如果甚至可能，如何调用类方法作为表单操作php mvc应用程序？

I'm trying to write a simple mvc application with php and mysql. I'm very new to mvc and relativly new to php aswell. I'm letting the user choose from different movies and then add the ones they want to their own list. But I can't figure out how to get the correct form action to insert the choosen movie into the db.

This is how my two model class methods looks like:

public function checkMovie() {
        // Check if movie exist in db.
        $stmt = $this->dbh->prepare("SELECT * FROM watchlist WHERE my_title='{$_POST['my_title']}'");
        $stmt->bindParam(':my_title', $_POST['my_title']);
        $stmt->execute();

        $rows = $stmt->fetchALL();
        $this->n = count($rows);
     }  

public function addMovie() {
        // Add choosen movie to db.
        $sql = $this->dbh->prepare("INSERT INTO watchlist(my_title, my_des, my_link) 
                                         VALUES ('{$_POST['my_title']}', '{$_POST['my_des']}', '{$_POST['my_link']}')"); 
        $sql->bindParam(':my_title', $_POST['my_title'], PDO::PARAM_STR);
        $sql->bindParam(':my_des', $_POST['my_des'], PDO::PARAM_STR);
        $sql->bindParam(':my_link', $_POST['my_link'], PDO::PARAM_STR);
        $sql->execute(array(':my_title' => $_POST['my_title'],':my_des' => $_POST['my_des'],':my_link' => $_POST['my_link']));
    }

As you can see I have the basic sql-code in here and then I call the methods from a method in my controller:

public function getAddMovie() {
        $this->addModel = new AddMovieModel();

        if (isset($_POST['submit'])) {

            //  Call checkmovie from addmoviemodel and check if movie allready is taken.
            $checkmovie = $this->addModel->checkMovie();
            if($this->n > 0) { // Should this logic perhaps be in my model?
                // Shows javascript-popup eg. 'movie allready added'.
                include 'view/viewscripterror.php';
            }
            else { // Call addMovie from addmoviemodel to insert movie to db.
                $addmovie = $this->addModel->addMovie();
                // Shows javascript-popup eg. 'movie is now added'.
                include 'view/viewscriptsuccess.php';
            }
        }
    }

I'm not sure if the if($this->n > 0) perhaps should be in my model aswell?

And here's the form, I can't figure out what to pass as form action? This problem has been driving me crazy for a while now and that's why I'm turning here in hope for some help.

echo '<form action="??" method="post">',
     '<input type="hidden" name="my_title" value="'.$title.'">',
     '<input type="hidden" name="my_des" value="'.$description.'">',
     '<input type="hidden" name="my_link" value="'.$link.'">',
     '<input type="submit" name="submit" value="Peppa!">',
     '</form></div>'; 

Try like

echo '<form action="http://site_url/getAddMovie" method="post">',

You need to pass the url of the function getAddMovie into the action,then after submitting it,it will post/get the params into that function.

And try to load the model like

$this->load->model('AddMovieModel');

And try to call it like

$checkmovie = $this->AddMovieModel->checkMovie();

Or even you can try like

$addModel = new AddMovieModel();

and call it like

$checkmovie = $addModel->checkMovie();