在php中使用undefined常量

Question

不知道更好的標題，但這是我的代碼。

我時，它的實例化，檢查形式的數據類用戶，但我得到下面的錯誤/通知：

Notice: Use of undefined constant username - assumed 'username' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7 

Notice: Use of undefined constant password - assumed 'password' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7 

Notice: Use of undefined constant passwordc - assumed 'passwordc' in C:\Users\Jinxed\Desktop\WebTrgovina\app\m\Register\User.m.php on line 7 

... and so on for every defined variable in user class. 

這裏的用戶等級：

class User { 
    function __construct(){ 
     $test = 'blah'; 
     $username; $password; $passwordc; $name; $surname; $address; 
     $this->checkInput(array(username=>20, password=>20, passwordc=>20, name=>20, surname=>40, address=>40)); 
    } 
    //array(formName=>CharacterLimit) 
    private function checkInput($fields){ 
     foreach($fields as $field=>$limit){ 
      if($_POST[$field]=='' || strlen($_POST[$field])>$limit) $this->error[] = "$field must be filled in, and it must be less than or $limit characters long."; 
      else $this->{$field} = $_POST[$field]; 
     } 
    } 
} 

我不太明白什麼是問題，我試着先創建變量，然後從主程序調用checkInput方法，但我得到同樣的錯誤。

Answer 1

你應該引用的是你作爲數組鍵使用字符串：

$this->checkInput(array(
    'username'=>20, 
    'password'=>20, 
    'passwordc'=>20, 
    'name'=>20, 
    'surname'=>40, 
    'address'=>40)); 

的documentation說：

If you use an undefined constant, PHP assumes that you mean the name of the constant itself, just as if you called it as a string (CONSTANT vs "CONSTANT"). An error of level E_NOTICE will be issued when this happens. See also the manual entry on why $foo[bar] is wrong (unless you first define() bar as a constant).

相關部分的一半的權利，例如HTTP的