【发布时间】:2017-01-04 03:41:33
【问题描述】:
我创建了一个帐户页面,我可以在其中查看数据库中的所有帐户。他们每个人都有一个名为查看更多的操作按钮。如果我按查看更多,来自数据库的其他详细信息将显示在文本框中。我想获取每一行的 ID 并将其传递给控制器以便查看它。但是,我似乎无法让它们显示在框中。
预期:Image
当前:Image
控制器:
function viewmore()
{
$userid = $this->input->get('userid', TRUE);
$data['view'] = $this->model_accounts->viewmore_user($userid);
$data['main_content'] = 'view_adminviewmore';
$this->load->view('includes/admin_viewmore_template', $data);
}
型号:
public function viewmore_user($userid)
{
$query= $this->db->select('*')->from('accounts')->where('userid', $userid)->get();
return $query->result();
}
操作按钮的视图:
<?php foreach($users as $row): ?>
<tr>
<td><?php echo $row->firstname; ?></td>
<td><?php echo $row->lastname; ?></td>
<td><?php echo $row->username; ?></td>
<td><?php echo $row->address; ?></td>
<td class="action-button mobile-important">
<a href="<?php echo base_url() ."admin_accounts/viewmore/?id=". $row->userid ?>"> <button type="button" class="btn btn-custom-3">View More</button></a>
</td>
<?php endforeach; ?>
</tr>
显示数据的视图:
<?php foreach($view as $row): ?>
<br>
<p> First Name </p>
<input class="form-control" id="sel1" type="text" placeholder="" value="<?php echo $row->firstname; ?>">
<br>
<p> Last Name </p>
<input class="form-control" id="sel1" type="text" placeholder="" value="<?php echo $row->lastname; ?>">
<br>
<p> Username </p>
<input class="form-control" id="sel1" type="text" placeholder="" value="<?php echo $row->username; ?>">
<br>
<p> Address </p>
<input class="form-control" id="sel1" type="text" placeholder="" value="<?php echo $row->address; ?>">
<br>
<p> E-mail Address </p>
<input class="form-control" id="sel1" type="email" placeholder="" value="<?php echo $row->email; ?>">
<br>
<p> Contact Number </p>
<input class="form-control" id="sel1" type="text" placeholder="" value="<?php echo $row->contactnum; ?>">
<br>
<?php endforeach; ?>
【问题讨论】:
标签: php html mysql codeigniter