click jquery button + send data without form - bookmark

Question

I'm working on a bookmarking function where the user clicks on a jQueryui button and certain information is sent to the database. But I'm not using a form, because there is no information for the user to enter.

I'm pulling the user's ID from the session data, and I'm sending a URI segment (portion of the URL)

Using codeigniter/php.

I'm trying to figure out what to put in the data portion of the ajax/post function, since there's no form/no data entered, and what to do about the "submit" part of the controller.

Controller

function addBookmark(){

        if ($this->input->post('submit')) {

            $id = $this->session->userdata('id');               
            $bookmark = $this->uri->segment(3, 0);

            $this->bookmarks_model->postBookmark($id, $bookmark);
        }

    }

Model

function postBookmark() {

     $data = array(
            'user_id' => $user_id,
            'bookmark_id' => $bookmark,
    );

    $this->db->insert('bookmarks', $data);

    }

HTML

<button class="somebutton">Add bookmark</button>

jQuery

$('.somebutton').click(function() { 

            $.ajax({
                url: 'controller/addBookmark',
                type: 'POST',
                data: ???,
                success: function (result) {
                  alert("Your bookmark has been saved");
                }
            });  

    });

Answer 1

Your problem is you are checking for a submit key in the POST args. You can either fake it by sending data: {submit:true} or by by removing your if statement and just processing a POST request

$('.somebutton').click(function() { 

        $.ajax({
            url: 'controller/addBookmark',
            type: 'POST',
            data: {'submit':true}, // An object with the key 'submit' and value 'true;
            success: function (result) {
              alert("Your bookmark has been saved");
            }
        });  

});