使用Ajax动态显示数据

毕竟有帮助和研究。这是此问题的运行代码。

在视图中：

<p id='state'><i ></i><span ><?php echo $countLike;?></span> likes &bull; <i ></i><?php echo $countDisLike;?> dislikes &bull;<i ></i><a href='<?php echo base_url();?>index.php/photoCheese/deleteUploadPic/<?php echo $row['uploadID'];?>'>Delete Picture</a></p><input type="button" onclick="getVal(this.value)"  name='like_name' id='like_id' value='<?php echo $link;?>' title='Like this post'><i ></i> Like</input>

Javascript：

<script type="text/javascript"> function getVal(value) { jQuery.ajax({type:"GET", url: "<?php echo base_url();?>index.php/photoCheese/like_total/",dataType:'json', data: {like_id : value}, error: function(result){$('.likeThis').append('<p>goodbye world</p>'); }, success: function(result){jQuery(".likeThis").html(result);  } }); } </script>

控制器：

public function like_total(){        $id = $this->session->userdata('userID');        $upload = $this->input->get('like_id');        $data = array('like' => 1,  'userID'=>$id,  'uploadID'=>$_GET['like_id']);        $result = $this->photoCheese_model->get_like_total($data,$upload);        $this->output->set_content_type('application/json');        $this->output->set_output(json_enpre($result));        return $result;    }

模型：

public function get_like_total($data,$upload){        $success = $this->db->insert('tbl_like',$data);        //Query the total likes        if($success){ $this->db->select()->from('tbl_like'); $this->db->where('uploadID',$upload); $this->db->where('like !=',2); $query = $this->db->get(); return $query->num_rows();        }        return 0;}

该代码现在可以完美运行。无论如何，谢谢您的帮助。