how to link my sweetalert to another page in php echo tag?

Question

I'm using sweet alert to send the user a warning message that show to them that the information insert is not valid in PHP echo. I want the "okay" button link to the desire page but it just could not work. my developer console shows the error of login.php:29 Uncaught TypeError: Cannot read properties of undefined (reading 'then') but I have no idea how can I make this work. I spent hours searching here and there but it just don't. Anyone who lend me a helping hand will be truly appreciated. Thanks.

Below are my codes:

if (isset($_POST['login'])) {
    $email = $_POST['email'];
    $user_password = $_POST['user_password'];


    $sql = "SELECT * FROM users WHERE user_phone = '$email' OR username = '$email' ";
    $query = $conn->query($sql);
if ($query->num_rows < 1) {
    echo '<script> swal({
            title: "Wrong information",
            text: "Sorry, your login information is not in our system. Please check your username/email correctly.",
            confirmButtonText: "Okay"}).then(function() {
                window.location = "../login/index.php";
            }); 
            </script>';

}