如何使用Codeigniter和AJAX从MySQL提交和获取数据时避免双输出

Problem:

Hi friend i am inserting data into mysql database table via textbox using ajax in codeIgniter and at the same time, i am displaying existing data in front end from database.

Once i insert data into database, i am getting two input textbox and two outputs. So plz tell me how to display one input textbox and output.

My Original Codes as follows:

sample_view:

    <form action="" method="post" accept-charset="utf-8">

        <table align="center">

            <tr>
                <td>Message :</td>
                <td>
                    <textarea name="message" id="message" placeholder="Write here the message"></textarea>
                </td>
            </tr>

            <tr>
                <td>&nbsp;</td>
                <td id="result"> </td>
            </tr>

            <tr>
                <td>&nbsp;</td>
                <td>
                    <button type="button" id="submit">Submit</button>
                </td>
            </tr>

        </table>

    </form>
 <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>

    <script type="text/javascript">
        $(function() {

            $('#submit').click(function() {

                //get input data as a array
                var post_data = {
                    'message': $("#message").val(),
                    '<?php echo $this->security->get_csrf_token_name(); ?>': '<?php echo $this->security->get_csrf_hash(); ?>'
                };

                $.ajax({
                    type: "POST",
                    url: "http://192.168.1.17/ci_civic_soft/index.php/sample_control/index",
                    data: post_data,
                    success: function(message) {
                        // return success message to the id='result' position
                        $("#result").html(message);
                    }
                });

            });

        });
    </script>

sample_control:

<pre>


if ( ! defined('BASEPATH')) exit('No direct script access allowed');

class Sample_control extends CI_Controller
{
     function __construct()
     {
       parent::__construct();
       $this->load->model('sample_model');
     }

    public function index()
    {
        $this->load->view('sample_view');

        $data['message'] = $this->input->post('message');

        $insert = $this->sample_model->insertDataToDB($data);

            if ($insert) 
            {

                $tree_menu = $this->sample_model->getLastEnrtyData(); //getting tree from DB
                foreach($tree_menu as $tv)
                {
                    echo $tv['message'];
                }

            }

    }


}
</pre>

sample_model:

<pre>
Class sample_model extends CI_Model
{/*
* Insert data to the content table
*/
    public function insertDataToDB($data) 
    {
        return $this->db->insert('content', $data);
    }


/*
* Get the Inserted data from content table
*/
    public function getLastEnrtyData() 
    {
        $query = $this->db->query("CALL content()"); 
        return $query->result_array();
    }
}

</pre>

Move your insert code into a separate function like saveMessage(). You are loading the view in the index thus it wil be returned by ajax.

public function saveMessage(){
   $data['message'] = $this->input->post('message');

    $insert = $this->sample_model->insertDataToDB($data);

        if ($insert) 
        {

            $tree_menu = $this->sample_model->getLastEnrtyData(); //getting tree from DB
            foreach($tree_menu as $tv)
            {
                echo $tv['message'];
            }

        }
}