Regular Expression to get a string between parentheses in Javascript

Question

I am trying to write a regular expression which returns a string which is between parentheses. For example: I want to get the string which resides between the strings "(" and ")"

I expect five hundred dollars ($500).

would return

$500

Found Regular expression to get a string between two strings in Javascript

I don't know how to use '(', ')' in regexp.

possible duplicate of Regular Expression to find a string included between two characters, while EXCLUDING the delimiters — Ro Yo Mi
– Ro Yo Mi, Commented Jul 22, 2013 at 4:27
you can use go-oleg's solution below, slightly modified to test for the dollar sign: var regExp = /$\$([^)]+)$/; — Dom Day
– Dom Day, Commented Jul 22, 2013 at 4:44

user1063287 · Accepted Answer · 2018-10-15 06:46:29Z

642

You need to create a set of escaped (with \) parentheses (that match the parentheses) and a group of regular parentheses that create your capturing group:

var regExp = /\(([^)]+)\)/;
var matches = regExp.exec("I expect five hundred dollars ($500).");

//matches[1] contains the value between the parentheses
console.log(matches[1]);

Breakdown:

\( : match an opening parentheses
( : begin capturing group
[^)]+: match one or more non ) characters
) : end capturing group
\) : match closing parentheses

Here is a visual explanation on RegExplained

edited Oct 15, 2018 at 6:46

user1063287

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answered Jul 22, 2013 at 4:11

go-oleg

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16 Comments

Praveen Over a year ago

+1 Nice, worked like a charm. I would like to add you to add this tinyurl.com/mog7lr3 in your for a visual explanation.

JayB Over a year ago

This doesn't work if there are parentheses within the parentheses you want to capture. Such as: "I expect five hundred dollars ($500 (but I know I'll never see that money again))."

silkAdmin Over a year ago

This regex works only if matching one instance per sting. Using the global /g flag wont work as expected.

sbru Over a year ago

@IanWarburton, matches[0] gives you the full string of characters matched '($500)', while matches[1] gives you just what's in the capturing group. Here is some more info on the return value of regExp.match(). developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

Mohamad Al Asmar Over a year ago

what is the solution if you have multiple strings inside parentheses like below: "I expect (five) hundred dollars ($500)" let's say I want an array of those special strings.

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Mr_Green · Accepted Answer · 2013-12-17 06:58:59Z

44

Try string manipulation:

var txt = "I expect five hundred dollars ($500). and new brackets ($600)";
var newTxt = txt.split('(');
for (var i = 1; i < newTxt.length; i++) {
    console.log(newTxt[i].split(')')[0]);
}

or regex (which is somewhat slow compare to the above)

var txt = "I expect five hundred dollars ($500). and new brackets ($600)";
var regExp = /\(([^)]+)\)/g;
var matches = txt.match(regExp);
for (var i = 0; i < matches.length; i++) {
    var str = matches[i];
    console.log(str.substring(1, str.length - 1));
}

edited Dec 17, 2013 at 6:58

answered Jul 22, 2013 at 5:06

Mr_Green

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1 Comment

Alex McMillan Over a year ago

Running that jsperf today in Chrome 59.0.3071, the regex was faster by 8%

Ali Soltani · Accepted Answer · 2020-03-22 02:33:55Z

26

Simple solution

Notice: this solution can be used for strings having only single "(" and ")" like string in this question.

("I expect five hundred dollars ($500).").match(/\((.*)\)/).pop();

Online demo (jsfiddle)

edited Mar 22, 2020 at 2:33

answered Dec 18, 2017 at 8:50

Ali Soltani

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1 Comment

dano Over a year ago

add optional chaining in case match doesnt exist will return undefined instead of resulting in uncaught exception ("I expect five hundred dollars ($500).").match(/((.*))/)?.pop();

Wiktor Stribiżew · Accepted Answer · 2020-07-01 15:10:26Z

25

To match a substring inside parentheses excluding any inner parentheses you may use

\(([^()]*)\)

pattern. See the regex demo.

In JavaScript, use it like

var rx = /\(([^()]*)\)/g;

Pattern details

\( - a ( char
([^()]*) - Capturing group 1: a negated character class matching any 0 or more chars other than ( and )
\) - a ) char.

To get the whole match, grab Group 0 value, if you need the text inside parentheses, grab Group 1 value.

Most up-to-date JavaScript code demo (using matchAll):

const strs = ["I expect five hundred dollars ($500).", "I expect.. :( five hundred dollars ($500)."];
const rx = /\(([^()]*)\)/g;
strs.forEach(x => {
  const matches = [...x.matchAll(rx)];
  console.log( Array.from(matches, m => m[0]) ); // All full match values
  console.log( Array.from(matches, m => m[1]) ); // All Group 1 values
});

Legacy JavaScript code demo (ES5 compliant):

var strs = ["I expect five hundred dollars ($500).", "I expect.. :( five hundred dollars ($500)."];
var rx = /\(([^()]*)\)/g;


for (var i=0;i<strs.length;i++) {
  console.log(strs[i]);

  // Grab Group 1 values:
  var res=[], m;
  while(m=rx.exec(strs[i])) {
    res.push(m[1]);
  }
  console.log("Group 1: ", res);

  // Grab whole values
  console.log("Whole matches: ", strs[i].match(rx));
}

edited Jul 1, 2020 at 15:10

answered Jul 12, 2018 at 8:21

Wiktor Stribiżew

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5 Comments

limitlessloop Over a year ago

This regex is great because it only captures pairs of () and their contents, even when side by side()().

Wiktor Stribiżew Over a year ago

@limitlessloop Since this seems a helpful answer, I also added a new matchAll based JavaScript code demo.

GeminiYellow Over a year ago

this answer is the best.

postgresnewbie Over a year ago

How to include parantheses, can you give an example about that too?

Wiktor Stribiżew Over a year ago

@postgresnewbie It is already there, console.log( Array.from(matches, m => m[0]) ); // All full match values. You do not need the group then, and can use var rx = /$[^()]*$/g;

Community · Accepted Answer · 2017-05-23 12:02:47Z

13

Ported Mr_Green's answer to a functional programming style to avoid use of temporary global variables.

var matches = string2.split('[')
  .filter(function(v){ return v.indexOf(']') > -1})
  .map( function(value) { 
    return value.split(']')[0]
  })

edited May 23, 2017 at 12:02

CommunityBot

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answered Dec 26, 2015 at 3:58

silkAdmin

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1 Comment

wobsoriano Over a year ago

What I was looking for. stackoverflow.com/questions/49614116/…

marcofriso · Accepted Answer · 2020-04-21 17:32:46Z

8

Alternative:

var str = "I expect five hundred dollars ($500) ($1).";
str.match(/\(.*?\)/g).map(x => x.replace(/[()]/g, ""));

→ (2) ["$500", "$1"]

It is possible to replace brackets with square or curly brackets if you need

edited Apr 21, 2020 at 17:32

answered Apr 21, 2020 at 17:13

marcofriso

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1 Comment

Ranjith Varatharajan Over a year ago

This is broken when str has no brackets

P0W · Accepted Answer · 2013-07-22 04:11:56Z

6

For just digits after a currency sign : $.+\s*\d+\s*$ should work

Or $.+$ for anything inside brackets

answered Jul 22, 2013 at 4:11

P0W

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3 Comments

Ro Yo Mi Over a year ago

This will work for the sample text, however if another ) appears later in the string like I expect five hundred dollars ($500). (but I'm going to pay). the greediness of the + will capture a lot more.

Tom Bird Over a year ago

@Denomales So how would you have it capture both instances of those strings. Ideally I am doing something similar but I would like to have: ($500) and (but I'm going to pay) be matched but in two separates matches rather than being considered a single matched item? Thanks in advance!

Ojchris Over a year ago

(.+) worked for my requirement. Needed to capture stuffs like "Dunhill (Bottle/Packet/Pot/Shot)" and return just "Dunhill"

Andran · Accepted Answer · 2019-10-14 20:40:20Z

4

let str = "Before brackets (Inside brackets) After brackets".replace(/.*\(|\).*/g, '');
console.log(str) // Inside brackets

answered Oct 14, 2019 at 20:40

Andran

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Comments

AJ Dhaliwal · Accepted Answer · 2015-11-20 13:06:52Z

2

var str = "I expect five hundred dollars ($500) ($1).";
var rex = /\$\d+(?=\))/;
alert(rex.exec(str));

Will match the first number starting with a $ and followed by ')'. ')' will not be part of the match. The code alerts with the first match.

var str = "I expect five hundred dollars ($500) ($1).";
var rex = /\$\d+(?=\))/g;
var matches = str.match(rex);
for (var i = 0; i < matches.length; i++)
{
    alert(matches[i]);
}

This code alerts with all the matches.

References:

search for "?=n" http://www.w3schools.com/jsref/jsref_obj_regexp.asp

search for "x(?=y)" https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/RegExp

answered Nov 20, 2015 at 13:06

AJ Dhaliwal

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Comments

Bryan Henrique Cruz · Accepted Answer · 2020-12-23 19:43:13Z

0

Simple: (?<value>(?<=$).*(?=$))

I hope I've helped.

answered Dec 23, 2020 at 19:43

Bryan Henrique Cruz

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