Troubleshooting 'undefined variable' error

I keep getting the following error Undefined variable: password on line 33 how do I correct this problem? So this error will stop showing.

Here is the php code.

$first_name = mysqli_real_escape_string($mysqli, $purifier->purify(htmlentities(strip_tags($_POST['first_name']))));
$password1 = mysqli_real_escape_string($mysqli, $purifier->purify(strip_tags($_POST['password1'])));
$password2 = mysqli_real_escape_string($mysqli, $purifier->purify(strip_tags($_POST['password2'])));




// Check for a password and match against the confirmed password:
if ($password1 == $password2) {
    $sha512 = hash('sha512', $password1);
    $password = mysqli_real_escape_string($mysqli, $sha512);
} else {
    echo '<p class="error">Your password did not match the confirmed password!</p>';
}



//If the table is not found add it to the database
if (mysqli_num_rows($dbc) == 0) {
        $mysqli = mysqli_connect("localhost", "root", "", "sitename");
        $dbc = mysqli_query($mysqli,"INSERT INTO users (user_id, first_name, password) 
                                     VALUES ('$user_id', '$first_name', '$password')");
}



//If the tabl开发者_JS百科e is in the database update each field when needed
if ($dbc == TRUE) {
        $dbc = mysqli_query($mysqli,"UPDATE users 
                                     SET first_name = '$first_name', password = '$password' 
                                     WHERE user_id = '$user_id'");

        echo '<p class="changes-saved">Your changes have been saved!</p>';

}

There's only one place where a value is assigned to $password

if ($password1 == $password2) {
    $sha512 = hash('sha512', $password1);
    $password = mysqli_real_escape_string($mysqli, $sha512);
}

So, if the condition isn't met there will be no $password. And in that case it doesn't make sense to perform the UPDATE query anyway.

At the top define

$password = '';

then change the DBC check to

if ($dbc == TRUE && $password != ''){

As you can see, the database insert is done whether the first if() was true or false. If it's false ($password1 and $password2 doesn't match), $password won't be defined.

If this condition fails:

 if ($password1 == $password2) {

$password will not get defined, raising an error in one of the lines it is used in later.

You don't raise an ERROR with an ELSE statement on the $password = ...... line so there is clearly an error there and it's not being defined. The top level if statement is fine, but the error is on the $password declaration line. Do you see how that works?

Instead of retrying the query if the insert fails (presumably because the user_id already exists - you've made that your primary key?), you could use the alternate INSERT INTO ... ON DUPLICATE KEY UPDATE syntax:

INSERT INTO users (user_id, first_name, password) 
VALUES ('$user_id', '$first_name', '$password')
ON DUPLICATE KEY UPDATE
    first_name=VALUES(first_name),
    password=VALUES(password)

On a nitpicky point, your comments say "if the table is not found" and "if the table...". You're not dealing with table creation/modification - you're working with records that are stored in a table.