我一直收到此错误"不在对象上下文中使用$this"
实例。
$axre = new Axre();
if (!filter_var($this->email, FILTER_VALIDATE_EMAIL)){
echo "yes";
} else {
echo"no";
}
如何进入对象上下文?
它$email受到保护,我需要保护它,因为它在表单上。
此框架中的大多数对象都是通过调用构造函数填充的,但是这个有各种各样的入口点。
class Axre extends Base {
protected $email;
protected $user_name;
protected $temp_token;
protected $sign_in_token;
protected $UserShoppingList;
function __construct($email = null) {
if (strpos($email, '@') === false) {
$this->sign_in_token = $email;
} else {
$this->email = $email;
}
}
$this 变量是对类的当前实例的特殊引用。您正在使用没有任何意义的类之外。
从 PHP 示例
class Vegetable {
var $edible;
var $color;
function Vegetable($edible, $color="green")
{
$this->edible = $edible; // $this refers to the instance of vegatable
$this->color = $color; // ditto
}
}
$this->variable; //This does not make sense because it's not inside an instance method
class Axre extends Base {
protected $email;
protected $user_name;
protected $temp_token;
protected $sign_in_token;
protected $UserShoppingList;
function __construct($email = null) {
if (strpos($email, '@') === false) {
$this->sign_in_token = $email;
} else {
$this->email = $email;
}
}
function isValidEmail() {
if (filter_var($this->email, FILTER_VALIDATE_EMAIL)) {
return true;
}
return false;
}
}
现在要使用它,您只需实例化类并调用 isValidEmail 方法来获得真/假结果。
$test = new Axre($email);
$validEmail = $test->isValidEmail();
您不能在类外部使用$this并期望它正常工作。