在视图中使用动态显示消息

Question

我正在尝试在$ data ['message']

中设置动态消息 例如，假设在登录表单上我想显示消息，请输入用户名和密码，但是在注销操作上显示你是succ。退出。

所以我在想

function login()
{
   $data['message'];
   if(!isset $data['message'])
   {
      $data['message'] = "please enter your username and password";
   }
   $this->load->view('login' $data);
}

function logout()
{
   unset session and .....
   $data['message'] = "You were succ. logged out!";
   $this->login($data);
}

但我在渲染登录方法

上遇到以下错误

Parse error: syntax error, unexpected T_VARIABLE, expecting '('

内线if(!isset $data['message'])

Answer 1

if(!isset($data['message']))
 { //etc...

isset需要括号

Answer 2

你可以这样试试.. 你忘记了isset

的括号

function login()
    {
       $data['message'];
       if(!isset($data['message']))
       {
          $data['message'] = "please enter your username and password";
       }
       $this->load->view('login' $data);
    }

    function logout()
    {
       //unset session and .....
       $data['message'] = "You were succ. logged out!";
       $this->login($data);
    }