Reputation: 33
I have a script which successfully writes to the database.
However, when I change the variable name $password1
or $password9
or $password34
and save the script, the script still works and a new entry in the DB appears with all the fields filled out including the password field.
$password0 = trim($_POST['password1']);
$hashed_passcode = password_hash($password4, PASSWORD_DEFAULT);
mysqli_stmt_bind_param($q, 'ssss', $first_name, $last_name, $email, $hashed_passcode);
Above is an example, where I changed the variable to $password0
. and when I hash it the variable is $password4
. I resubmit the form and it still works. How can this be?
Upvotes: 3
Views: 686
Reputation: 74217
Reason being is that it will create a hash alright, it just won't store the hash associated with the actual password you want to store.
Having all error reporting set though, would have signaled an undefined variable.
Note: password_hash()
only hashes passwords and does not show any type of error, that isn't its job to do that, it's the error handling on the server that does.
Upvotes: 3