php - 如何在codeigniter中连接2个表

Question

数据存储在数据库中，但无法在页面中显示。请帮助我

-表项目

项目_ID，项目名称，项目_描述，用户ID

-表用户

用户 ID、用户名

这是我在模型中的代码

public function get_all_project()  { 

$this->db->select('*');
$this->db->from('user');
$this->db->join('project','project.User_ID = user.User_ID');
$query = $this->db->get();  
return $query->result();  

} 

这是我在 Controller 中的代码

    public function list_all_project() {

    $data['projectadmin_list'] = $this->projectadmin_model->get_all_project();
    $this->load->view('projectadmin_list',$data);
    $this->load->model('projectadmin_model');
  }

这是我的代码

    <?php 

        foreach ($projectadmin_list as $data){ ?> 

     <tr> 
      <td><?php echo $data->Project_ID; ?></td> 
      <td><?php echo $data->Project_Name; ?></td>  
      <td><?php echo $data->Project_Desc; ?></td>
      <td><?php echo $data->Project_Total; ?></td>
      <td><?php echo $data->User_ID; ?></td>

      <td width="60" align="left" ><a href="#" onClick="show_confirm('edit',<?php echo $data->Project_ID;?>)">Edit</a></td>
      <td width="60" align="left" ><a href="#" onClick="show_confirm('delete_project',<?php echo $data->Project_ID;?>)">Delete </a></td>

     </tr>  
    <?php }?>  

Answer 1

在使用模型方法之前加载模型

public function list_all_project() {
    $this->load->model('projectadmin_model'); // should be load here
    $data['projectadmin_list'] = $this->projectadmin_model->get_all_project();
    $this->load->view('projectadmin_list',$data);

  }