[英]Codeigniter: Passing Data to View Undefined variable
when i pass $data
from controller to view, show me this error: 当我从控制器传递
$data
到视图时,告诉我这个错误:
Severity: Notice
严重性:注意
Message: Undefined variable: data
消息:未定义的变量:数据
Filename: views/admin.php
文件名:views / admin.php
Line Number: 10
行号:10
My Controller 我的控制器
class Home extends CI_Controller
{
public function index()
{
if($this->is_logged_in())
{
$data['login'] = $this->session->userdata('login');
$this->load->view('admin',$data['login']);
}else
{
$this->load->view('login');
}
}
public function login()
{
$this->load->model('usuario');
$resultado = $this->usuario->validate();
if($resultado)
{
$dados = array(
'login' => $this->input->post('login'),
'is_logged_in' => true
);
$this->session->set_userdata($dados);
$this->index();
}
else
{
redirect('fadas');
}
}
public function is_logged_in()
{
return $this->session->all_userdata();
}
and my View 和我的观点
<!DOCTYPE html>
<html>
<head>
<title>Admin Panel - Sistema de Controle de Indicações de Vereadores</title>
<link rel="stylesheet" href="<?php echo asset_url();?>css/style.css">
</head>
<body>
<header>
<nav>
<?php echo $data['login']; ?> </p>
</nav>
</header>
</body>
</html>
ty :) ty :)
Just pass $data, not $data['login']. 只需传递$ data,而不是$ data ['login']。 Then access it via $login.
然后通过$ login访问它。
Code below: 代码如下:
Controller: 控制器:
$this->load->view('admin',$data);
View: 视图:
<?php echo $login ?> </p>
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