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();

Just out of curiosity, do you have error reporting set to E_ALL? — MitMaro
– MitMaro, Commented Jul 25, 2009 at 6:38

Pradeep · Accepted Answer · 2018-05-10 10:00:01Z

8

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);
}

edited May 10, 2018 at 10:00

Pradeep

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answered Jul 25, 2009 at 6:01

Andrew Moore

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3 Comments

deceze Over a year ago

+1, especially for the downvote part. This is turning into a kindergarten thread.

JasonDavis Over a year ago

Another way I found is by passing the variable into the function on call ipban($bannedIPs);

Andrew Moore Over a year ago

@jasondavis: Yes, but then you are making your function call more verbose for nothing. It's code smell.

Pradeep · Accepted Answer · 2018-05-10 10:00:33Z

2

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.

edited May 10, 2018 at 10:00

Pradeep

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answered Jul 25, 2009 at 5:49

deceze♦

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1 Comment

JasonDavis Over a year ago

I tried moving the variable inside the function and it still has the same error

Tim · Accepted Answer · 2009-07-25 05:41:53Z

1

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

answered Jul 25, 2009 at 5:41

Tim

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Comments

Tyler Carter · Accepted Answer · 2009-07-25 06:02:49Z

1

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.

edited Jul 25, 2009 at 6:02

answered Jul 25, 2009 at 5:42

Tyler Carter

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5 Comments

Tyler Carter Over a year ago

Which means that it is a problem with $_SERVER

Tyler Carter Over a year ago

And for some reason some idiots decided to mark down my answer ... really?

deceze Over a year ago

@Chacha102: Sorry, yes, because "try switching arguments" and "trim $_SERVER" is totally unfounded non-sense.

Tyler Carter Over a year ago

The entire problem was unfound nonsense. He even commented that for some reason it works now.

deceze Over a year ago

The original question was a perfectly valid scope problem, as the beginning of your answer says. The rest of your answer is bad advise. Feel free to edit it.

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