如何在不指定基类名的情况下从sub-sub-base-class静态方法引用基类的父类

Question

在PHP中，我想从子子类中调用父类的父类中的静态方法，而不引用父类的父类的名称（请参阅中的注释）代码如下）：

class Base {

  public static function helloStatic() {

    return "Hello base!\n";

  }

}

class Foo extends Base {

  private $fooMember;

  public static function helloStatic() {

    return "Hello foo!\n";

  }

  private function __construct() {

    $this->fooMember = "hello";

  }

  public function getFooMember() {

    return $this->fooMember;

  }

}

class Bar extends Foo {

  private $barMember;

  public static function helloStatic() {

    // I want to write the equivalent of:
    //echo Base::helloStatic();
    // here *without specifying any class names*

    echo get_parent_class(get_parent_class())::helloStatic();

  }

}

echo Bar::helloStatic();

预期输出：

Hello base!

输出：

<br />
<b>Parse error</b>:  syntax error, unexpected T_PAAMAYIM_NEKUDOTAYIM, expecting ',' or ';' on line <b>45</b><br />

Answer 1

将父类名存储在变量中，并使用该变量调用静态方法。像这样：

$parentClassName = get_parent_class(get_parent_class());
echo $parentClassName::helloStatic();