click jquery button + send data without form - bookmark

问题

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");
                }
            });  

    });

回答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");
            }
        });  

});

回答2:

Instead of using .ajax() use either .get() or .post()

Using .get()

 $.get('controller/addBookmark',function(data){ 
     alert('Your bookmark has been saved'); 
 });

Using .post()

 $.post('controller/addBookmark', function(data) {
     alert('Your bookmark has been saved, The contents of the page were:' + data); 
 });

回答3:

from the documentation of jQuery.ajax()

Data to be sent to the server. It is converted to a query string, if not already a string. It's appended to the url for GET-requests.

If you do not know what to put for the data, probably you should remove that option? In your controller function addBookmark(), you can reduce the code by remove the if check. Try this and see if it works for you.

来源：https://stackoverflow.com/questions/10406571/click-jquery-button-send-data-without-form-bookmark

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jquery

ajax

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