ajax - 使用 Ajax 动态显示数据

Question

在此代码中，单击“赞”按钮后，数据已添加到数据库中。我现在想要做的是我想在添加数据后，我会查询所选项目的总数并显示它而不加载页面。

这是我现在的代码:

我的看法:

<p id='state'><i class='fa fa-thumbs-up'></i><span id="likeThis"><?php echo $countLike;?></span> likes &bull; <i class='fa fa-thumbs-down'></i><?php echo $countDisLike;?> dislikes &bull;<i class='fa fa-thumbs-down'></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)" class='detailButton1' name='like_name' id='like_id' value='<?php echo $link;?>' title='Like this post'><i class='fa fa-thumbs-up'></i> Like</input>

Javascript:
function getVal(value) { jQuery.ajax({ type:"GET", url: "<?php echo base_url();?>index.php/photoCheese/like_total/", dataType:'json', data: {like_id : value}, success: function(res){ alert(res.no_likes); if(res){ jQuery("#likeThis").html(res.no_likes); }
} });
Controller :

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);


        return json_encode($result);
    }

模型:

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();
    }

此代码不会显示 total_likes。这个有什么问题吗？

Answer 1

这对我来说似乎是一个不错的代码

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);


    return json_encode($result);
}

试一试..改变你的

return json_encode($result)

到

echo json_encode($result)

这个例子以后可能对你有帮助jquery ajax php example