如何从其他表中获取数据？

威尔金斯·皮埃尔

我想显示另一个表中的数据。这是CodeIgniter框架。

我尝试了许多代码，但是它们不起作用。

这是我的代码：

<?php
foreach ($users->result_array() as $key => $user): ?>
   <tr>
      <td> <?php echo $key+1; ?></td>
      
      <td>
        <img src="<?php echo $this->user_model->get_user_image_url($user['id']);?>" alt="" height="50" width="50" class="img-fluid rounded-circle img-thumbnail">
      </td>

      <td> <?php echo $user['first_name'].' '.$user['last_name']; ?>
        <?php if($user['status'] != 1): ?>
          <small><p><?php echo get_phrase('status'); ?>: <span class="badge badge-danger-lighten"><?php echo get_phrase('unverified'); ?></span></p></small>
        <?php endif; ?>
      </td>

      <td>
        <?php echo $user['email']; ?>
      </td>

      <td>
        <?php echo $user['classcategory_id']; ?>
      </td>

      <td>
        <?php $enrolled_courses = $this->crud_model->enrol_history_by_user_id($user['id']);?>
          <ul>
            <?php foreach ($enrolled_courses->result_array() as $enrolled_course):
            $course_details = $this->crud_model->get_course_by_id($enrolled_course['course_id'])->row_array();?>
              <li><?php echo $course_details['title']; ?></li>
           <?php endforeach; ?>
         </ul>
     </td>

我该如何改变

 <td>
   <?php echo $user['classcategory_id']; ?>
 </td>

要从表名称“ classcategory”字段中的数据“名称”

我的控制器，用于用户和课程

public function get_courses() {
        if ($this->session->userdata('admin_login') != true) {
            redirect(site_url('login'), 'refresh');
        }
...
$data = array();
        if(!empty($courses)) {
            foreach ($courses as $key => $row) {
                $instructor_details = $this->user_model->get_all_user($row->user_id)->row_array();
                $category_details = $this->crud_model->get_category_details_by_id($row->sub_category_id)->row_array();
                
                $sections = $this->crud_model->get_section('course', $row->id);
                $lessons = $this->crud_model->get_lessons('course', $row->id);
                $enroll_history = $this->crud_model->enrol_history($row->id);
...

public function users($param1 = "", $param2 = "") {
        if ($this->session->userdata('admin_login') != true) {
            redirect(site_url('login'), 'refresh');
        }
        if ($param1 == "add") {
            $this->user_model->add_user();
            redirect(site_url('admin/users'), 'refresh');
        }
        elseif ($param1 == "edit") {
            $this->user_model->edit_user($param2);
            redirect(site_url('admin/users'), 'refresh');
        }
        elseif ($param1 == "delete") {
            $this->user_model->delete_user($param2);
            redirect(site_url('admin/users'), 'refresh');
        }

        $page_data['page_name'] = 'users';
        $page_data['page_title'] = get_phrase('student');
        $page_data['users'] = $this->user_model->get_user($param2);
        $this->load->view('backend/index', $page_data);
    }

我的课程模型和已注册的课程

public function get_course_by_id($course_id = "")
    {
        return $this->db->get_where('course', array('id' => $course_id));
    }

public function enrol_history_by_user_id($user_id = "")
    {
        return $this->db->get_where('enrol', array('user_id' => $user_id));
    }

而我的数据库就是这样的表-> CLASSCATEGORY：列：ID +名称表-> USER：列：ID + classcategory_id

艾哈迈德·谢沙阿（Ahmed Sheashaa）

您可以使用CLASSCATEGORY和SELECT CLASSCATEGORY.name AS类别名称，然后使用，在模型中加入USER表echo $user['categoryname']。这是如何在CodeIgniter中联接两个表的示例

    $this->db->select('USER.*, CLASSCATEGORY.name AS categoryname')
         ->from('users')
         ->join('CLASSCATEGORY', 'CLASSCATEGORY.id = USER.classcategory_id')
         ->get()
         ->result();

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