Javascript Regex to Extract a Number

Question

I have strings such as

(123)abc
defg(456)
hijkl
(999)
7

I want to match these strings one at a time with a regular expression to extract any number from the string where the string starts with a '(' has a number in between and then a ')' followed by zero or more characters. So, in the above 5 examples I would match 123 in the first case, nothing in the second and third case, 999 in the fourth case and nothing in the fifth case.

I have tried this

var regex = new RegExp("^\((\d+)\)", "gm");
var matches = str.match(regex);

But matches always comes out as null. What am I doing wrong?

I have tried this regex at regex101 and it seems to work so I am at a loss why the code doesn't work.

• Jonathan Lonowski: When using the RegExp constructor, you'll need to provide escape sequences for both the pattern and the string literal -- new RegExp("^\\((\\d+)\\)", "gm").

Answer 1

You need to push the result of the capturing group to an array, for this use the exec() method.

var str = '(123)abc\ndefg(456)\nhijkl\n(999)\n7'
var re = /^\((\d+)\)/gm, 
matches = [];
while (m = re.exec(str)) {
 matches.push(m[1]);
}
console.log(matches) //=> [ '123', '999' ]

Answer 2

I don't see anything wrong with your regex, try this code generated from regex101:

var re = /^\((\d+)\)/gm; 
var str = '(123)abc\ndefg(456)\nhijkl\n(999)\n7';
var m;
while ((m = re.exec(str)) !== null) {
 if (m.index === re.lastIndex) {
 re.lastIndex++;
 }
 // View your result using the m-variable.
 // eg m[0] etc.
}

Working demo

Btw, as jonathan lonowski pointed in his comment you have to escape backslashes when you use RegExp:

new RegExp("^\\((\\d+)\\)", "gm")

Answer 3

You can use this regex:

var regex = /^\((\d+)(?=\))/gm;

and use captured group #1

RegEx Demo

Using regex constructor (note double escaping):

var regex = new RegExp("^\\((\\d+)(?=\\))", "gm");