AJAX使用笨

Question

<button id="survey_act" method="post" class="tiny ui blue button" type="button" value="<?php echo $surv['id']; ?>" >Activate Survey</button> 

這是我在點擊按鈕 -

<script> 
$(document).ready(function(){ 
     $(document).on("click","#survey_act", function(){ 
      alert(this.value); 
       idx = this.value; 

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

     }).done(function(msg){ 
      if(msg=="success"){ 
         alert('You Successfully Activated the Survey!'); 
      } 

        });      
      }); 
    }); 

    </script> 

這是我的javascript -

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

    }else{ 

     echo "invalid"; 
    } 
} 

這是我的控制器 -

  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; 

       } 

    } 


    } 

這是我的模特 -

問題：當我點擊激活調查時，它不會更改/更新調查的詳細信息。我真的非常需要這方面的幫助。謝謝 。 。 。

Answer 1

變化$就功能類似下面

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

和控制器像下面

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"; 
    } 
}