我有这样的类似按钮，当我点击这一块，不喜欢的会自动更改。显示按钮上方喜欢的数目。要做到这一点，我已经使用AJAX来实现此功能。

I have this like button, when I click this one, the no of likes will change automatically. The number of likes is displayed above the button. To do this, I've used ajax to implement this function.

这是我的code。在查看：

This is my code in the View:

<p id='state'>
   <i class='fa fa-thumbs-up'></i> <?php echo $countLike;?> likes &bull; 
   <i class='fa fa-thumbs-down'></i> <?php echo $countDisLike;?> dislikes</p>

   <button class='detailButton' name='like_name' id='click_like' value='<?php echo $link;?>' title='Like this post'><i class='fa fa-thumbs-up'></i> Like</button>

Javascript的code中的上述观点：

Javascript code above the view:

<script type="javascript/text">
$(document).ready(function){
  $("#click_like").click(function(event){
    event.preventDefault();
    var id = document.getElementById("click_name").value;
    jQuery.ajax({
      type:"GET",
      url: "<?php echo base_url();?>index.php/photoCheese/like_total/",
      dataType:'json',
      data: {like_id = id},
      success: function(res){
        if(res){
          jQuery("div#state").html(res.no_likes);
        }      
      }
    });
  });
});

我的控制器：

public function like_total(){
        $id = $this->session->userdata('userID');
        $upload = $_GET['like_id'];
        $data = array('like' => 1,
                        'userID'=>$id,
                        'uploadID'=>$_GET['like_id']);

        $this->photoCheese_model->get_like_total($data,$upload);

        return json_encode($data);
    }

型号：

public function get_like_total($data,$uplaod){
        $success = $this->db->insert('tbl_like',$data);

        if($success){
            $this->db->select('uploadID,SUM(`like`) as no_likes',false);
            $this->db->where('uploadID',$upload);
            $this->db->where('like !=',2);

            $query = $this->db->get();


        }
        return $query->result_array();
    }

当我尝试运行此code，它没有任何效果。我的意思是，它表明没有结果的。我是新来的阿贾克斯，这是我第一次也是为了实现这一个。请帮我这一个。非常感谢。

When I try to run this code, it has no effect. I mean, it shows no result at all. I'm new to ajax and it is my first time also to implement this one. Please, help me with this one. Thanks a lot.

推荐答案

象下面这样使用更改数据：到位 =

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>

jQuery.ajax({
      type:"GET",
      url: "<?php echo base_url();?>index.php/photoCheese/like_total/",
      dataType:'json',
      data: {like_id : id},
      success: function(res){

          $("#like").html(res);

      }
    });

<p id='state'>
   <i class='fa fa-thumbs-up'></i> <b id="like"><?php echo $countLike;?></b> likes &bull; 
   <i class='fa fa-thumbs-down'></i> <?php echo $countDisLike;?> dislikes</p>