Question

我正在codeigniter中创建一个应用程序，用于我们希望国家/地区的形式下拉国家/地区的更改选择控制器就像

public function user()
{
        $data   = array();
        $this->load->model('enquiry');
        $this->load->helper('url');
        $data['result'] = $this->enquiry->get_reqtutor();
        $data['countries'] = $this -> country_model -> get_countries();
        $this->load->view('Table\DemoHistory',$data);
}

function get_states($country) {
     $this->load->helper('url');
     $this->load->model('state_model');
     header('Content-Type: application/x-json; charset=utf-8');
     echo(json_encode($this->state_model->get_states($country)));
}

和模型就好 state_model `

<?php
class State_model extends CI_Model {
    public function __construct() {
         $this -> load -> database();
    }
    function get_states($country = null){
         $this->db->select('state_id, state_name');
         if($country != NULL){
             $this->db->where('country_id', $country);
         }
         $query = $this->db->get('state_master');
         $states = array();
         if($query->result()){
             foreach ($query->result() as $state) {
                 $cities[$state->state_id] = $state->state_name;
             }
             return $states;
         } else {
             return FALSE;
         }
    }

}
?>

country_model

<?php
class Country_model extends CI_Model {

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

    function get_countries() {
         $this -> db -> select('country_id, country_name');
         $query = $this -> db -> get('country_master');

        $countries = array();

        if ($query -> result()) {
             foreach ($query->result() as $country) {
                 $countries[$country -> country_id] = $country -> country_name;
             }
             return $countries;
        } else {
             return FALSE;
        }
     }

}
?>

Ajax就像

 <script type="text/javascript">// <![CDATA[
     $(document).ready(function(){
         $('#country').change(function(){ //any select change on the dropdown with id country trigger this code
             $("#cities > option").remove(); //first of all clear select items
             var country_id = $('#country').val(); // here we are taking country id of the selected one.
             $.ajax({
                 type: "POST",
                 url: /mtb/Schedule/get_states/"+country_id, //here we are calling our user controller and get_cities method with the country_id

                 success: function(cities) //we're calling the response json array 'cities'
                 {
                     $.each(cities,function(state_id,state_name) //here we're doing a foeach loop round each city with id as the key and city as the value
                     {
                         var opt = $('<option />'); // here we're creating a new select option with for each city
                         opt.val(state_id);
                         opt.text(state_name);
                         $('#cities').append(opt); //here we will append these new select options to a dropdown with the id 'cities'
                     });
                 }

             });

         });
     });
 // ]]>
</script>

并且查看

<?php $countries['#'] = 'Please Select'; ?>

<label for="country">Country: </label><?php echo form_dropdown('country_id', $countries, '#', 'id="country"'); ?>
 <?php $cities['#'] = 'Please Select'; ?>
<label for="city">State: </label><?php echo form_dropdown('state_id', $cities, '#', 'id="cities"'); ?>

我正在获取国家/地区名称，而我的状态值不会在下拉列表中显示，请帮助........

Answer 1

您需要考虑的代码：

你在这里有一个错误网址：/mtb/Schedule/get_states/"+country_id，在开始时缺少双引号。始终使用调试工具，如firebug，它会显示javascript错误。

另一件事，代码

if($country != NULL){

应该是

if(!is_null($country)){

看看is_null function。

另一件事，你应该使用application/x-json而不是application/json json标题。

另一件事，代码

echo(json_encode($this->state_model->get_states($country)));

应该是

echo(json_encode($this->state_model->get_states($country)));
exit;

否则可能会将意外结果附加到最终的json。 Example: CodeIgniter分析。

另一件事，代码

$countries = array();

if ($query -> result()) {
     foreach ($query->result() as $country) {
         $countries[$country -> country_id] = $country -> country_name;
     }
     return $countries;
} else {
     return FALSE;
}

应该是

$countries = array();

foreach ($query->result() as $country) {
    $countries[$country -> country_id] = $country -> country_name;
}
return $countries;

因为没有国家/地区且返回FALSE会在form_dropdown函数中出现错误，请查看其中。 get_states函数也是如此。

这是我在代码的初步看法中看到的内容。

codeigniter - 在下拉列表中显示来自数据库的数据

1 个答案: