How can I invert a regular expression in JavaScript?

I have a string A and want to test if another string B is not part of it. This is a very simple regex whose result can be inverted afterwards.

I could do:

/foobar/.test('[email protected]')

and invert it afterwards, like this:

!(/foobar/).test('[email protected]') 

The problem I have is, that I need to do it within the regular expression and not with their result. Something like:

/!foobar/.test('[email protected]')

(which does not work)

In other words: the regular expression should test for a non-existence and return true in that case.

Is this possible with JavaScript?



will match a string that does not contain any of word1, word2, or word3 (and you can extend the list indefinitely). But this also matches null strings. To reject nulls use


If what you're searching for really isn't more complicated than a simple string like "foobar":

if (yourString.indexOf("foobar") === -1) {
  // ...



/^(?!.*foobar)/.test('[email protected]')

A (short) explanation:

^          # start of the string 
(?!        # start negative look-ahead
  .*       # zero or more characters of any kind (except line terminators)
  foobar   # foobar
)          # end negative look-ahead

So, in plain English, that regex will look from the start of the string if the string 'foobar' can be "seen". If it can be "seen" there is no* match.

* no match because it's negative look-ahead!

More about this look-ahead stuff: But Note that JavaScript only supports look-aheads, no look-behinds!


Here's an example of an inequality. First I isolate the operator '<', later the operands 'a' and 'b'. Basically, I take the direct expression, include it into right parentheses, invert the latter by '^' and finally embed the resulting expression into square brackets, 'cause the '^' at the beginning would be interpreted differently.

var _str = "a < b" ;
var _op = /</g ;
var _no_op = /[^(<|\ )]/g ;
console.log( _str, _str.match( _op ) ); // get '<'
console.log( _str, _str.match( _no_op ) ); // get 'a', 'b'

P.s.: I just added the blank space in the inverse expression, in order to retrieve exact matching for the operands.


Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.