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copied!<p>You have run into a little problem with your variable scoping.</p> <p>Any variables outside a function in PHP is not accessible inside. There are multiple ways to overcome this.</p> <p>You could either declare <code>$bannedIPs</code> inside your function as such:</p> <pre><code>function ipban() { $bannedIPs = array('127.0.0.1','72.189.218.85'); if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) { echo 'test'; } } </code></pre> <p>Tell your function to access <code>$bannedIPs</code> outside the function using the <strong><code>global</code></strong> keyword:</p> <pre><code>$bannedIPs = array('127.0.0.1','72.189.218.85'); function ipban() { global $bannedIPs; if (in_array($_SERVER['REMOTE_ADDR'], $bannedIPs)) { echo 'test'; } } </code></pre> <p>Or, use the <code>$GLOBALS</code> super global:</p> <pre><code>$bannedIPs = array('127.0.0.1','72.189.218.85'); function ipban() { if (in_array($_SERVER['REMOTE_ADDR'], $GLOBALS['bannedIPs'])) { echo 'test'; } } </code></pre> <p>I recommend you read the manual page on Variable scope:</p> <p><a href="http://www.php.net/manual/en/language.variables.scope.php" rel="nofollow noreferrer">PHP: Variable Scope</a></p> <hr> <p><strong>If it's still not working, you have another problem in your code. In which case, you might want to consider using a <code>var_dump()</code> to check what datatype is <code>$bannedIPs</code> before down voting us all.</strong></p> <pre><code>function ipban() { global $bannedIPs; var_dump($bannedIPs); } </code></pre>

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