CodeIgniter：尝试根据第一个选择选项的选择从数据库表中加载选项值

我正在使用 CodeIgniter 并且有一个包含 2 个选择选项的表单。第一个选择选项是汽车制造，第二个选择选项是制造。如果我从 As 'BMW' 中选择 Car Make，则第二个选择选项中的值应更改并显示所有由 BMW 制造的车型。

**WelcometoDemoCar.php (View)**

*//to get the Car Make List Box*

<input type = "text" name = "car_list" list="car_dalalist" id = "car_list" class = "inktext inklarge" placeholder = "Type of Car" required = "" autocomplete="off" />
<datalist id="car_dalalist">
 <?php foreach($carlist as $row_carlist){?>
<?php //echo $row_carlist->Make . " " .$row_carlist->Model ." " .$row_carlist->Year ;?>
<option value="<?php echo $row_carlist->Make;?>"> <?php echo $row_carlist->Make;?></option>
<?php }?> 
</datalist>


*//to get the value in the Make Select List Box*


<input type = "text" name = "car_model" list="car_model_dalalist" id = "car_model" class = "inktext inklarge" placeholder = "Car Model" required = <datalist id="car_model_dalalist">
<?php foreach($carModel as $row_carModel){?>
<?php //echo $row_carlist->Make . " " .$row_carlist->Model ." " .$row_carlist->Year ;?>
<option value="<?php echo $row_carModel->Model;?>"><?php echo $row_carModel->Model;?> </option>
<?php }?> 
</datalist> 


**Welcome.php (Controller)**

$this->data['carlist'] = $this->PostModel->getCarDetails(); 
$this->data['carModel'] = $this->PostModel->getCarModel();

**PostModel.php (Model)**



*//to get car make*

 function getCarDetails(){
  $this->db->Distinct();
  $this->db->select("Make"); 
  $this->db->from('carlist');
  $carListquery = $this->db->get();
  return $carListquery->result();
 }


 *// to get car model*


 function getCarModel(){
  $make = $this->input->post('car_list');   

  $this->db->Distinct();
  $this->db->select("Model"); 
  $this->db->from('carlist');
  $this->db->where('Make' . $make);
  $carmodelquery = $this->db->get();
  return $carmodelquery->result();
 }


public function get_data()
 {
      $value = $this->input->post("value");
      $data = $this->PostModel->get_data($value);
      $option ="";
      foreach($data as $d)
      {
         $option .= "<option value='".$d->id."' >".$d->Model."</option>";
      }
       echo $option;
 }

我尝试使用 ajax 在各种网站上发布的几个解决方案，但我认为我的值没有发布到控制器。

阿贾克斯代码

    $("#car_list").on("change",function(){ 
    var value = $(this).val(); 
    $.ajax({ url : "welcome/get_data", 
    type: "post", 
    data: {"value":'value'}, 

    success : function(data){ 
    $("#car_model").html(data); 
    }, 
   });
  });

非常感谢您的时间和帮助。

预先感谢。