将多个URL函数映射到同一视图

Question

In working with codeigniter, I have a controller for exiting a page. I'd like a user to be able to go to the url site.com/example/copy, site.com/example/revise, and the same page loads each time with a different variable passed, as so:

class example extends Secure_Controller {

    public function revise()
    {
        $id = $_GET["id"];
        $show_for_edit("revise", $id);
    }

    public function copy()
    {
        $id = $_GET["id"];
        $show_for_edit("copy", $id);
    }

    public function show_for_edit($edit_type, $id)
    {
        //A lot of database queries that I don't
        //want duplicated under revise() and copy()
        $data["id"] = $id;
        $data["edit"] = $edit_type;
        $this->load->view('model', $data);
    }
}

But I receive an error:

Fatal error: Call to undefined function show_for_edit()

Any ideas?

Answer 1

In PHP, this error occurs when the object that invokes the method does not have a reference, ie it is null.

In your case, the object that calls the method "show_for_edit" is null, and so it appears that mistake.

Check the variable that calls this method and make sure it is not null and is an instance of the "example" class.

Moreover, I am not sure that these calls $show_for_edit ("check", $id) and $show_for_edit ("copy", $id) are correct. I think they should be:

$this-> show_for_edit ("check", $ id);

And

$this-> show_for_edit ("copy", $ id);