Regex with global modifier to capture words within lines - regex

The Input:
Let's consider this string below
* key : foo bar *
* big key : bar*bar
* healthy : cereal bar *
sadly : without star *
The Output:
I would like to retrieve the key:value pairs for each match.
'key', 'foo bar'
'big key', 'bar*bar'
'healthy', 'cereal bar'
'sadly', 'without star'
The Regex:
My first success was achieved with this Regex (PCRE/Perl):
/(\n?)([^\* ].*[^ *])\s+:\s+([^\* ].*[^ *])[\s\*]+(?|\n)/g
Here the DEMO.
My question
I really find my regex pretty ugly. The main reason is because I can't use /^ and $/ in a global regex and I had to play with /(\n?)...(?|\n)/g.
Is there any possibility to shorten the above regex ?
The optional challenge
Actually this was the easy part. My string is supposed to be embedded in a C comment and I have to make sure I am not trying to match something outside a comment block.
(I not really need an answer to this second tricky question because if I write a script I can first match all the comments blocks, then find all the key:values patterns).
/********************************
* key : foo bar *
* big key : bar*bar
* healthy : /*cereal bar *
sadly : without star *
********************************/
not a key : this key

You can add the m -flag to the regexp to make anchors ^ and $ match beginnings and ends of each line within the string, i.e:
/^\s*\*?\s*([^:]+?)\s*:\s*(.*?)\s*\*?\s*$/gm
Note the use of non-greedy quantifiers (+? and *?) to not eat up characters that can be matched after the quantifier, i.e. the first capture group will not include the optional trailing whitespace before the colon, and the second capture group will not include trailing whitespace and an optional asterisk at the end of a line.

http://regex101.com/r/oJ8uW4/1
the regex I used is: /^\s*[*]*\s+(.*)\s+:\s+(.*?)\s+[*]*\s*$/gm
It works for your exemple as the not a key : this key has no space after it, so it would miss comments which do not close whith * and get values with trailing spaces too.
The point you're looking for is the modifiers after the last /
m to says it's multiline so ^ and $ are usable and g to rematch on each line.
The drawback is you can't rely on having /* and */ on lines around when using ^ and $
But Avinash will prove me wrong I bet :) (he's far better than me with regexes)

Related

Regex to replace block comment with line comment

There are tons of examples to do the conversion from C-style line comment to 1-line block comment. But I need to do the opposite: find a regex to replace multi-line block comment with line comments.
From:
This text must not be touched
/*
This
is
random
text
*/
This text must not be touched
To
This text must not be touched
// This
// is
// random
// text
This text must not be touched
I was thinking if there's a way to represent "each line" concept in regex, then just add // in front of each line. Something like
\/\*\n(?:(.+)\n)+\*\/ -> // $1
But the greediness nature of the regex engine makes $1 just match the last line before */. I know Perl and other languages have some advanced regex features like recursion, but I need to do this in a standard engine. Is there any trick to accomplish this?
EDIT: To clarify, I'm looking for pure regex solution, not involving any programming language. Should be testable on sites like https://regex101.com/.
If you are interested in a single regex pass in the modern JavaScript engine (and other regex engines supporting infinite length patterns in lookbehinds), you can use
/(?<=^(\/)\*(?:(?!^\/\*)[\s\S])*?\r?\n)(?=[\s\S]*?^\*\/)|(?:\r?\n)?(?:^\/\*|^\*\/)/gm
Replace with $1$1, see the regex demo.
Details
(?<=^(\/)\*(?:(?!^\/\*)[\s\S])*?\r?\n) - a positive lookbehind that matches a location that is immediately preceded with
^(\/)\* - /* substring at the start of a line (with / captured into Group 1)
(?:(?!^\/\*)[\s\S])*? - any char, zero or more occurrences, as few as possible, not starting a /* char sequence that appears at the start of a line
\r?\n - a CRLF or LF ending
(?=[\s\S]*?^\*\/) - a positive lookahead that requires any 0 or more chars as few as possible followed with */ at the start of a line, immediately to the right of the current location
| - or
(?:\r?\n)? - an optional CRLF or LF linebreak
(?:^\/\*|^\*\/) - and then either /* or */ at the start of a line.
As usual in such cases, two regular expressions—the second applied to the matches of the first—can do what one cannot achieve.
const txt = `This text must not be touched
/*
This
is
random
text
*/
This text must not be touched`;
const to1line = str => str.replace(
/\/\*\s*(.*?)\s*\*\//gs,
(_, comment) => comment.replace( /^/mg, '//')
);
console.log( to1line( txt ));

Regular expression for a string that does not start with a /*

I use EditPad Pro text editor.
I need read string into code, but I need to ignore ones that start with the label "/*" or tab + /*, for example:
/**
* Light up the dungeon using "claravoyance"
*
* memorizes all floor grids too.
**/
/** This function returns TRUE if a "line of sight" **/
#include "cave.h"
(tab here) /* Vertical "knights" */
if (g->multiple_objects) {
/* Get the "pile" feature instead */
k_ptr = &k_info[0];
}
put_str("Text inside", hgt - 1, (wid - COL_MAP) / 2);
/* More code*** */
I like to return:
"Text inside"
I have try this (reading Regular expression for a string that does not start with a sequence), but not work for me:
^(?! \*/\t).+".*"
any help?
Edit: I used:
^(?!#| |(\t*/)|(/)).+".*"
And it return:
put_str("Text inside"
I'm close to finding the solution.
EditPad obviously supports variable-length lookbehind in pro version 6 and lite version 7 since it's flavor is indicated as "JGsoft": Just Great Software regular expression engine.
Knowing this and without the use of capture groups, you could combine two variable length lookbehinds:
(?<!^[ \t]*/?[*#][^"\n]*")(?<=^[^"\n]*")[^"]+
(?<!^[ \t]*/?[*#][^"\n]*") The negative lookbehind for avoiding the quoted part to be preceded by [ \t]*/?[*#] any comments, which could be preceded by any amount of space/tab. Made the / optional, as a multi-line comment can also start with *.
(?<=^[^"\n]*") The positive lookbehind for assuring, that there's any amount of [^"\n], characters, that are no quotes or newlines followed by one quote before.
[^"]+ As supposed to be always balanced quoting, now it should be convenient, to match the non-quotes after the first double-quote (which is inside the lookbehind)
If a single " may occur in any line (not balanced), change the end: [^"]+ to [^"\n]+(?=")
Possibly there are different solutions for the problem. Hope it helps :)
Here's one approach: ^(?!\t*/\*).*?"(.+?)"
Breakdown:
^(?!\t*/\*) This is a negative lookahead anchored to the beginning of the line,
to ensure that there is no `/*` at the beginning (with or
without tabs)
.*?" Next is any amount of characters, up to a double-quote. It's lazy
so it stops at the first quote
(.+?)" This is the capture group for everything between the quotes, again
lazy so it doesn't slurp other quotes
You can use this regex:
/\*.*\*/(*SKIP)(*FAIL)|".*?"
Working demo
Edit: if you use EditPad then you can use this regex:
"[\w\s]+"(?!.*\*/)

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!";
}
}

Regular Expression issue with * laziness

Sorry in advance that this might be a little challenging to read...
I'm trying to parse a line (actually a subject line from an IMAP server) that looks like this:
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
It's a little hard to see, but there are two =?/?= pairs in the above line. (There will always be one pair; there can theoretically be many.) In each of those =?/?= pairs, I want the third argument (as defined by a ? delimiter) extracted. (In the first pair, it's "Here is som", and in the second it's "e text.")
Here's the regex I'm using:
=\?(.+)\?.\?(.*?)\?=
I want it to return two matches, one for each =?/?= pair. Instead, it's returning the entire line as a single match. I would have thought that the ? in the (.*?), to make the * operator lazy, would have kept this from happening, but obviously it doesn't.
Any suggestions?
EDIT: Per suggestions below to replace ".?" with "[^(\?=)]?" I'm now trying to do:
=\?(.+)\?.\?([^(\?=)]*?)\?=
...but it's not working, either. (I'm unsure whether [^(\?=)]*? is the proper way to test for exclusion of a two-character sequence like "?=". Is it correct?)
Try this:
\=\?([^?]+)\?.\?(.*?)\?\=
I changed the .+ to [^?]+, which means "everything except ?"
A good practice in my experience is not to use .*? but instead do use the * without the ?, but refine the character class. In this case [^?]* to match a sequence of non-question mark characters.
You can also match more complex endmarkers this way, for instance, in this case your end-limiter is ?=, so you want to match nonquestionmarks, and questionmarks followed by non-equals:
([^?]*\?[^=])*[^?]*
At this point it becomes harder to choose though. I like that this solution is stricter, but readability decreases in this case.
One solution:
=\?(.*?)\?=\s*=\?(.*?)\?=
Explanation:
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
\s* # Match spaces.
=\? # Literal characters '=?'
(.*?) # Match each character until find next one in the regular expression. A '?' in this case.
\?= # Literal characters '?='
Test in a 'perl' program:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group 1 -> %s\nGroup 2 -> %s\n], $1, $2 if m/=\?(.*?)\?=\s*=\?(.*?)\?=/;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?=
Running:
perl script.pl
Results:
Group 1 -> utf-8?Q?Here is som
Group 2 -> utf-8?Q?e text.
EDIT to comment:
I would use the global modifier /.../g. Regular expression would be:
/=\?(?:[^?]*\?){2}([^?]*)/g
Explanation:
=\? # Literal characters '=?'
(?:[^?]*\?){2} # Any number of characters except '?' with a '?' after them. This process twice to omit the string 'utf-8?Q?'
([^?]*) # Save in a group next characters until found a '?'
/g # Repeat this process multiple times until end of string.
Tested in a Perl script:
use warnings;
use strict;
while ( <DATA> ) {
printf qq[Group -> %s\n], $1 while m/=\?(?:[^?]*\?){2}([^?]*)/g;
}
__DATA__
=?utf-8?Q?Here is som?= =?utf-8?Q?e text.?= =?utf-8?Q?more text?=
Running and results:
Group -> Here is som
Group -> e text.
Group -> more text
Thanks for everyone's answers! The simplest expression that solved my issue was this:
=\?(.*?)\?.\?(.*?)\?=
The only difference between this and my originally-posted expression was the addition of a ? (non-greedy) operator on the first ".*". Critical, and I'd forgotten it.

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)
"(([^"\\]?(\\\\)?)|(\\")+)+"