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Old March 10th, 2006, 11:53 AM
Gotaka4 Gotaka4 is offline
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let's change these lines first:

$q1 = "SELECT StaffID
               FROM fault_form
               WHERE StaffID = 'user';";


$ql = "SELECT StaffID FROM fault_form WHERE StaffID = '$user'";

try that and tell me what error you are getting exactly as it appears on your browser.

Never bother to learn something not knowing which does not do you any harm, and never neglect to learn something whose negligence will increase your ignorance - Imam Jafar Sadeq
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