lua demo
Here, the pattern matches:
param= - literal substring param=
[^&]* - 0 or more chars other than & as many as possible
([^&]) - Group 1 capturing a symbol other than & (here, backtracking will occur, as the previous pattern grabs all such chars other than & and then the engine will take a step back and place the last char from that chunk into Group 1).
There are probably other ways to do this, but here is one:
somestring = "param=valueZ&456"
local stringToPrint = (somestring):gsub("(param=).-([^&]&)", "%1hello%2", 1)
print(stringToPrint)
The thing here is that I match the shortest string that ends with a character that is not & and a character that is &. Then I add the two ending characters to the replaced part.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.