Codeigniter Ajax - Help Needed with Error

Question

Hey I'm new to CI and have scoured the internet for a tutorial that will work but for some reason it won't work. Can someone help me with the code please:

What's the right edit to the code to submit an entry to the database via ajax without reloading the page?

Controller:

public function index(){
$this->load->helper('url');
$this->load->view('template');

}

function create()
{
    $this->load->library('form_validation');
    $this->load->helper('form');
    $this->load->model('dbmodel');

    $response = array();
    $this->load->model('dbmodel');
    $result = $this->dbmodel->addnew_row();     
}

Model:

public function addnew_row() {

    $data = array(
        'title' => $this->input->post('title'),
        'description' => $this->input->post('description'),
    );
    $result = $this->db->insert('books', $data);
    return $result;
}

View Form:

<h2>Create</h2> 
<?php echo form_open('welcome/create', array('id'=>'form'));?>


<p>Title: <?php echo form_input('title') ?></p>
<p>Description: <?php echo form_input('description') ?></p> 

<?php echo form_submit('btn_submit','Save'); ?>


<?php echo form_close();?>

View AJAX:

 // $.POST Example Request
$('#form').submit(function(eve){
    eve.preventDefault();

    var title = $('#title').val();
    var description = $('#description').val();

    url = '<?php echo site_url('welcome/create');?>'
    $.ajax({
        url: url,
        type: 'POST',
        data:{title:title, description:description
        }

        $(#dynamic_container).html(response.output);
    });
});

Well, at the very least your url variable isn't defined properly nor is it terminated. var url = '<?php echo site_url('welcome/create');?>'; — Serg Chernata
– Serg Chernata, Commented May 25, 2014 at 17:15
could you post the error message please? You can find it in the developer console (F12->Network->Trigger ajax request, and then select the ajax request in the list) — newBee
– newBee, Commented May 25, 2014 at 17:16

Runcorn · Accepted Answer · 2014-05-25 18:23:14Z

1

Ok,At first you need to briefly go through the syntax of jQuery.ajax() before using it.

Now going though the AJAX code you mentioned in the question , this block of code is not suppose to be there

$(#dynamic_container).html(response.output);

AJAX provides Callback Function Queues to manipulate response before/after an AJAX call has been successfully completed , and in your case using success will resolve the issue :

  $.ajax({
        url: url,
        type: 'POST',
        data:{title:title, description:description
        },
        success : function(response){
        $(#dynamic_container).html(response.output);
        }
    });

And you might be interested in using jQuery.post().

answered May 25, 2014 at 18:23

Runcorn

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