键入错误的用户名时没有任何反应。 我的代码有什么问题？

I'm trying to understand what's went wrong with my code. it's a login proccess to cp. you do get the right outcome when writing the correct username (only username yet, just for that test). but you get nothing at all when you don't - nothing I've programmed it to do.

I'm REALLY desperate, tried to solve this the whole night. please help me. here's the code.

calculations:

if (isset($_POST['connection_made'])) { // is a connection been made? using a hidden input

$form_user = forms_filter($_POST['form_user']); // filter any tags or any unwanted actions first
$form_pass = forms_filter($_POST['form_pass']); // filter any tags or any unwanted actions first

if (strlen($form_user) > 5 AND strlen($form_user) < 16 AND strlen($form_pass) > 5 AND strlen($form_pass) < 16) {
        if (login_blank_filter($form_user, $form_pass) == true) { // are those values length shorter than the minimal value or longer? 

            $user_name = $form_user;
            $raw_password = $form_pass;
            $user_pass = password_hash($raw_password, PASSWORD_DEFAULT); // generating a hashed and salted password 

                $cp_validate_login = mysqli_query($data_connection, "SELECT * FROM `admins` WHERE `username` = '$user_name' ");
                if (!$cp_validate_login) { die('error: ' . mysql_error()); }

                while ($admin_row = mysqli_fetch_array($cp_validate_login)) {

                    if (mysqli_num_rows($cp_validate_login) == 1) {
                        echo "you made it.";
                        echo mysqli_num_rows($cp_validate_login);
                    }
                    else {
                        echo "not yet there.";
                        echo mysqli_num_rows($cp_validate_login);
                    }

                }
        }
        else {
            header('Location: index.php?login_status=2');
        }
}

}

Form:

        <?php 
    if (isset($_GET['login_status'])) {
        switch ($_GET['login_status']) {

            case 1:
            echo "wrong username or password";
            break;      

            case 2:
            echo "the inputs has to be filled";
            break;

            case 3:
            echo "username\passwords are too short or too long";
            break;

            default:
            echo "unknown error";
        }
    }
    else {
        echo "welcome. log in to the control panel please";
    }
    ?>
    </div>

    <form name="loginform" method="post" action="index.php" onSubmit="return validateBlank()">
    <input type="hidden" name="connection_made" value="">

    <div class="login_layout">

        <div class="login_right_layout">
        user:
        </div>

        <div class="login_left_layout">
        <input class="login_input" name="form_user" type="text">
        </div>

        <div class="login_right_layout">
        password:
        </div>

        <div class="login_left_layout">
        <input class="login_input" name="form_pass" type="password">
        </div>

    </div>

    <input type="submit" value="כניסה" class="login_submit">
    </form>


</center>

PLEASE help me. thank you.

First, you are wide open to SQL injection. You need to use prepared statements, rather than concatenating variables into your query. See How can I prevent SQL injection in PHP?.

Second, your code fails because you never make it into the while loop if the username is invalid. That is, if you type a bogus username, this condition is never satisfied:

while ($admin_row = mysqli_fetch_array($cp_validate_login))

So, your if/else logic is never executed.

The solution here is not to try to improve your existing code. It's to stop what you're doing and use an existing authentication (login) library. Security is hard, and proper authentication and authorization is no exception. You should not roll your own security code.