Question

<?PHP
$bannedIPs = array('127.0.0.1','72.189.218.85');

function ipban() 
{
    if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) 
    { 
        echo 'test';
    }   
}

ipban();

?>

The output of this script is:

Warning: in_array() [function.in-array]: Wrong datatype for second
argument in C:\webserver\htdocs\test\array.php on line 93

Can someone tell me why? I don’t get it

And yes $_SERVER['REMOTE_ADDR'] is displaying 127.0.0.1

UPDATE

As suggested I tried this now but still get the same error

function ipban() {
    $bannedIPs = array('127.0.0.1','72.189.218.85');
    if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) { 
        echo 'test';
    }   
}
ipban();

Answer 1

You have run into a little problem with your variable scoping.

Any variables outside a function in PHP is not accessible inside. There are multiple ways to overcome this.

You could either declare $bannedIPs inside your function as such:

function ipban() {
    $bannedIPs = array('127.0.0.1','72.189.218.85');
    if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) { 
        echo 'test';
    }   
}

Tell your function to access $bannedIPs outside the function using the global keyword:

$bannedIPs = array('127.0.0.1','72.189.218.85');

function ipban() {
    global $bannedIPs;

    if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) { 
        echo 'test';
    }
}

Or, use the $GLOBALS super global:

$bannedIPs = array('127.0.0.1','72.189.218.85');

function ipban() {
    if (in_array($_SERVER['REMOTE_ADDR'], $GLOBALS['bannedIPs'])) { 
        echo 'test';
    }
}

I recommend you read the manual page on Variable scope:

PHP: Variable Scope

---

If it’s still not working, you have another problem in your code. In which case, you might want to consider using a var_dump() to check what datatype is $bannedIPs before down voting us all.

function ipban() {
    global $bannedIPs;

    var_dump($bannedIPs);
}