php in array error

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

What happens if you move $bannedIPs inside the function declaration? It's possible PHP thinks it's out of scope.

Answer 2

You need to global $bannedIPs;

This works for me:

function ipban() {
    $bannedIPs = array('127.0.0.1','72.189.218.85');
    $ip = '127.0.0.1';
    if (in_array($ip, $bannedIPs)) { 
        echo 'test';
    }   
}
ipban();

So, you might want to see if that works, substitute in the IP address, and then finally replace it with the SERVER variable.

Answer 3

Your variable $bannedIPs is out-of-scope inside the function. Read up on variables scope: http://php.net/variables.scope

$var = 'xyz';

function abc() {
    // $var does not exist here

    $foo = 'abc';
}

// $var exists here

// $foo does not exist here

RE: Update:

Moving the variable inside the function works, your code snippet executes fine. There's got to be something else in your code.

Answer 4

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);
}