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Ronbo51

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Everything posted by Ronbo51

  1. Thank you very much requinix for your quick reply and help. I changed the code to what is shown below and now it works. $sql = "UPDATE missionaries SET missionary=?, active=?, country=?, city=?, state=?, zip=?, street=?, phone=?,email=?, website=?, facebook=?, photo=? WHERE ID=?"; $stmt = $con->prepare($sql); $stmt->bind_param("ssssssssssssi", $missionary,$active,$country,$city,$state,$zip,$street,$phone,$email,$website,$facebook,$photo,$id); I definitely do not to change the ID. The ID input statement is type=hidden so it can't be changed. I thought I had to include it but I now see that I don't have to. This prepared statement programing is all new to me and you have helped me a lot. Thanks again.
  2. I wrote a php script to update information in an mysqli database table and and am getting the error: The number of variables must match the number of parameters in a prepared statement. This is the code snippet with the error line in bold red. I don't see anything wrong with it but hopefully some of you sharp eyed people can help me. include "churchdb_data.php"; $con=mysqli_connect('localhost',$username,$password,$database); if (mysqli_connect_errno()) { die ("Failed to connect to MySQL: " . mysqli_connect_error()); } $sql = "UPDATE missionaries SET ID=?, missionary=?, active=?, country=?, city=?, state=?, zip=?, street=?, phone=?,email=?, website=?, facebook=?, photo=? WHERE ID=?"; $stmt = $con->prepare($sql); $stmt->bind_param("issssssssssss", $id, $missionary,$active,$country,$city,$state,$zip,$street,$phone,$email,$website,$facebook,$photo); if ($stmt->execute()) { echo "Record updated successfully!"; } else { echo "Error updating record: " . $stmt->error; } $stmt->close(); $con->close(); ?>
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