Why don't I see any value from autocomplete search in jQuery - codeigniter

Question

Why can't I see any value suggestion from text input. I already use autocomplete code with JSON and jQuery. But no value is displayed in the text input
bookstore/views/admin/auto_complete.php
<script>
$('#swSearch').keypress(function () {
var dataObj = $(this).closest('form').serializeArray();
$.ajax({
url: 'http://localhost/boostore/admin_d_book_groups/search',
data: dataObj,
dataType: 'json',
success: function (data) {
$("#suggestion_tab").html('');
$.each(data.name, function (a, b) {
$("#suggestion_tab").append('<li>' + data.b + '</li>');
});
// Display the results
///alert(data);
},
"error": function (x, y, z) {
// callback to run if an error occurs
alert("An error has occured:\n" + x + "\n" + y + "\n" + z);
}
});
});
</script>
<div id="swSearch">
<form>
<input type="text" value="" id="swSearch" class="swSearch" />
</form>
<div class="suggestion_tab" id="suggestion_tab"></div>
</div>
Admin_d_book_groups controller
function search(){
$searchterm = $this->input->post('search_hotel');
echo json_encode($this->d_book_groups->sw_search($searchterm));
}
d_book_groups_model
function sw_search($searchterm)
{
$query = $this->db->order_by("bg_id", "desc")->like('bg_name', $searchterm, 'after')->get('d_book_groups');
$data = array();
foreach ($query->result() as $row)
{
$data[] = $row->bg_name;
}
return $data;
//return mysql_query("select * from hotel_submits where name LIKE '".$searchterm."'");
}

Answer 1

Is $config['compress_output'] = TRUE; in your application/config/config.php?
If it is, you can't output content directly from the controller methods. You must use a view. You can create a view as simple as:
<?php echo $response;
and just pass the json data like this:
function search()
{
$searchterm = $this->input->post('search_hotel');
$data['response'] = json_encode($this->d_book_groups->sw_search($searchterm));
$this->load->view('ajax/json_response', $data);
}

Answer 2

I dont know why, but you make a mistake in this code:
$.each(data.name, function (a, b) {
$("#suggestion_tab").append('<li>' + data.b + '</li>');
});
You can not use "data.b" in $.each method, b is an object of data.name
for more help, actually "b" equal to data.name[a]
(data.name[a] == b)
So data.b is false, (data.name[a] == b) && data.name[a] != data.b
Use console.log(b) in $.each loop to see your array object (you need firefox and firebug extention)
$.each(data.name, function (a, b) {
console.log(b);
});
I think you want to put each name, in li element, if its true, use this code:
$.each(data, function (a, b) {
$("#suggestion_tab").append('<li>' + b.name + '</li>');
});
// here b.name == data[a].name