Question

是否可以在扩展类中使用 类实例 调用静态方法？在扩展它时，我不知道从我的课程中获得$this->access来自__construct()的{{1}}，所以也许PHP专家可以帮助我。

namespace Folder\Subfolder;

class ClassA {  // <-- this class is a custom helper

  public static function accessA() {
    return 'Class A\'s Access';
  }

  public static function accessB() {
    return 'Class B\'s Access';
  }

}

在其他文件中：

use Folder\SubFolder\ClassA;

class ParentClass {

  public function __construct() {
    $this->access = new \Folder\Subfolder\ClassA;
  }

  public function admin() {
   $ac = $this->access::accessA(); 

   //When I do this it gives me an error syntax error, unexpected '::' (T_PAAMAYIM_NEKUDOTAYIM)

   $ac = $this->access->accessA();

   //When I also do this it gives me an error Call to a member function accessA() on null

   var_dump($ac);
   exit;
  }
}

class ChildClass extends ParentsClass { 

 public function __construct() {
  parent::__construct();
 }

 public function employee() {
  $ac = $this->access::accessB();
  var_dump($ac);
  exit;
 }
}

class ChildClass2 extends ParentClass {
  public function __construct() {
   parent::__construct();
  }
} //And so on and so forth

注意：我不想使用new关键字来实例化另一个类对象。我想仅使用此代码$this->access;

可以在php的其他类中调用实例化的静态方法吗？

0 个答案: