php - Prepared statement fails when it shouldn't

I have a signup form on my website and the following code should run whenever a new user signs up:

$sql = "SELECT * FROM users WHERE uidUsers=?;";
$stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt)) {
    header("Location: ../signup.php?error=sqlerror");
    exit();

}
else {
 mysqli_stmt_bind_param($stmt, "s", $username);
 mysqli_stmt_execute($stmt);
 $resultCheck = mysqli_stmt_num_rows($stmt);
if ($resultCheck > 0) {
    header("Location: ../signup.php?error=usertaken");
    exit();
}
else {

    $hashedPwd = password_hash(PASSWORD_DEFAULT, $password);
    $sql = "INSERT INTO users (`uidUsers`, `pwdUsers`, `phraseUsers`) VALUES(?, 
?, ?)";
    $stmt = mysqli_stmt_init($conn);
if (!mysqli_stmt_prepare($stmt)) {
    header("Location: ../signup.php?error=sqlerror");
    exit();
}
else {
    mysqli_stmt_bind_param($stmt, "sss", $username, $hashedPwd, 

    $securityphrase);
    mysqli_stmt_execute($stmt);
    header("Location: ../login.php?signup=success");
    exit();
}

The !mysqli_stmt_prepare error handler is triggered when it shouldn't be given the database circumstance, as well as the correct INSERT statement. Therefore I don't understand why it's triggered and I'm asking why?

There is a similar question here on

Answer

Thank you for your help, the problems where the password_hash order, it should be like this password_hash($var, PASSWORD_DEFAULT);. The second mistake I made was not including my $sql statements in the mysqli_stmt_prepare, it should be like this !mysqli_stmt_prepare($stmt, $sql).