[MySql code help needed]

I am trying to code a query whereby the query checks for duplicate items in a table and if found sends an error back to jquery. However, what is happening, is that jquery keeps giving me a typeError and I cannot see why. In my posted code, the code that is commented out, /* if ($box == 'DEMO111') works fine but not my query.

All mysql connections are working and I can connect to the db.

I would be grateful if someone could check my code and show me where I have gone wrong. Thanks

 

DB Config

<?php

function runSQL($rsql) {
    $hostname = "localhost";
    $username = "root";
    $password = "";
    $dbname   = "logistor_logistor";
    $connect = mysql_connect($hostname,$username,$password) or die ("Error: could not connect to database");
    $db = mysql_select_db($dbname);
    $result = mysql_query($rsql) or die (mysql_error());
    return $result;
    mysql_close($connect);

?>

<?php

$array = split('[,]', $_POST['box_add']);
$boxerrortext = 'No duplicate boxes';
?>

<?php

if (isset($_POST['submit']))     {
            $error = array();
            foreach ($array as $box) {
            $sql = "SELECT * FROM act WHERE item = '" . $box . "'";
            $result = runSQL($sql) or die(mysql_error());
            $num_rows = mysql_num_rows($result);
            
            if ($num_rows) {
            //trigger_error('It exists.', E_USER_WARNING);
            
            $error[] = array('boxerror'=>$boxerrortext,
                            'box'=>$box);
            $result = json_encode($error);

            echo $result;
            return;
            }
            }

            /* if ($box == 'DEMO111')
            
            {
            //echo 'There was an error somewhere';
            //$box = 'ERROR';

            //$error = array('boxerror'=>$boxerrortext, 'box'=>$box);
            //$output = json_encode($error);

            //echo $output;
            //return;
            } */
            
            else {
    
            $form = array();

            foreach ($array as $box) {

            $form[] = array('dept'=>$dept,
            'company'=>$company,
            'address'=>$address,
            'service'=>$service,
            'box'=>$box,
            'destroydate'=>$destroydate,
            'authorised'=>$authorised,
            'submit'=>$submit);

             $sql = "INSERT INTO `temp` (service, activity, department, company, address, user, destroydate, date, item, new) VALUES ('$service', '$activity', '$dept', '$company', '$address', '$requested', '$destdate', NOW(), '$box', 1)";
             $result = runSQL($sql) or die(mysql_error());
            } 
           }
          }

             $result=json_encode($form);

             echo $result;

?>