454

I need a regular expression to select all the text between two outer brackets.

Example:
START_TEXT(text here(possible text)text(possible text(more text)))END_TXT
^ ^

Result:
(text here(possible text)text(possible text(more text)))

3
  • 7
    This question is very poor because it's not clear what it's is asking. All of the answers interpreted it differently. @DaveF can you please clarify the question? Commented Dec 17, 2012 at 18:25
  • 2
    Answered in this post: stackoverflow.com/questions/6331065/… Commented Dec 6, 2013 at 22:47
  • Not all regular expression processing engines are the same. Knowing the context, for example, what language hosts the reg ex processor is important, which is why bobble bubbles answer includes so many answers that depend on the host. Commented Sep 23, 2024 at 14:52

23 Answers 23

296

I want to add this answer for quickreference. Feel free to update.


.NET Regex using balancing groups:

\((?>\((?<c>)|[^()]+|\)(?<-c>))*(?(c)(?!))\) 

Where c is used as the depth counter.

Demo at Regexstorm.com


PCRE using a recursive pattern:

\((?:[^)(]+|(?R))*+\) 

Demo at regex101; Or without alternation:

\((?:[^)(]*(?R)?)*+\) 

Demo at regex101; Or unrolled for performance:

\([^)(]*+(?:(?R)[^)(]*)*+\) 

Demo at regex101; The pattern is pasted at (?R) which represents (?0).

Perl, PHP, Notepad++, R: perl=TRUE, Python: PyPI regex module with (?V1) for Perl behaviour.
(the new version of PyPI regex package already defaults to this → DEFAULT_VERSION = VERSION1)


Ruby using subexpression calls:

With Ruby 2.0 \g<0> can be used to call full pattern.

\((?>[^)(]+|\g<0>)*\) 

Demo at Rubular; Ruby 1.9 only supports capturing group recursion:

(\((?>[^)(]+|\g<1>)*\)) 

Demo at Rubular  (atomic grouping since Ruby 1.9.3)


JavaScript  API :: XRegExp.matchRecursive or regex-recursion

XRegExp.matchRecursive(str, '\\(', '\\)', 'g'); 

Java: An interesting idea using forward references by @jaytea.


Without recursion up to 3 levels of nesting:
(JS, Java and other regex flavors)

To prevent runaway if unbalanced, with * on innermost [)(] only.

\((?:[^)(]|\((?:[^)(]|\((?:[^)(]|\([^)(]*\))*\))*\))*\) 

Demo at regex101; Or unrolled for better performance (preferred).

\([^)(]*(?:\([^)(]*(?:\([^)(]*(?:\([^)(]*\)[^)(]*)*\)[^)(]*)*\)[^)(]*)*\) 

Demo at regex101; Deeper nesting needs to be added as required.

// JS-Snippet to generate pattern function generatePattern() { // Set max depth & pattern type let d = document.getElementById("maxDepth").value; let t = document.getElementById("patternType").value; // Pattern variants: 0=default, 1=unrolled (more efficient) let p = [['\\((?:[^)(]|',')*\\)'], ['\\([^)(]*(?:','[^)(]*)*\\)']]; // Generate and display the pattern console.log(p[t][0].repeat(d) + '\\([^)(]*\\)' + p[t][1].repeat(d)); } generatePattern();
Max depth = <input type="text" id="maxDepth" size="1" value="3"> <select id="patternType" onchange="generatePattern()"> <option value="0">default pattern</option> <option value="1" selected>unrolled pattern</option> </select> <input type="submit" onclick="generatePattern()" value="generate!">


Reference - What does this regex mean?

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17 Comments

When you repeat a group with a possessive quantifier, it's useless to make that group atomic since all backtracking positions in that group are deleted at each repetition. So writing (?>[^)(]+|(?R))*+ is the same than writing (?:[^)(]+|(?R))*+. Same thing for the next pattern. About the unrolled version, you can put a possessive quantifier here: [^)(]*+ to prevent backtracking (in case there's no closing bracket).
@CasimiretHippolyte Thank you! I adjusted the PCRE patterns and for Ruby 1.9, do you mean the whole pattern to be like this? Please feel free to update yourself. I understand what you mean, but not sure if there is much improvement.
Thank you, this answer is amazing, all the effort with possible demos allowed me to pick the one I use (regex101.org). I'm very grateful for your work
In case anyone needs a curly bracket version of this for .NET: \{(?>\{(?<c>)|[^{}]+|\}(?<-c>))*(?(c)(?!))\}
For the recursion, instead of \((?:[^)(]+|(?R))*+\) I would recommend (\((?:[^)(]+|(?1))*+\)) (or ?2, ?3, etc, depending on what number group it is). ?R always recurses back to the very beginning of the expression. Which, if you're using this alone, is fine. But for example, if you're finding logical comparisons following an if statement if \((?:[^)(]+|(?R))*+\) won't match anything because the if would also have to be repeated to match, not just the parentheses. if (\((?:[^)(]+|(?1))*+\)) however, will only check for if once and then recursively check the first group.
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187

Regular expressions are the wrong tool for the job because you are dealing with nested structures, i.e. recursion.

But there is a simple algorithm to do this, which I described in more detail in this answer to a previous question. The gist is to write code which scans through the string keeping a counter of the open parentheses which have not yet been matched by a closing parenthesis. When that counter returns to zero, then you know you've reached the final closing parenthesis.

11 Comments

.NET's implementation has [Balancing Group Definitions msdn.microsoft.com/en-us/library/… which allow this sort of thing.
I disagree that regular expressions are the wrong tool for this for a few reasons. 1) Most regular expression implementations have a workable if not perfect solution for this. 2) Often you are trying to find balanced pairs of delimiters in a context where other criteria well suited to regular expressions are also in play. 3) Often you are handing a regular expression into some API that only accepts regular expressions and you have no choice.
Regex is the RIGHT tool for the job. This answer is not right. See rogal111's answer.
Absolutely agree with the answer. Although there are some implementations of recursion in regexp, they are equal to finite-state machines and are not supposted to work with nested structures, but Context Free Grammars do this. Look at Homsky's hierarcy of Formal Grammars.
Frank is right, Context free grammars cannot be described by regular expressions. That's the key point to this answer.
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144

You can use regex recursion:

\(([^()]|(?R))*\) 

13 Comments

An example would be really useful here, I can't get this to work for things like "(1, (2, 3)) (4, 5)".
@AndyHayden this is because "(1, (2, 3)) (4, 5)" has two groups separated with space. Use my regexp with global flag: /(([^()]|(?R))*)/g. Here is online test: regex101.com/r/lF0fI1/1
In .NET 4.5 I get the following error for this pattern: Unrecognized grouping construct.
Awesome! This is a great feature of regex. Thank you for being the only one to actually answer the question. Also, that regex101 site is sweet.
As a side note, this is kind of misnamed because a real regex isn't recursive.
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35
[^\(]*(\(.*\))[^\)]* 

[^\(]* matches everything that isn't an opening bracket at the beginning of the string, (\(.*\)) captures the required substring enclosed in brackets, and [^\)]* matches everything that isn't a closing bracket at the end of the string. Note that this expression does not attempt to match brackets; a simple parser (see dehmann's answer) would be more suitable for that.

4 Comments

the bracket inside the class does not need to be escaped. Since inside it is not a metacharacted.
This expr fails against something like "text(text)text(text)text" returning "(text)text(text)". Regular expressions can't count brackets.
@ChristianKlauser this /\(([^()]|(?R))*\)/g works great, check this demo.
Solution pointed out by @AmineKOUIS will not work in languages like javascript(I am aware of the fact that op has nowhere mentioned that he/she is using js) as they don't support recursive regex. thought of sharing this as it might be helpful to someone.
30

This answer explains the theoretical limitation of why regular expressions are not the right tool for this task.


Regular expressions can not do this.

Regular expressions are based on a computing model known as Finite State Automata (FSA). As the name indicates, a FSA can remember only the current state, it has no information about the previous states.

FSA

In the above diagram, S1 and S2 are two states where S1 is the starting and final step. So if we try with the string 0110 , the transition goes as follows:

 0 1 1 0 -> S1 -> S2 -> S2 -> S2 ->S1 

In the above steps, when we are at second S2 i.e. after parsing 01 of 0110, the FSA has no information about the previous 0 in 01 as it can only remember the current state and the next input symbol.

In the above problem, we need to know the no of opening parenthesis; this means it has to be stored at some place. But since FSAs can not do that, a regular expression can not be written.

However, an algorithm can be written to do this task. Algorithms are generally falls under Pushdown Automata (PDA). PDA is one level above of FSA. PDA has an additional stack to store some additional information. PDAs can be used to solve the above problem, because we can 'push' the opening parenthesis in the stack and 'pop' them once we encounter a closing parenthesis. If at the end, stack is empty, then opening parenthesis and closing parenthesis matches. Otherwise not.

5 Comments

There are several answers here, which prooves, it IS possible.
@Marco This answer talks about regular expressions in theoretical perspective. Many regex engines now a days does not only rely on this theoretical model and uses some additional memory to do the job!
@JiříHerník: those are not regular expressions in the strict sense: not defined as regular expressions by Kleene. Some regular expression engines indeed have implemented some extra capabilities, making them parse more than only regular languages.
This one should be an accepted answer. Unfortunately many "developers" do not have a proper Comp Sc/Eng education and unaware of such topics as Halting problem, Pumping lemma, etc...
28
(?<=\().*(?=\)) 

If you want to select text between two matching parentheses, you are out of luck with regular expressions. This is impossible(*).

This regex just returns the text between the first opening and the last closing parentheses in your string.


(*) Unless your regex engine has features like balancing groups or recursion. The number of engines that support such features is slowly growing, but they are still not a commonly available.

6 Comments

What do the "<=" and "=" signs mean? What regexp engine is this expression targeting?
This is look-around, or more correctly "zero width look-ahead/look-behind assertions". Most modern regex engines support them.
According to the OP's example, he wants to include the outermost parens in the match. This regex throws them away.
@Alan M: You are right. But according to the question text, he wants everything between the outermost parens. Pick your choice. He said he'd been trying for hours, so didn't even consider "everything including the outermost parens" as the intention, because it is so trivial: "(.*)".
@ghayes The answer is from 2009. That is a long time ago; regular expression engines that allow some form of recursion have been more uncommon than they are now (and they still are pretty uncommon). I'll mention it in my answer.
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15

It is actually possible to do it using .NET regular expressions, but it is not trivial, so read carefully.

You can read a nice article here. You also may need to read up on .NET regular expressions. You can start reading here.

Angle brackets <> were used because they do not require escaping.

The regular expression looks like this:

< [^<>]* ( ( (?<Open><) [^<>]* )+ ( (?<Close-Open>>) [^<>]* )+ )* (?(Open)(?!)) > 

Comments

14

I was also stuck in this situation when dealing with nested patterns and regular-expressions is the right tool to solve such problems.

/(\((?>[^()]+|(?1))*\))/ 

3 Comments

As a user looking for help on a similar topic, I have no idea what that regex does specifically and how I can use it to apply it to my own problem. Perhaps this is a good answer but given the nature of regex being cryptic, I would have to look up every part of it just to see if this would help me. Given that there are so many answers with this type of "solution", I don't think I will.
regex101.com is a good explainer tool to exlpain this regex.
This PCRE solution using ?1 works with (*SKIP)(*FAIL) when the ?R method posted by bobble bubble does not. Use this solution if you need to find things that ARE NOT in parenthesis. I created an example here: regex101.com/r/xkVzVP/1
6

This is the definitive regex:

\( (?<arguments> ( ([^\(\)']*) | (\([^\(\)']*\)) | '(.*?)' )* ) \) 

Example:

input: ( arg1, arg2, arg3, (arg4), '(pip' ) output: arg1, arg2, arg3, (arg4), '(pip' 

note that the '(pip' is correctly managed as string. (tried in regulator: http://sourceforge.net/projects/regulator/)

1 Comment

I like this technique if there's no nesting or you only care about the innermost group. It doesn't rely on recursion. I was able to use it to extract an argument that contained parenthesis. I made a working example at Regex101
6

Adding to bobble bubble's answer, there are other regex flavors where recursive constructs are supported.

Lua

Use %b() (%b{} / %b[] for curly braces / square brackets):

  • for s in string.gmatch("Extract (a(b)c) and ((d)f(g))", "%b()") do print(s) end (see demo)

Raku (former Perl6):

Non-overlapping multiple balanced parentheses matches:

my regex paren_any { '(' ~ ')' [ <-[()]>+ || <&paren_any> ]* } say "Extract (a(b)c) and ((d)f(g))" ~~ m:g/<&paren_any>/; # => (「(a(b)c)」 「((d)f(g))」) 

Overlapping multiple balanced parentheses matches:

say "Extract (a(b)c) and ((d)f(g))" ~~ m:ov:g/<&paren_any>/; # => (「(a(b)c)」 「(b)」 「((d)f(g))」 「(d)」 「(g)」) 

See demo.

Python re non-regex solution

See poke's answer for How to get an expression between balanced parentheses.

Java customizable non-regex solution

Here is a customizable solution allowing single character literal delimiters in Java:

public static List<String> getBalancedSubstrings(String s, Character markStart, Character markEnd, Boolean includeMarkers) { List<String> subTreeList = new ArrayList<String>(); int level = 0; int lastOpenDelimiter = -1; for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == markStart) { level++; if (level == 1) { lastOpenDelimiter = (includeMarkers ? i : i + 1); } } else if (c == markEnd) { if (level == 1) { subTreeList.add(s.substring(lastOpenDelimiter, (includeMarkers ? i + 1 : i))); } if (level > 0) level--; } } return subTreeList; } } 

Sample usage:

String s = "some text(text here(possible text)text(possible text(more text)))end text"; List<String> balanced = getBalancedSubstrings(s, '(', ')', true); System.out.println("Balanced substrings:\n" + balanced); // => [(text here(possible text)text(possible text(more text)))] 

1 Comment

See an online Java demo for a proof it works with multiple matches.
5

The regular expression using Ruby (version 1.9.3 or above):

/(?<match>\((?:\g<match>|[^()]++)*\))/ 

Demo on rubular

Comments

5

I have written a little JavaScript library called balanced to help with this task. You can accomplish this by doing

balanced.matches({ source: source, open: '(', close: ')' }); 

You can even do replacements:

balanced.replacements({ source: source, open: '(', close: ')', replace: function (source, head, tail) { return head + source + tail; } }); 

Here's a more complex and interactive example JSFiddle.

Comments

2
""" Here is a simple python program showing how to use regular expressions to write a paren-matching recursive parser. This parser recognises items enclosed by parens, brackets, braces and <> symbols, but is adaptable to any set of open/close patterns. This is where the re package greatly assists in parsing. """ import re # The pattern below recognises a sequence consisting of: # 1. Any characters not in the set of open/close strings. # 2. One of the open/close strings. # 3. The remainder of the string. # # There is no reason the opening pattern can't be the # same as the closing pattern, so quoted strings can # be included. However quotes are not ignored inside # quotes. More logic is needed for that.... pat = re.compile(""" ( .*? ) ( \( | \) | \[ | \] | \{ | \} | \< | \> | \' | \" | BEGIN | END | $ ) ( .* ) """, re.X) # The keys to the dictionary below are the opening strings, # and the values are the corresponding closing strings. # For example "(" is an opening string and ")" is its # closing string. matching = { "(" : ")", "[" : "]", "{" : "}", "<" : ">", '"' : '"', "'" : "'", "BEGIN" : "END" } # The procedure below matches string s and returns a # recursive list matching the nesting of the open/close # patterns in s. def matchnested(s, term=""): lst = [] while True: m = pat.match(s) if m.group(1) != "": lst.append(m.group(1)) if m.group(2) == term: return lst, m.group(3) if m.group(2) in matching: item, s = matchnested(m.group(3), matching[m.group(2)]) lst.append(m.group(2)) lst.append(item) lst.append(matching[m.group(2)]) else: raise ValueError("After <<%s %s>> expected %s not %s" % (lst, s, term, m.group(2))) # Unit test. if __name__ == "__main__": for s in ("simple string", """ "double quote" """, """ 'single quote' """, "one'two'three'four'five'six'seven", "one(two(three(four)five)six)seven", "one(two(three)four)five(six(seven)eight)nine", "one(two)three[four]five{six}seven<eight>nine", "one(two[three{four<five>six}seven]eight)nine", "oneBEGINtwo(threeBEGINfourENDfive)sixENDseven", "ERROR testing ((( mismatched ))] parens"): print "\ninput", s try: lst, s = matchnested(s) print "output", lst except ValueError as e: print str(e) print "done" 

3 Comments

Thank you for this very valuable example code. ChatGPT is still horrible in creating such code. If you don't mind, I would like to publish it as a GitHub gist. The only change I've made to your original code in order to make it more robust is to allow the matching terms as a function parameter (therefore a separate function for recursive calls): gist.github.com/dmikushin/a47e1444c7a32f208db4d8579ce0a13d
I know I am being a putz for saying this, but I am uncomfortable with the Phrase "Published by Gene Olson". I think you should take credit by saying "Adapted from ...".
I would have written your version to use two functions. One would generate the pattern and the dictionary. The other would match the text. I prefer not to repeatedly generate the same pattern in functions that might be called many times. This is especially true in recursive functions. Pattern generation is an expensive operation.
1

The answer depends on whether you need to match matching sets of brackets, or merely the first open to the last close in the input text.

If you need to match matching nested brackets, then you need something more than regular expressions. - see @dehmann

If it's just first open to last close see @Zach

Decide what you want to happen with:

abc ( 123 ( foobar ) def ) xyz ) ghij 

You need to decide what your code needs to match in this case.

2 Comments

This is not an answer.
Yes, the demand for a change in the question should be given as a commentary,
1

You need the first and last parentheses. Use something like this:

str.indexOf('('); - it will give you first occurrence

str.lastIndexOf(')'); - last one

So you need a string between,

String searchedString = str.substring(str1.indexOf('('),str1.lastIndexOf(')'); 

Comments

1

As Uniform answer:

This would be much more efficient and portable to most languages if you use the regex to do that it building for - fast pattern matching. (not logic by it) some languages has nested regex features, but even in it this be less functionality, then in Stack algorithm working over the regex base.

Even if you realy need ony top level brace select you better check stack solution (cause of portability regex code, and efficient)

Here is the example by JS:

let[struct]='a{aa},b{},c{ca,cb},d{da{daa,dab},db}' .matchAll(/([^,\{\}]+)(\{?)(\}?)(,?)/g).reduce(($,[_,key,brO,brC,com])=>{ let val=brO?{}:$.n++; // define nested or primitive value // primitive can set by more complex regex logic above just 'key' grp // for example primitive set just natural index (starts from zero) $.at(-1)[key]=val; // set key=val to the current level of object if(brO)$.push(val); // push stack level if need if(brC)$.pop(); // pop stack level if need return $; // return stack },Object.assign([{}],{n:0})) console.log("struct:",struct) 

you can reduce this version

  • for only selecting top level baraket chank (but this step most likely will also only the part of you job, already or soon)

or you can code above:

  • to chank the key part more complex (e.g. split it to 'key:type' to add some primitive Type aliases)
  • to create some custom JSON alternative format
  • to test in parallel some other structure by string pattern

Comments

0

because js regex doesn't support recursive match, i can't make balanced parentheses matching work.

so this is a simple javascript for loop version that make "method(arg)" string into array

push(number) map(test(a(a()))) bass(wow, abc) $$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups) 
const parser = str => { let ops = [] let method, arg let isMethod = true let open = [] for (const char of str) { // skip whitespace if (char === ' ') continue // append method or arg string if (char !== '(' && char !== ')') { if (isMethod) { (method ? (method += char) : (method = char)) } else { (arg ? (arg += char) : (arg = char)) } } if (char === '(') { // nested parenthesis should be a part of arg if (!isMethod) arg += char isMethod = false open.push(char) } else if (char === ')') { open.pop() // check end of arg if (open.length < 1) { isMethod = true ops.push({ method, arg }) method = arg = undefined } else { arg += char } } } return ops } // const test = parser(`$$(groups) filter({ type: 'ORGANIZATION', isDisabled: { $ne: true } }) pickBy(_id, type) map(test()) as(groups)`) const test = parser(`push(number) map(test(a(a()))) bass(wow, abc)`) console.log(test) 

the result is like

[ { method: 'push', arg: 'number' }, { method: 'map', arg: 'test(a(a()))' }, { method: 'bass', arg: 'wow,abc' } ] 
[ { method: '$$', arg: 'groups' }, { method: 'filter', arg: '{type:\'ORGANIZATION\',isDisabled:{$ne:true}}' }, { method: 'pickBy', arg: '_id,type' }, { method: 'map', arg: 'test()' }, { method: 'as', arg: 'groups' } ] 

Comments

0

While so many answers mention this in some form by saying that regex does not support recursive matching and so on, the primary reason for this lies in the roots of the Theory of Computation.

Language of the form {a^nb^n | n>=0} is not regular. Regex can only match things that form part of the regular set of languages.

Read more @ here

Comments

0

I didn't use regex since it is difficult to deal with nested code. So this snippet should be able to allow you to grab sections of code with balanced brackets:

def extract_code(data): """ returns an array of code snippets from a string (data)""" start_pos = None end_pos = None count_open = 0 count_close = 0 code_snippets = [] for i,v in enumerate(data): if v =='{': count_open+=1 if not start_pos: start_pos= i if v=='}': count_close +=1 if count_open == count_close and not end_pos: end_pos = i+1 if start_pos and end_pos: code_snippets.append((start_pos,end_pos)) start_pos = None end_pos = None return code_snippets 

I used this to extract code snippets from a text file.

Comments

0

This do not fully address the OP question but I though it may be useful to some coming here to search for nested structure regexp:

Parse parmeters from function string (with nested structures) in javascript

Match structures like:
Parse parmeters from function string

  • matches brackets, square brackets, parentheses, single and double quotes

Here you can see generated regexp in action

/** * get param content of function string. * only params string should be provided without parentheses * WORK even if some/all params are not set * @return [param1, param2, param3] */ exports.getParamsSAFE = (str, nbParams = 3) => { const nextParamReg = /^\s*((?:(?:['"([{](?:[^'"()[\]{}]*?|['"([{](?:[^'"()[\]{}]*?|['"([{][^'"()[\]{}]*?['")}\]])*?['")}\]])*?['")}\]])|[^,])*?)\s*(?:,|$)/; const params = []; while (str.length) { // this is to avoid a BIG performance issue in javascript regexp engine str = str.replace(nextParamReg, (full, p1) => { params.push(p1); return ''; }); } return params; }; 

Comments

0

A Vim regex that achieve that is:

:%s/\(.\{-}\)\((\(.\{-}(\)\+.\{-}\().\{-}\)\+)\)\(.*\)/\2/g 

Comments

-2

This might help to match balanced parenthesis.

\s*\w+[(][^+]*[)]\s* 

Comments

-4

This one also worked

re.findall(r'\(.+\)', s) 

Comments

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