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;
?>