如何在PHP中输入提示__call()魔术方法？

Question

出于测试目的，我正在实现一个Adapter pattern，我想用两个不同类的方法来输入提示适配器的返回，我该如何实现呢？

class Foo {
  public function somethingOnlyFooHave() {};
}

class Adapter {
  protected $handler;

  public function __construct(object $handler) {
    $this->handler = $handler;
  }

  public function __call($name, ...$args) {
    $this->handler->$name($args);
  }

  public function somethingOnlyTheAdapterHave() {}
}

$foo = new Adapter(new Foo);

// How can I get type-hinting for both the Adapter and Foo?
$foo->somethingOnlyFooHave();
$foo->somethingOnlyTheAdapterHave();

Answer 1

似乎PHP本身不能做到这一点，但是如果您使用的是PHPStorm 2018.3+，那么只要它不在__construct上，您就可以使用联合类型(或交集类型)

class Foo {
    public function somethingOnlyFooHave() {}
}

class Adapter {
    protected $handler;

    public function __construct(object $handler) {
        $this->handler = $handler;
    }

    public function __call($name, ...$args) {
        $this->handler->$name($args);
    }

    /**
     * @return Adapter|Foo
     */
    public function get() {
        return $this;
    }

    public function somethingOnlyAdapterHave() {}
}

$foo_adapter = ( new Adapter(new Foo) )->get();

// The following methods receives type-hinting:
$foo_adapter->somethingOnlyFooHave();
$foo_adapter->somethingOnlyAdapterHave();

有用的链接：

• https://blog.jetbrains.com/phpstorm/2018/09/phpstorm-2018-3-eap-183-2635-12/#intersection-types
• https://medium.com/@ondrejmirtes/union-types-vs-intersection-types-fd44a8eacbb