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 \"(\"

Answer 1

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

Answer 2

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

Answer 3

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));
}

Answer 4

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)

Answer 5

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));
}

Answer 6

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