如何在不指定基類名的情況下從子基類靜態方法引用基類的父類

Question

在PHP中，我想在子類的父類的父類中調用一個靜態方法，類，不參照父類的父類的名稱（請參閱下面的代碼中的註釋）：

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(); 

預期輸出：

Hello base! 

OUTPUT：

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

Answer 1

將父類名稱存儲在變量中，並使用該變量調用靜態方法。像這樣：

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