Question

mysql表中的多个id如何在codeigniter的另一个mysql表中加载并获取特定的id值，

页面查看已添加的图片，

查看页面

   `<?php foreach($company_branch as $loop){  ?>
    <tr> 
    <td></td>
    <td><?=$loop->branch_name ?></td>
    <td><?=$loop->branch_head ?></td>
    <td><?=$loop->departments_list_id ?></td>
    <td><?=$loop->write_date ?></td>
     </tr>
    <?php } ?>`

控制器页面

      `public function company_settings() {
        $data['company_branch'] = $this->settings_model->company_branch();
      $this->load->view('settings/company_settings',$data);`

模型页面

   `function company_branch(){    
    $this->db->select('company_branch.*,
       company_departments.department_name as departments_name  ')-
    >from('company_branch');
     $this->db- >join('company_departments',' 
    company_branch.departments_list_id = company_departments.id');
    $query = $this->db->get();      
    return $query->result();
    }`


  How to display the `departments_list_id ` to department names in view 
  page,

The image is company_branch and company_department mysql table view，

browser View page，

Answer 1

查看页面

          `<?php foreach($company_branch as $loop){  ?>
             <tr> 
            <td></td>
            <td><?=$loop->branch_name ?></td>
            <td><?=$loop->branch_head ?></td>
            <td>
               <?php   
                  foreach(explode(",",$loop->departments_list_id) as $department)
                  {  $depart=$this->db->query("select department_name from company_departments where id=".$department)->row()->department_name;
                  echo $depart.',<br />';
                 }
                ?>
            </td>
            <td><?=$loop->write_date ?></td>
             </tr>
             <?php } ?>`

Answer 2

您无法在1个字段中输入一些ID。您可以将一些字段添加到另一个ID。示例字段：departement_id_1，departement_id_2，departement_id_4，departement_id_5等取决于需要。

您可以将sql select更改为：

select branc.*,
(select dept.departement_name as departement_name_1 from company_departement as dept where branc.departement_id_1=dept.id),
(select dept.departement_name as departement_name_2 from company_departement as dept where branc.departement_id_2=dept.id),
(select dept.departement_name as departement_name_3 from company_departement as dept where branc.departement_id_2=dept.id)
from company_branc as b

然后您可以使用表单中的脚本

    <?php foreach($company_branch as $loop){  ?>
    <tr> 
    <td></td>
    <td><?=$loop->branch_name ?></td>
    <td><?=$loop->branch_head ?></td>
    <td><?=$loop->departments_name_1 ?></br>
<?=$loop->departments_name_2 ?></br>
<?=$loop->departments_name_3 ?>
</td>
    <td><?=$loop->write_date ?></td>
     </tr>
    <?php } ?>

我希望这有帮助

sql中的多个id如何在codeigniter中进行expload并获取id值

2 个答案: