I want to remove not only spaces, but certain characters, as well from the beginning or end of a JavaScript string.

function trim(str, characters) {
  var c_array = characters.split('');
  var result  = '';

  for (var i=0; i < characters.length; i++)
    result += '\\' + c_array[i];

  return str.replace(new RegExp('^[' + result + ']+|['+ result +']+$', 'g'), '');

The function can be used as follows:

trim(' - hello**   +', '+*- ');

The escaping is required as + * . etc. would be used in the regular expression. Is there a better way to trim any character in JavaScript? Or a better way to do the escaping, as not every character needs it?


1 Answer 1


There are many different ways to tackle trimming , a is probably the most popular, and there are lots of blogs and performance tests out there, (one such blog), that you can look at and decide which is best.

So continuing along the lines that you have demonstrated, then this may be of some help to you.


/*jslint maxerr: 50, indent: 4, browser: true */

var trim = (function () {
    "use strict";

    function escapeRegex(string) {
        return string.replace(/[\[\](){}?*+\^$\\.|\-]/g, "\\$&");

    return function trim(str, characters, flags) {
        flags = flags || "g";
        if (typeof str !== "string" || typeof characters !== "string" || typeof flags !== "string") {
            throw new TypeError("argument must be string");

        if (!/^[gi]*$/.test(flags)) {
            throw new TypeError("Invalid flags supplied '" + flags.match(new RegExp("[^gi]*")) + "'");

        characters = escapeRegex(characters);

        return str.replace(new RegExp("^[" + characters + "]+|[" + characters + "]+$", flags), '');

/*jslint devel: true */

console.log(trim(" - hello** +", "+*- "));

On jsfiddle

Of course, you can customise which characters you wish to escape, or flags to test for, as per your own needs. Regular Expressions

  • \$\begingroup\$ Good solution, but please use following function to determine if the parameter is a string: var isString = function(str) { return typeof str == 'string' || str instanceof String; }; If you try to use a string like this new String('123') it won't work... \$\endgroup\$
    – algorhythm
    Commented May 6, 2014 at 8:52
  • \$\begingroup\$ If that was a concern then I would probably go for Object.prototype.toString.call(str) === '[object String]' as a safer option. \$\endgroup\$
    – Xotic750
    Commented May 6, 2014 at 14:52

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