$employee_stmt=$con->prepare("INSET INTO employee (emp_name, emp_desig, emp_salary, years_exp, emp_status, emp_adr, city, pin_code, emp_phone, emp_email, emergency_contact, emergency_phone, blood_group, known_health_issue)VALUES(?,?,?,?,?,?,?,?,?,?,?,?,?,?)");
$employee_stmt->bind_param('ssssssssssssss',$_POST['employee_name'],$_POST['designation'],$_POST['salary'],$_POST['exp'],$_POST['emp_status'],$_POST['address'],$_POST['city'],$_POST['pin'],$_POST['phone'],$_POST['emp_email'],$_POST['emr_contact'],$_POST['emr_phone'],$_POST['blood_group'],$_POST['health_issue']);
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Alive to die - Anant
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gotocartik
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Please refer to https://stackoverflow.com/help/how-to-ask to encourage people to answer your question. – Camilo Go Jr. Jul 27 '17 at 05:19
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https://stackoverflow.com/a/4488100/4248328 – Alive to die - Anant Jul 27 '17 at 05:23
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I have found sometimes one get errors if you pass in a string instead of a variable storing the string, have you tried that ? – MrK Jul 27 '17 at 07:12
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Looks like your statement can not be prepared. You should check SQL for errors. (For example in post INSET used instead of INSERT).
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