Regular Expression to get a string between parentheses in Javascript

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

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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));
}

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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)

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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

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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]
  })

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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

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

Or \(.+\) for anything inside brackets

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

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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

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

I hope I've helped.

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