我想以我附加的特定格式显示数据。教师添加的每个对象,标题和坐标应显示在一行教师面前。这就是我想要的。这就是我目前获取数据的方式
视图和控制器的代码在这里
<table id="example2" class="table table-bordered table-hover">
<thead>
<tr>
<th>#</th>
<th>Instructor</th>
<th>Title</th>
<th>Coordinates</th>
<th>Actions</th>
</tr>
</thead>
<tbody>
<?php $i=1; foreach ($res as $value) { ?>
<tr>
<td><?php echo $i++; ?></td>
<td><?php echo $value['first_name'] . " " . $value['last_name']; ?></td>
<td><?php echo $value['description']; ?></td>
<td><?php echo $value['coordinates']; ?></td>
<td>
<?php if($edit_delete['edit']) {?>
<a href="<?php echo $base_url; ?>Objects/view/<?php echo $value['id']; ?>"><i class="fa fa-eye"></i></a>
<?php } ?>
</td>
</tr>
<?php } ?>
</tbody>
</table>
public function index()
{
$this->user_model->check_permissions('Objects/index');
$data['edit_delete']=$this->user_model->checkEditDelete('Objects/index');
$rl_id= $this->rol_id;
$str = implode('-',$rl_id);
if($str==2)
{
$data['menu']=$this->load_model->menu_inst();
}else
{
$data['menu']=$this->load_model->menu();
}
$data['base_url'] = base_url();
$data['userInfo'] = $this->userInfo;
$rl_id= $this->rol_id;
$super = implode('-',$rl_id);
if($super==1)
{
$this->db->select('object.id, object.description, object.coordinates, admin.first_name, admin.last_name');
$this->db->from('object');
$this->db->join('admin', 'admin.id = object.instructor_id');
$this->db->where('admin.is_delete',0);
$this->db->where('admin.role_id',2);
$this->db->where('object.is_delete',0);
$data['res'] = $this->db->get()->result_array();
}
else
{
$in_id= $this->insret_id;
$insrtuctur = implode('-',$in_id);
$data['data'] = $this->db->where('is_delete',0)->where('instructor_id',$insrtuctur)->get('object')->result_array();
}
$data['page'] = "objects/objects";
$this->load->view('Template/main', $data);
}