使用Ajax动态显示数据

一尘不染

使用Ajax动态显示数据

ajax

在此代码中，单击“赞”按钮后，数据已添加到数据库中。我现在想做的是添加数据后，我将查询所选项目的总数，并在不加载页面的情况下显示它。

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); } } });

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);


        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();
    }

此代码将不会显示total_likes。这是怎么了

阅读 253

2020-07-26

共1个答案

一尘不染

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 class='fa fa-thumbs-up'></i><span class="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：

<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');
        $this->output->set_output(json_encode($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。

2020-07-26