Javascript equivalent of Perl’s \Q … \E or quotemeta()

In Perl regular expressions, you can surround a subexpression with \Q and \E to indicate that you want that subexpression to be matched as a literal string even if there are metacharacters in there. You also have the quotemeta function that inserts exactly the right number of backslashes in a string so that if you subsequently interpolate that string into a regular expression, it will be matched literally, no matter what its contents were.

Does Javascript (as deployed in major browsers) have any built in equivalent? I can write my own just fine, but I would like to know if I don’t have to bother.

There is no such built-in feature.

Rather than implementing your own, I advise you look into the multitude of regex escape functions available on the internet.

That page proposes the following solution (by Colin Snover):

RegExp.escape = function(text) {
    return text.replace(/[-[\]{}()*+?.,\\^$|#\s]/g, "\\$&");

or advises to use the XRegExp library.

Quotemeta isn’t implemented natively as far as I know, but I’ve used this a few months ago for just this:

function quotemeta (str) {
  // +   original by: Paulo Freitas
  // *     example 1: quotemeta(". + * ? ^ ( $ )");
  // *     returns 1: '\. \+ \* \? \^ \( \$ \)'
  return (str + '').replace(/([\.\\\+\*\?\[\^\]\$\(\)])/g, '\\$1');


There’s also a quotemeta npm module, which you can use in node.js or in the browser. The implementation is to quote all non-word characters, (short for [^a-zA-Z0-9_]).

String(str).replace(/(\W)/g, '\\$1');

This works because all the characters that need escaping are non-words, while the other characters that end getting escape are harmless. For example, here the percent character gets escaped, but it still matches normally in the RegExp, although it didn’t need to be escaped:

if ("Hello%".match(RegExp(String("%").replace(/(\W)/g,'\\$1')))) { console.log("matched!"); } 


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Someone has forked the quotemeta module and noted that the capturing parens aren’t needed, so the regex can be further simplified like this:

String(str).replace(/\W/g, '\\$&');

The answers/resolutions are collected from stackoverflow, are licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 .

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