Ajax获取索引页而不是值

Question

我正在尝试使用codeigniter 2.1.4中的ajax从州获取城市。我面临的问题是取代整个html索引页面的数据。无法解决问题请帮忙。我的代码如下：

控制器

class Pages extends CI_Controller{

    public function __construct(){
        parent::__construct();
        $this->load->model('home_model');
        $this->load->model('sub_cat_model');
        $this->load->model('state_model');
        $this->load->model('city_model');
        $this->load->model('location_model');
        $this->load->library('email');
    }

    public function index(){
        $data['state'] = $this->home_model->get_state();
        $data['title'] = 'Rimi Classified - Home';

        $this->load->view('templates/header', $data);
        $this->load->view('index', $data);
        $this->load->view('templates/footer', $data);
    }

    public function sign_up(){
        $data['states'] = $this->state_model->get_states();

        $data['error'] = '';
        $data['title'] = 'Rimi Classified - Sign up';
        $this->load->view('templates/header1', $data);
        $this->load->view('sign-up', $data);
        $this->load->view('templates/footer', $data);
    }

    public function get_cities($state){
        header('Content-Type: application/x-json; charset=utf-8');
        die(json_encode($this->city_model->get_cities($state)));
    }
}

城市模型

class City_model extends CI_Model{

    public function __construct(){
        $this -> load -> database();
        //$this->output->enable_profiler(TRUE);
    }

    function get_cities($state){
        if($state != NULL){
            $this->db->where('state_id', $state);

            $query = $this->db->get('city');
            $cities = array();
            if($query->result()){
                foreach($query->result() as $city){
                    $cities[$city->id] = $city->city;
                }
                return $cities;
            }else{
                return FALSE;
            }
        }else{
            $cities[] = "-- Select City --";
            return $cities;
        }
    }
}

州模式

class State_model extends CI_Model{

    public function __construct(){
        $this -> load -> database();
    }

    function get_states() {
        $query = $this->db->order_by('state', 'ASC')->get('state');

        $states = array();

        if($query -> result()){
            foreach($query->result() as $state){
                $states[$state -> id] = $state -> state;
            }
            return $states;
        }else{
            return FALSE;
        }
    }
}

视图

<div id="innerdiv1">
    <label>State</label>
    <br />
    <?php
        $states['#'] = '-- Select State --';

        echo form_dropdown('state_id', $states, '#', 'id=state_id');
    ?>
</div>

<div id="innerdiv2">
    <label>City</label>
    <br />
    <div id="city">
        <select name="city_id" id="city_id">
            <option value="">-- Select City-- </option>
        </select>
    </div>

AJAX

<script type="text/javascript" src="http://code.jquery.com/jquery-latest.min.js"></script>
<script type="text/javascript">
    $(document).ready(function (){
        $('#states-dropdown select').change(function (){
            var selState = $(this).attr('value');
            console.log(selState);
            $.ajax({
                url: "pages/get_cities", 
                async: false, 
                type: "POST", 
                data: "state="+selState, 
                dataType: "html",

                success: function(data) {
                    //data is the html of the page where the request is made.
                    $('#city').html(data);
                }
            })
        });
    });
</script>

Answer 1

在评论中发布的routes.php中，更改

$route['(:any)'] = 'pages/category_list/$1';

进入

$route['pages/category_list/(:any)'] = 'pages/category_list/$1';

请注意我在代码中注意到的两件事是错的，但我不确定它们是否是原因：

1-不要使用“die”调用echo或打印数据，仅使用die进行错误

2- ajax调用应该是“GET”而不是“POST”：

  $.ajax({   
        url: "pages/get_cities/"+selState,
    async: false,
    type: "GET", 
    success: function(data) {
            //data is the html of the page where the request is made.
    $('#city').html(data);
    },
    error: function(response) {
    //Do Something on Error
    }
     });

Answer 2

尝试这样的事情

var url = "pages/get_cities/" + $(this).attr('value');
 $.ajax({
    url: url,
    async: false,
    type: "GET",
    dataType: "html",
    success: function(data) {
        $('#city').html(data);
    }
})

php code

public function get_cities($state){
    header('Content-Type: application/x-json; charset=utf-8');
    echo json_encode($this->city_model->get_cities($state));
    return ;
}