笨更新特定值

Question

我在MySQL数据库有POST_ID，post_number，post_like

我使用MVC的笨从MySQL表获取所有数据，并在页面上显示它的表（此部分已经完成）

我需要添加一个按钮，每个POST_ID，增加+1至post_like在mysql表，对应于POST_ID。

所以，如果我点击了+1按钮POST_ID 7，post_like应该只增加POST_ID 7

这里是我到目前为止！

view_page.php

 <?php foreach($members as $value): ?> 
     <tr> 
      <td><?php echo $value['post_id']; ?></td> 
      <td><?php echo $value['post_like']; ?></td> 
      <td> <a href="plusone/<?php echo $value['post_id']; ?>">Add +1</a></td> 
     </tr> 
     <?php endforeach; ?> 

controller_page.php

<?php 


    function plusone($post_id){ 

     $this->load->model('model_page'); 
     $this->model_page->data_addition($post_id); 
     redirect('controller_page/viewdata'); 

    } 

    ?> 

Model_page.php

 <?php 

    function data_addition($post_id){ 

     //trying to add 1 the post_like and update it in the db 
     $add_one_to_data = $post_like+ 1; 

     $data = array(
      'post_like' => $add_one_to_data +1 
     ); 

     $this->db->where('post_id', $post_id); 
     $this->db->update('post_tbl', $data); 

    } 

    ?> 

请不要“T嘲笑我的阵列在上面添加的$ add_one_to_data，我不知道自己还能做些什么

在此先感谢

Answer 1

，你需要首先得到ifno像：

function data_addition($post_id){ 
     $post_like = $this->db->get_where('post_tbl', array('post_id' => $post_id))->row(); 

     //trying to add 1 the post_like and update it in the db 
     $add_one_to_data = (int) $post_like->post_like + 1; 

     $data = array(
      'post_like' => $add_one_to_data +1 
     ); 

     $this->db->where('post_id', $post_id); 
     $this->db->update('post_tbl', $data); 

    } 

Answer 2

更换您的data_addition功能

function data_addition($post_id){ 
    $this->db->set('post_like', 'post_like+1', FALSE); 
    $this->db->where('post_id', $post_id); 
    $this->db->update('post_tbl'); 
}