带有codeigniter的Jquery AJAX，总是返回错误

Question

I am trying to write a script that will add the video currently being viewed to a database of favourites. However every time it runs, an error is returned, and nothing is stored in the database.

Here is the JQuery

$(document).ready(function() {

$("#addfav").click(function() {

 var form_data = {heading: $("#vidheading").text(), embed : $("#vidembed").text()};

  jQuery.ajax({
      type:"POST",
      url:"localhost/stumble/site/add_to_fav.php",
      dataType: "json",
      data: form_data,
      success: function (data){
          console.log(data.status);
          alert("This Video Has Been Added To Your Favourites")

      },
      error: function (data){
        console.log(data.status); 
        alert("You Must Be Logged In to Do That")
      }
  });
})

})

The add_to_fav.php is this...

public function add_to_fav(){

      $this->load->model('model_users');
      $this->model_users->add_favs();    

}

And the add_favs function is below

public function add_favs(){
    if($this->session->userdata('username')){
    $data = array(
        'username' => $this->session->userdata('username'),
      'title' => $this->input->post('heading'),
      'embed' => $this->input->post('embed')
      );

      $query = $this->db->insert('fav_videos',$data);
        if($query){
      $response_array['status'] = 'success';
      echo json_encode($response_array);
        }}else {

          $response_array['status'] = 'error';
          echo json_encode($response_array);
      }

  }   

Thank you for the input, this has me stuck but I am aware it may be something relatively simple, my hunch is that it is something to do with returning success or error.

Answer 1

Try

$(document).ready(function() {

$("#addfav").click(function() {

 var form_data = {heading: $("#vidheading").text(), embed : $("#vidembed").text()};

  jQuery.ajax({
      type:"POST",
      url:"http://localhost/stumble/Site/add_to_fav",
      dataType: "json",
      data: form_data,
      success: function (data){
          console.log(data.status);
          alert("This Video Has Been Added To Your Favourites")

      },
      error: function (data){
        console.log(data.status); 
        alert("You Must Be Logged In to Do That")
      }
  });
})

})

Also to use base_url in javascript. In your template view :-

<script>
 window.base_url = "<?php echo base_url(); ?>";
</script>

Now you can use base_url in all your ajax scripts.