Regex pattern with different rule at end of string? - regex

The pattern should match word with capictal letter .
words are seprated with [ ]+
But the last word should not have [ ] after it.
there is no limit of number of the words.
I have managed to do : (http://regexr.com?32s1h)
^([A-Z]{1}[a-z]+([a-z]+)?[ ]+)+$
which is Working for Xav Tvc Dcc_ //notice the last space
but not for Xav Tvc Dcc
How can i fix my regex ?

If space behind the last word is optional, use regex pattern
^(?:[A-Z][a-z]*(?:[ ]+|$))+$
...and if there should not be a space behind the last word, then go with
^(?:[A-Z][a-z]*(?:[ ]+(?=.)|$))+$

Require a word not followed by a space at the end:
^([A-Z]{1}[a-z]+([a-z]+)?[ ]+)*[A-Z]{1}[a-z]+([a-z]+)?$
Quick PowerShell test:
PS Home:> 'Xav Tvc Dcc ','Xav Tvc Dcc' -match '^([A-Z]{1}[a-z]+([a-z]+)?[ ]+)*[A-Z]{1}[a-z]+([a-z]+)?$'|%{"<$_>"}
<Xav Tvc Dcc>
If you're worried about possible errors introduced by changing the regex you could always construct it on the fly:
var word = "([A-Z]{1}[a-z]+([a-z]+)";
var regex = string.Format("^({0}?[ ]+){0}$", word);
Or similar for whatever language you use.

Why not do something like this:
^([A-Z]{1}[a-z]+\s?)*
Show Here
And then run a trim operation to remove the trailing spaces?

Related

Regex: Separate a string of characters with a non-consistent pattern (Oracle) (POSIX ERE)

EDIT: This question pertains to Oracle implementation of regex (POSIX ERE) which does not support 'lookaheads'
I need to separate a string of characters with a comma, however, the pattern is not consistent and I am not sure if this can be accomplished with Regex.
Corpus: 1710ABCD.131711ABCD.431711ABCD.41711ABCD.4041711ABCD.25
The pattern is basically 4 digits, followed by 4 characters, followed by a dot, followed by 1,2, or 3 digits! To make the string above clear, this is how it looks like separated by a space 1710ABCD.13 1711ABCD.43 1711ABCD.4 1711ABCD.404 1711ABCD.25
So the output of a replace operation should look like this:
1710ABCD.13,1711ABCD.43,1711ABCD.4,1711ABCD.404,1711ABCD.25
I was able to match the pattern using this regex:
(\d{4}\w{4}\.\d{1,3})
It does insert a comma but after the third digit beyond the dot (wrong, should have been after the second digit), but I cannot get it to do it in the right position and globally.
Here is a link to a fiddle
https://regex101.com/r/qQ2dE4/329
All you need is a lookahead at the end of the regular expression, so that the greedy \d{1,3} backtracks until it's followed by 4 digits (indicating the start of the next substring):
(\d{4}\w{4}\.\d{1,3})(?=\d{4})
^^^^^^^^^
https://regex101.com/r/qQ2dE4/330
To expand on #CertainPerformance's answer, if you want to be able to match the last token, you can use an alternative match of $:
(\d{4}\w{4}\.\d{1,3})(?=\d{4}|$)
Demo: https://regex101.com/r/qQ2dE4/331
EDIT: Since you now mentioned in the comment that you're using Oracle's implementation, you can simply do:
regexp_replace(corpus, '(\d{1,3})(\d{4})', '\1,\2')
to get your desired output:
1710ABCD.13,1711ABCD.43,1711ABCD.4,1711ABCD.404,1711ABCD.25
Demo: https://regex101.com/r/qQ2dE4/333
In order to continue finding matches after the first one you must use the global flag /g. The pattern is very tricky but it's feasible if you reverse the string.
Demo
var str = `1710ABCD.131711ABCD.431711ABCD.41711ABCD.4041711ABCD.25`;
// Reverse String
var rts = str.split("").reverse().join("");
// Do a reverse version of RegEx
/*In order to continue searching after the first match,
use the `g`lobal flag*/
var rgx = /(\d{1,3}\.\w{4}\d{4})/g;
// Replace on reversed String with a reversed substitution
var res = rts.replace(rgx, ` ,$1`);
// Revert the result back to normal direction
var ser = res.split("").reverse().join("");
console.log(ser);

Split string by multiple delimiters and keep the delimiter in result [duplicate]

How to split string with Regex.Split and keep all separators?
I have a string:"substring1 delimeter1 substring2" , where delimeter+substring2 is a part of address.
Also i have 2 and more delimeters: delim1,delim2 wich are equivalent in meaning;
And i want to get string array like this:
arr[0]="subsctring1";
arr[1]="delim1 subsctring2";
or,
arr[1]="delim2 subsctring2;
I have a pattern:
addrArr= Regex.Split(inputText, String.Concat("(?<=",delimeter1, "|",delimeter2, ")"), RegexOptions.None);
But it not works well.
Can you help me to create a valid pattern to to that?
You need a pattern with a lookahead only:
\s+(?=delim1|delim2)
The \s+ will match 1 or more whitespaces (since your string contains whitespaces). In case there can be no whitespaces, use \s* (but then you will need to remove empty entries from the result). See the regex demo. If these delimiters must be whole words, use \b word boundaries: \s+(?=\b(?:delim1|delim2)\b).
In C#:
addrArr = Regex.Split(inputText, string.Format(#"\s+(?={0})", string.Join("|", delimeters)));
If the delimiters can contain special regex metacharacters, you will need to run Regex.Escape on your delimiters list.
A C# demo:
var inputText = "substring1 delim1 substring2 delim2 substr3";
var delimeters = new List<string> { "delim1", "delim2" };
var addrArr = Regex.Split(inputText,
string.Format(#"\s+(?={0})", string.Join("|", delimeters.Select(Regex.Escape))));
Console.WriteLine(string.Join("\n", addrArr));
I think you need to use a lookahead, not a lookbehind, for this to work (haven't tried it though).
Also, you have to be careful with the separators; they must be escaped to work correctly as patterns in the regex.
Try this:
addrArr= Regex.Split(inputText, string.Format("(?={0}|{1})", Regex.Escape(delimeter1), Regex.Escape(delimeter2)), RegexOptions.None);

Regex matching all characters from the beginning of the string to the first underscore

I am trying to substring elements of a vector to only keep the part before the FIRST underscore. I am a bit of a newbie with taking substrings and don't fully understand all regex yet. I am close to the answer, I can get the part that I want to delete but still don't see how to get the opposite part. Any help and/or explanation of regex is appreciated!
my vector looks like the following, with multiple underscores in some elements
v = c("WL_Alk", "LQ_Frac_C_litter_origin", "MI_Nr_gat", "SED_C_N", "WL_CO2", "WL_S")
my desired output looks like
v_short = c("WL", "LQ", "MI", "SED", "WL", "WL")
The code that gets me the part I want to delete is sub("^[^_]*", "", v). I think I have to do something with $ in regex because sub("[_$]", "", v) deletes the first underscore, but I can't get it to delete the part behind it. Even with the regex helpfile I don't fully understand the meaning of ^, $ and * yet, so explanation on those is also appreciated!
You can use
> v = c("WL_Alk", "LQ_Frac_C_litter_origin", "MI_Nr_gat", "SED_C_N", "WL_CO2", "WL_S")
> sub("_.*", "", v)
[1] "WL" "LQ" "MI" "SED" "WL" "WL"
The "_.*" pattern matches the first _ and .* matches any 0+ characters up to the end of string greedily (that is, grabs them at one go).
With stringr str_extract, you can use your pattern:
> library(stringr)
> v_short = str_extract(v, "^[^_]*")
> v_short
[1] "WL" "LQ" "MI" "SED" "WL" "WL"
The ^[^_]* pattern matches the beginning of the string and 0 or more characters other than _.
If I understood correctly
gsub("(.*?)(_.*)","\\1",v, perl = TRUE)
Explanation:
(.*?) the first capturing group;
(_.*) the second capturing group;
\\1 return the first capturing group;
There are two ways to do it.
Either use ^[^_]+ and match string before first _. Regex101 Demo
OR
Select the part after first _ using \_.+$ and eliminate it. Regex101 Demo

Regular Expressions: querystring parameters matching

I'm trying to learn something about regular expressions.
Here is what I'm going to match:
/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
My expression should "grabs" abc123 and def456.
And now just an example about what I'm not going to match ("question mark" is missing):
/parent/child/firstparam=abc123&secondparam=def456
Well, I built the following expression:
^(?:/parent/child){1}(?:^(?:/\?|\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?
But that doesn't work.
Could you help me to understand what I'm doing wrong?
Thanks in advance.
UPDATE 1
Ok, I made other tests.
I'm trying to fix the previous version with something like this:
/parent/child(?:(?:\?|/\?)+(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?)?$
Let me explain my idea:
Must start with /parent/child:
/parent/child
Following group is optional
(?: ... )?
The previous optional group must starts with ? or /?
(?:\?|/\?)+
Optional parameters (I grab values if specified parameters are part of querystring)
(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*)?
End of line
$
Any advice?
UPDATE 2
My solution must be based just on regular expressions.
Just for example, I previously wrote the following one:
/parent/child(?:[?&/]*(?:firstparam=([^&]*)|secondparam=([^&]*)|[^&]*))*$
And that works pretty nice.
But it matches the following input too:
/parent/child/firstparam=abc123&secondparam=def456
How could I modify the expression in order to not match the previous string?
You didn't specify a language so I'll just usre Perl. So basically instead of matching everything, I just matched exactly what I thought you needed. Correct me if I am wrong please.
while ($subject =~ m/(?<==)\w+?(?=&|\W|$)/g) {
# matched text = $&
}
(?<= # Assert that the regex below can be matched, with the match ending at this position (positive lookbehind)
= # Match the character “=” literally
)
\\w # Match a single character that is a “word character” (letters, digits, and underscores)
+? # Between one and unlimited times, as few times as possible, expanding as needed (lazy)
(?= # Assert that the regex below can be matched, starting at this position (positive lookahead)
# Match either the regular expression below (attempting the next alternative only if this one fails)
& # Match the character “&” literally
| # Or match regular expression number 2 below (attempting the next alternative only if this one fails)
\\W # Match a single character that is a “non-word character”
| # Or match regular expression number 3 below (the entire group fails if this one fails to match)
\$ # Assert position at the end of the string (or before the line break at the end of the string, if any)
)
Output:
This regex will work as long as you know what your parameter names are going to be and you're sure that they won't change.
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam)\=([\w]+)&?)?
Whilst regex is not the best solution for this (the above code examples will be far more efficient, as string functions are way faster than regexes) this will work if you need a regex solution with up to 3 parameters. Out of interest, why must the solution use only regex?
In any case, this regex will match the following strings:
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789
It will now only match those containing query string parameters, and put them into capture groups for you.
What language are you using to process your matches?
If you are using preg_match with PHP, you can get the whole match as well as capture groups in an array with
preg_match($regex, $string, $matches);
Then you can access the whole match with $matches[0] and the rest with $matches[1], $matches[2], etc.
If you want to add additional parameters you'll also need to add them in the regex too, and add additional parts to get your data. For example, if you had
/parent/child/?secondparam=def456&firstparam=abc123&fourthparam=jkl01112&thirdparam=ghi789
The regex will become
\/parent\/child\/?\?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?(?:(?:firstparam|secondparam|thirdparam|fourthparam)\=([\w]+)&?)?
This will become a bit more tedious to maintain as you add more parameters, though.
You can optionally include ^ $ at the start and end if the multi-line flag is enabled. If you also need to match the whole lines without query strings, wrap this whole regex in a non-capture group (including ^ $) and add
|(?:^\/parent\/child\/?\??$)
to the end.
You're not escaping the /s in your regex for starters and using {1} for a single repetition of something is unnecessary; you only use those when you want more than one repetition or a range of repetitions.
And part of what you're trying to do is simply not a good use of a regex. I'll show you an easier way to deal with that: you want to use something like split and put the information into a hash that you can check the contents of later. Because you didn't specify a language, I'm just going to use Perl for my example, but every language I know with regexes also has easy access to hashes and something like split, so this should be easy enough to port:
# I picked an example to show how this works.
my $route = '/parent/child/?first=123&second=345&third=678';
my %params; # I'm going to put those URL parameters in this hash.
# Perl has a way to let me avoid escaping the /s, but I wanted an example that
# works in other languages too.
if ($route =~ m/\/parent\/child\/\?(.*)/) { # Use the regex for this part
print "Matched route.\n";
# But NOT for this part.
my $query = $1; # $1 is a Perl thing. It contains what (.*) matched above.
my #items = split '&', $query; # Each item is something like param=123
foreach my $item (#items) {
my ($param, $value) = split '=', $item;
$params{$param} = $value; # Put the parameters in a hash for easy access.
print "$param set to $value \n";
}
}
# Now you can check the parameter values and do whatever you need to with them.
# And you can add new parameters whenever you want, etc.
if ($params{'first'} eq '123') {
# Do whatever
}
My solution:
/(?:\w+/)*(?:(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)?|\w+|)
Explain:
/(?:\w+/)* match /parent/child/ or /parent/
(?:\w+)?\?(?:\w+=\w+(?:&\w+=\w+)*)? match child?firstparam=abc123 or ?firstparam=abc123 or ?
\w+ match text like child
..|) match nothing(empty)
If you need only query string, pattern would reduce such as:
/(?:\w+/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)
If you want to get every parameter from query string, this is a Ruby sample:
re = /\/(?:\w+\/)*(?:\w+)?\?(\w+=\w+(?:&\w+=\w+)*)/
s = '/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789'
if m = s.match(re)
query_str = m[1] # now, you can 100% trust this string
query_str.scan(/(\w+)=(\w+)/) do |param,value| #grab parameter
printf("%s, %s\n", param, value)
end
end
output
secondparam, def456
firstparam, abc123
thirdparam, ghi789
This script will help you.
First, i check, is there any symbol like ?.
Then, i kill first part of line (left from ?).
Next, i split line by &, where each value splitted by =.
my $r = q"/parent/child
/parent/child?
/parent/child?firstparam=abc123
/parent/child?secondparam=def456
/parent/child?firstparam=abc123&secondparam=def456
/parent/child?secondparam=def456&firstparam=abc123
/parent/child?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child?thirdparam=ghi789
/parent/child/
/parent/child/?
/parent/child/?firstparam=abc123
/parent/child/?secondparam=def456
/parent/child/?firstparam=abc123&secondparam=def456
/parent/child/?secondparam=def456&firstparam=abc123
/parent/child/?thirdparam=ghi789&secondparam=def456&firstparam=abc123
/parent/child/?secondparam=def456&firstparam=abc123&thirdparam=ghi789
/parent/child/?thirdparam=ghi789";
for my $string(split /\n/, $r){
if (index($string,'?')!=-1){
substr($string, 0, index($string,'?')+1,"");
#say "string = ".$string;
if (index($string,'=')!=-1){
my #params = map{$_ = [split /=/, $_];}split/\&/, $string;
$"="\n";
say "$_->[0] === $_->[1]" for (#params);
say "######next########";
}
else{
#print "there is no params!"
}
}
else{
#say "there is no params!";
}
}

Regex for quoted string with escaping quotes

How do I get the substring " It's big \"problem " using a regular expression?
s = ' function(){ return " It\'s big \"problem "; }';
/"(?:[^"\\]|\\.)*"/
Works in The Regex Coach and PCRE Workbench.
Example of test in JavaScript:
var s = ' function(){ return " Is big \\"problem\\", \\no? "; }';
var m = s.match(/"(?:[^"\\]|\\.)*"/);
if (m != null)
alert(m);
This one comes from nanorc.sample available in many linux distros. It is used for syntax highlighting of C style strings
\"(\\.|[^\"])*\"
As provided by ePharaoh, the answer is
/"([^"\\]*(\\.[^"\\]*)*)"/
To have the above apply to either single quoted or double quoted strings, use
/"([^"\\]*(\\.[^"\\]*)*)"|\'([^\'\\]*(\\.[^\'\\]*)*)\'/
Most of the solutions provided here use alternative repetition paths i.e. (A|B)*.
You may encounter stack overflows on large inputs since some pattern compiler implements this using recursion.
Java for instance: http://bugs.java.com/bugdatabase/view_bug.do?bug_id=6337993
Something like this:
"(?:[^"\\]*(?:\\.)?)*", or the one provided by Guy Bedford will reduce the amount of parsing steps avoiding most stack overflows.
/(["\']).*?(?<!\\)(\\\\)*\1/is
should work with any quoted string
"(?:\\"|.)*?"
Alternating the \" and the . passes over escaped quotes while the lazy quantifier *? ensures that you don't go past the end of the quoted string. Works with .NET Framework RE classes
/"(?:[^"\\]++|\\.)*+"/
Taken straight from man perlre on a Linux system with Perl 5.22.0 installed.
As an optimization, this regex uses the 'posessive' form of both + and * to prevent backtracking, for it is known beforehand that a string without a closing quote wouldn't match in any case.
This one works perfect on PCRE and does not fall with StackOverflow.
"(.*?[^\\])??((\\\\)+)?+"
Explanation:
Every quoted string starts with Char: " ;
It may contain any number of any characters: .*? {Lazy match}; ending with non escape character [^\\];
Statement (2) is Lazy(!) optional because string can be empty(""). So: (.*?[^\\])??
Finally, every quoted string ends with Char("), but it can be preceded with even number of escape sign pairs (\\\\)+; and it is Greedy(!) optional: ((\\\\)+)?+ {Greedy matching}, bacause string can be empty or without ending pairs!
An option that has not been touched on before is:
Reverse the string.
Perform the matching on the reversed string.
Re-reverse the matched strings.
This has the added bonus of being able to correctly match escaped open tags.
Lets say you had the following string; String \"this "should" NOT match\" and "this \"should\" match"
Here, \"this "should" NOT match\" should not be matched and "should" should be.
On top of that this \"should\" match should be matched and \"should\" should not.
First an example.
// The input string.
const myString = 'String \\"this "should" NOT match\\" and "this \\"should\\" match"';
// The RegExp.
const regExp = new RegExp(
// Match close
'([\'"])(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))' +
'((?:' +
// Match escaped close quote
'(?:\\1(?=(?:[\\\\]{2})*[\\\\](?![\\\\])))|' +
// Match everything thats not the close quote
'(?:(?!\\1).)' +
'){0,})' +
// Match open
'(\\1)(?!(?:[\\\\]{2})*[\\\\](?![\\\\]))',
'g'
);
// Reverse the matched strings.
matches = myString
// Reverse the string.
.split('').reverse().join('')
// '"hctam "\dluohs"\ siht" dna "\hctam TON "dluohs" siht"\ gnirtS'
// Match the quoted
.match(regExp)
// ['"hctam "\dluohs"\ siht"', '"dluohs"']
// Reverse the matches
.map(x => x.split('').reverse().join(''))
// ['"this \"should\" match"', '"should"']
// Re order the matches
.reverse();
// ['"should"', '"this \"should\" match"']
Okay, now to explain the RegExp.
This is the regexp can be easily broken into three pieces. As follows:
# Part 1
(['"]) # Match a closing quotation mark " or '
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
# Part 2
((?: # Match inside the quotes
(?: # Match option 1:
\1 # Match the closing quote
(?= # As long as it's followed by
(?:\\\\)* # A pair of escape characters
\\ #
(?![\\]) # As long as that's not followed by an escape
) # and a single escape
)| # OR
(?: # Match option 2:
(?!\1). # Any character that isn't the closing quote
)
)*) # Match the group 0 or more times
# Part 3
(\1) # Match an open quotation mark that is the same as the closing one
(?! # As long as it's not followed by
(?:[\\]{2})* # A pair of escape characters
[\\] # and a single escape
(?![\\]) # As long as that's not followed by an escape
)
This is probably a lot clearer in image form: generated using Jex's Regulex
Image on github (JavaScript Regular Expression Visualizer.)
Sorry, I don't have a high enough reputation to include images, so, it's just a link for now.
Here is a gist of an example function using this concept that's a little more advanced: https://gist.github.com/scagood/bd99371c072d49a4fee29d193252f5fc#file-matchquotes-js
here is one that work with both " and ' and you easily add others at the start.
("|')(?:\\\1|[^\1])*?\1
it uses the backreference (\1) match exactley what is in the first group (" or ').
http://www.regular-expressions.info/backref.html
One has to remember that regexps aren't a silver bullet for everything string-y. Some stuff are simpler to do with a cursor and linear, manual, seeking. A CFL would do the trick pretty trivially, but there aren't many CFL implementations (afaik).
A more extensive version of https://stackoverflow.com/a/10786066/1794894
/"([^"\\]{50,}(\\.[^"\\]*)*)"|\'[^\'\\]{50,}(\\.[^\'\\]*)*\'|“[^”\\]{50,}(\\.[^“\\]*)*”/
This version also contains
Minimum quote length of 50
Extra type of quotes (open “ and close ”)
If it is searched from the beginning, maybe this can work?
\"((\\\")|[^\\])*\"
I faced a similar problem trying to remove quoted strings that may interfere with parsing of some files.
I ended up with a two-step solution that beats any convoluted regex you can come up with:
line = line.replace("\\\"","\'"); // Replace escaped quotes with something easier to handle
line = line.replaceAll("\"([^\"]*)\"","\"x\""); // Simple is beautiful
Easier to read and probably more efficient.
If your IDE is IntelliJ Idea, you can forget all these headaches and store your regex into a String variable and as you copy-paste it inside the double-quote it will automatically change to a regex acceptable format.
example in Java:
String s = "\"en_usa\":[^\\,\\}]+";
now you can use this variable in your regexp or anywhere.
(?<="|')(?:[^"\\]|\\.)*(?="|')
" It\'s big \"problem "
match result:
It\'s big \"problem
("|')(?:[^"\\]|\\.)*("|')
" It\'s big \"problem "
match result:
" It\'s big \"problem "
Messed around at regexpal and ended up with this regex: (Don't ask me how it works, I barely understand even tho I wrote it lol)
"(([^"\\]?(\\\\)?)|(\\")+)+"