php - 将数据库内容从一个表复制到 codeigniter 中的另一个表

Question

我需要单击存档按钮将表内容从表:user_openletter 复制到表:国家/地区。如何编写查询以将 from、to、title、openletter、open_id 从 user_openletter 复制到另一个表国家/地区。这是一个 codeigniter 应用程序。

我的 Controller 代码是:

 public function store_user_data_archieve()       
 {
    $match = $this->input->post('submit');
    echo $match;
    if($match == "Archieve")
    {
        $data = array(
         'open_id' => $this->input->post('open_id'),
         'featured' => '1',
         'from' => $this->input->post('from'),
             'to' => $this->input->post('to'),
             'title' => $this->input->post('title'),

             'archieve' => '1',
             'latest' => '0',
             'sponsor' => 'images/sponsor.png'

           );
      $this->load->database();  
         //load the model  
         $this->load->model('select');  
         //load the method of model  
         $data['r']=$this->select->store_user_data_archieve($data);  
        echo "success";
        }
        else if(!$match == "new")
        {
            $data = array(
         'open_id' => $this->input->post('open_id'),
         'featured' => '1',
         'from' => $this->input->post('from'),
             'to' => $this->input->post('to'),
             'title' => $this->input->post('title'),
             'openletter' => $this->input->post('openletter'),
             'archieve' => '0',
             'latest' => '1',
             'sponsor' => 'images/sponsor.png'

           );
      $this->load->database();  
         //load the model  
         $this->load->model('select');  
         //load the method of model  
         $data['r']=$this->select->store_user_data_archieve($data);  
        echo "success";

        }
           else if(!$match == "Discard")
        {

        echo "failure";

        }

      }

我的查看代码是:

<?php  
         foreach ($r->result() as $row)  
         {  
        ?>

              <table border="1" cellpadding="4" cellspacing="0">
                <tr>
                    <td>from</td>
                    <td>to</td>
                    <td>title</td>
                    <td>openletter</td>
                    <td>Date & Time</td>
                    <td>open_id</td>
                </tr>

                    <tr>
                 <form action="/index.php/welcome/store_user_data_archieve" method="post">
                        <td><input type="text" name="from" value="<?php echo $row->from;?>" /></td>
                    <td><input type="text" name="to" value="<?php echo $row->to;?>" /></td>
                    <td><input type="text" name="title" value="<?php echo $row->title;?>" /></td>
                    <td><input type="text" name="openletter" value="<?php echo $row->openletter;?>" /></td>
                    <td><input type="text" name="datetime" value="<?php echo $row->datetime;?>" /></td>
                    <td><input type="text" name="open_id" value="<?php echo $row->open_id;?>" /></td>
                    <td><div><input type="submit" name="submit" value="Archieve" /></div></td>
                    <td><div><input type="submit" name="new" value="new" /></div></td>

                </form>
                </tr>
              </table>

         <?php } ?>

我的模型代码是:

public function store_user_data_archieve($data)  
{  
 //data is retrieved from this query
$this->db->insert('country', $data);
$this->db->set('openletter');       
$this->db->select('openletter');
$this->db->where('open_id', $data[open_id]); 
$this->db->from('user_openletter');

// Produces: INSERT INTO mytable (title, name, date) VALUES ('{$title}', '{$name}', '{$date}')
}  

Answer 1

根据我们的讨论，您需要将 from,to,title,openletter,open_id 从表 user_openletter 复制到 country 表

$this->db->select('from,to,title,openletter,open_id');// select your filed
$q = $this->db->get('user_openletter')->result(); // get result from table
foreach ($q as $r) { // loop over results
        $this->db->insert('country', $r); // insert each row to country table
    }