How can I invert a regular expression in JavaScript?

Question

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:

Answer 1

Try:

/^(?!.*foobar)/.test('foobar@bar.de')

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: http://www.regular-expressions.info/lookaround.html But Note that JavaScript only supports look-aheads, no look-behinds!

Answer 2

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

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

http://www.w3schools.com/jsref/jsref_indexOf.asp

Answer 3

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.

Answer 4

^(?!.*(word1|word2|word3))

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

^(?!$)(?!.*(word1|word2|word3))