Javascript 使用codeigniter更新数据库中数据的ajax

这是我的控制器-

   public function act_surveyby_id($id){
    $this->load->model('survey_m');
    if($this->survey_m->insert_activate($id)){                 
                echo "success";

    }else{

        echo "invalid";
    }   
}

这是我的模型-

---

问题：当我单击“激活调查”时，它不会更改/更新调查的详细信息。关于这件事，我真的非常需要帮助。谢谢

更改$.ajax函数，如下所示

         public function insert_activate($id){

                $date = date('m-d-Y',now());
                $stat = 'Active';

                        $data = array(
                                   'issued_date' =>  $date ,
                                   'status' => $stat            
        );

                     $this->db->update('survey', $data)->where('survey_id', $id);
    if($this->db->affected_rows()>0){
            return true;
        }else{ 
                    return false;

                }

    }


  }

$.ajax({
       url: '<?php  echo base_url(); ?>index.php/admin/survey/act_surveyby_id',
       type: "POST",
       data: {
             idx : idx;
             },

         public function insert_activate($id){

                $date = date('m-d-Y',now());
                $stat = 'Active';

                        $data = array(
                                   'issued_date' =>  $date ,
                                   'status' => $stat            
        );

                     $this->db->update('survey', $data)->where('survey_id', $id);
    if($this->db->affected_rows()>0){
            return true;
        }else{ 
                    return false;

                }

    }


  }

$.ajax({
       url: '<?php  echo base_url(); ?>index.php/admin/survey/act_surveyby_id',
       type: "POST",
       data: {
             idx : idx;
             },

public function act_surveyby_id(){
    $id=$_POST['idx'];
    $this->load->model('survey_m');
    if($this->survey_m->insert_activate($id))
    {                 
      echo "success";

    }else{

        echo "invalid";
    }   
}