有人可以幫我弄這個嗎？我不知道如何在視圖檔案上顯示結果。我嘗試使用$this->load->view('list-logs',$data);但它顯示錯誤Call to a member function view() on null。
控制器
public function getLogs()
{
helper(['form']);
$EstMod = new EstMod();
$data['est_data'] = $EstMod->orderBy('e_id', 'DESC')->findAll();
$rep_est = $this->request->getPost('rep_est');
$rep_date = $this->request->getPost('rep_date');
$builder = $this->db->table("logs as l");
$builder->select('l.l_id, u.fname, u.mname, u.lname, u.dept, u.user_type, u.u_grdsec, e.e_name, l.date_added, DATE(l.date_added), date_format(l.date_added, "%H:%i") as "time"');
$builder->join('users as u', 'l.u_id = u.u_id');
$builder->join('establishment as e', 'l.e_id = e.e_id');
$builder->where('DATE(l.date_added)', $rep_date);
$builder->groupBy('date_format(l.date_added, "%H:%i"), l.u_id');
$data['logs_data'] = $builder->get()->getResult();
$this->load->view('list-logs',$data);
//echo "<pre>";
//print_r($data);
}
看法
<?php
if ($logs_data) {
foreach ($logs_data as $row) {
echo '
<tr>
<td>' . $row['date_added'] . '</td>
<td>' . $row['lname'] . ", " . $row['fname'] . ", " . $row['mname'] . '</td>';
if ($row['user_type'] == 'S') {
echo '<td>STUDENT</td>';
} else if ($row['user_type'] == 'E') {
echo '<td>EMPLOYEE</td>';
} else if ($row['user_type'] == 'P') {
echo '<td>VISITOR</td>';
}
if ($row['dept'] == 'GS') {
echo '<td>GRADE SCHOOL</td>';
} else if ($row['dept'] == 'JHS') {
echo '<td>JUNIOR HIGH SCHOOL</td>';
} else if ($row['dept'] == 'SHS') {
echo '<td>SENIOR HIGH SCHOOL</td>';
}
echo '</td>
<td>' . $row['u_grdsec'] . '</td>';
}
}
?>
uj5u.com熱心網友回復：
如果您使用 CodeIgniter 4，那么您必須像這樣呼叫視圖 ont 控制器：
echo view('my_view', $data)
轉載請註明出處，本文鏈接：https://www.uj5u.com/caozuo/469013.html
標籤：php mysql 代码点火器 codeigniter-4