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Php 如何从其他表中获取数据?_Php_Codeigniter - Fatal编程技术网

Php 如何从其他表中获取数据?

Php 如何从其他表中获取数据?,php,codeigniter,Php,Codeigniter,我想显示另一个表中的数据。这是CodeIgniter框架 我试过很多代码,但都不起作用 这是我的代码: <?php foreach ($users->result_array() as $key => $user): ?> <tr> <td> <?php echo $key+1; ?></td> <td> <img src="<?p

我想显示另一个表中的数据。这是CodeIgniter框架

我试过很多代码,但都不起作用

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>
我的课程和注册课程模型

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));
    }
我的数据库是这样的 表->类别类别:列:ID+名称
table->USER:column:ID+classcategory\u ID

您可以将模型中的用户表与classcategory连接起来,并选择classcategory.name作为categoryname,然后
echo$USER['categoryname']
。下面是一个如何在CodeIgniter中连接两个表的示例

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

您可以将模型中的用户表与CLASSCATEGORY联接,并选择CLASSCATEGORY.name作为categoryname,然后
echo$USER['categoryname']
。下面是一个如何在CodeIgniter中连接两个表的示例

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

请分享您的相关控制器和型号代码、您想要的数据以及您从查询中获得的数据。此外,请提供表的架构。请共享相关控制器和模型代码、所需数据以及从查询中获得的数据。另外,提供表的模式。
    $this->db->select('USER.*, CLASSCATEGORY.name AS categoryname')
         ->from('users')
         ->join('CLASSCATEGORY', 'CLASSCATEGORY.id = USER.classcategory_id')
         ->get()
         ->result();
public function getallRecords($id,$product_id)
    {
        $this->db->select('*');
        $this->db->from('products');
        $this->db->where('id', $id);
        $this->db->where('product_id', $product_id);
        $this->db->where('show', 'no');
        $this->db->where('Results !=', '0');
        $this->db->order_by('added', 'DESC');
        $this->db->limit(30);
        $getallRecords=$this->db->get();
        $resultllRecords=$getallRecords->result();
        return $resultllRecords;
    }