在这里,我需要向数据库提交数据并使用ajax获取id

here is my view

<html>
<body>
    <form method="post" name="myForm1" id="myForm1" enctype="multipart/form-data" >
    Email: <input type="text" name="email" id="email">
    Question: <input type="text" name="qText" id="qText">
    <input id="submitbutton" type="submit">
    </form>
  <div id="abc"></div>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
    <script>   //no need to specify the language
       $(document).ready(function() {

       $('#myForm1').on("submit",function(e) {

            //var form = $(this);
            //dataString = $("#myForm1").serialize();
            e.preventDefault();

            $.ajax({
                type: "POST",
                url: "<?php echo site_url('form_controller/insert_into_db'); ?>",
                data: $(this).serialize(),
                //dataType: "json",
                success: function(data){
                   // top.location.href = "<?php echo site_url('form_controller/callform'); ?>";
                    //$.each(data.results, function(){
                   // $("#abc").append('<div><b>' + id.id + '</b></div><hr />');

                    //});
                    /*var site_url = "<?php// echo site_url('form_controller/callform/') ?>";
                    site_url = site_url +"/" + id;
                    $("#abc").load(site_url);*/
                    <?php //foreach(): ?>

                    var site_url = "<?php echo site_url('form_controller/callform'); ?>";
                    var mydata=window.JSON.stringify(data.trim());
                    alert(mydata);
                    //site_url = site_url +"/" +data.id;
                    alert(site_url);
                    $("#abc").load(site_url);
                    //$('#abc').html(data);
                    var item = data;
                    alert(item.id);

                }//,
                //error: function() { alert("Error posting feed."); }
           });

        });
        });
    </script>   
</body>
</html>

controller

function index(){
    $this->load->view('submit_form');
}

function callform($id){
    $this->load->view('view_form',$id);

}

public function insert_into_db(){
    $this->load->helper('url');
    $this->load->model('form_model');
    $data= $this->form_model->insertQ();

    $this->output->set_output(json_encode($data));

}

} 

model

 <?php

 class Form_model extends CI_Model{

function insertQ(){
    $email = $this->input->post('email');
    $text = $this->input->post('qText');
    $this->db->query("insert into form (email,text) values('$email','$text')");
    $this->load->database();
    $query = $this->db->query("SELECT MAX(id) AS id FROM form");
    return $query->result();

 }
 }

when insert record into database there is a auto increment id. I need to get that particular id from database and return it to the success function in ajax. then I need to load another page into a div and print that id on newly loaded content.here the problem is in view I couldn't get data into variable.for site_url

You can get the inserted id

    $id = $this->db->insert_id();