PHP preg_match？ 如何返回不匹配的字符？

Question

Lets say i have:

$string = 'qwe1ASD@';

if(preg_match('/^[a-zA-Z]+$/', $string))
{
    echo 'OK';
}
else
{
    echo 'BAD';
}

Now, is there simple solution, to find all characters from $string which don't match expression? So in return, in place of "BAD" i want to have ex. "BAD. You can't use following characters: 1@"

Thanks in advance for any simple hints! :)

---

Thank you Floern, your answer suit best my needs. It have only one "preg" so it's also good for performance. Thank you again. I implemented it for now as follw:

if(preg_match_all('/[^a-zA-Z0-9]/s', $string, $forbidden))
{
    $forbidden = implode('', array_unique($forbidden[0]));

    echo 'BAD. Your string contains forbidden characters: '.htmlentities($forbidden).'';
}

Answer 1

$tmpstring=preg_replace('~[A-Za-z]~','',$string);
if(strlen($tmpstring))
    //bad chars: $tmpstring

Answer 2

You could use preg_match_all() :

if(preg_match_all('/[^a-zA-Z]/s', $string, $matches)){
    var_dump($matches); // BAD
}
else{
    echo 'OK';
}

Answer 3

$string = 'qwe1ASD@';

if(preg_match('/^[a-zA-Z]+$/', $string))
{
    echo 'OK';
}
else
{
    echo 'BAD.  You cannot use the following characters: ' + preg_replace('/[a-zA-Z]/', '', $string);
}

Answer 4

There are different ways. I find this one nice:

$check = preg_prelace('/[a-zA-Z]/', '', $string);

if ($check) echo 'BAD ' . $check;

UPDATE:

if (strlen($check)) echo 'BAD ' . $check;