Regex to find multiple words that don't match (Negative words matching)

Question

I need to allow only a specific word per time per string, but exists multiple valid words.

Using the below regex i am not getting the expected result:

preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'false'); // Returns an empty string, but i expect 'false'

preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'test') // Returns 'test', but i expect an empty string

Someone knows what's wrong?

EDIT 1

An example for a long input:

preg_replace('/\b^(true|false|TRUE|FALSE)\b/', '', 'This regex allow only boolean, such as true, false, TRUE and FALSE');

It prints:

// This regex allow only boolean, such true, false, TRUE and FALSE

But i expect an empty string, because only a sigle word should be a valid match

Answer 1

In this case you don't need a regex, you can use in_array:

$arr = array('true', 'false', 'TRUE', 'FALSE');

$result = (in_array($str, $arr, true)) ? $str : '';

Answer 2

Your regex is wrong for what you're trying to do. You will need negative lookahead.

Try this code:

$re = '/\b(?!(?:true|false|TRUE|FALSE))\w+\b/'; 
$str = 'I need true to allow only false specific FALSE words in a TRUE string'; 

$repl = preg_replace($re, "", $str);
//=>   true    false  FALSE    TRUE 

Online Demo

Answer 3

Not sure I well understand your needs, but is that OK for you:

if (preg_match('/^(true|false|TRUE|FALSE)$/', $string, $match)) {
    echo "Found : ",$m[1],"\n";
} else {
    echo "Not found\n";
}

Answer 4

Is this what you want?

(?=.)(true|false|TRUE|FALSE|\s*)(?!.)

Debuggex Demo

$regex = '/(?=.)(true|false|TRUE|FALSE|\\s{0,})(?!.)/';
$testString = ''; // Fill this in
preg_match($regex, $testString, $matches);
// the $matches variable contains the list of matches