Question

我正在为“PickupPoints”创建一个界面，每个拾取点都需要能够返回找到的所有拾取点和拾取点详细信息，并可能在将来提供更多信息。这可以使用下面的代码：

<?php

interface iPickupPoint
{
    public function getPickupPoints($countryCode, $postalCode, $city);
    public function getPickupPointDetails($pickupPointId);
}


class PickupPoint1 implements iPickupPoint{
    ...
}

class PickupPoint2 implements iPickupPoint{
    ...
}

问题在于我不想调用PickupPoint1，PickupPoint2，它们自己的类。我想要一个像PickupPoint（PickupPointType）这样的类，所以我只想给出拾取点类型，类型需要是PickupPoint1，PickupPOint2，..

如何做到这一点？这甚至可能吗？这是最佳做法吗？

Answer 1

您所描述的是工厂模式，所以是的，这是最佳实践。

http://www.devshed.com/c/a/PHP/Design-Patterns-in-PHP-Factory-Method-and-Abstract-Factory/

您的工厂本身不需要实现该接口。它可以是静态类，甚至是函数：

class PickupPointFactory{
  public static function create($type){
    /* your creation logic here */
    switch ($type) {
      case "PickupPoint1" : 
        $obj = new PickupPoint1(); 
      break;
      case "PickupPoint2" :
        $obj = new PickupPoint2();  
      break;
    }
    return $obj;
  }
}

$newPoint = PickupPointFactory::create("PickupPoint2");

创建逻辑可以更通用，以避免每次向应用程序添加类时更改工厂：

class PickupPointFactory{
  public static function create($type, $options){
    /* your creation logic here */
    if(file_exists(dirname(__FILE__).'/'.$type.'.class.php')) {
      require_once(dirname(__FILE__).'/'.$type.'.class.php');
      $obj = new $type($options);
      return $obj;
    } else {
      throw new Exception('Unknown PickupPoint type: '.$type);
    }
  }
}

$newPoint = PickupPointFactory::create("PickupPoint2", array());

这假设您在工厂所在的同一目录中创建名为“PickupPoint1.class.php”的文件中的类，并且您的类中的构造函数只需要一个参数。

我没有测试这段代码，因此可能会出现一些错误。

Answer 2

我会创建1个拾取点功能，您可以在其中指定带有if语句的拾取点编号。

Pickup Point ($pickuppointnum) {

  if ($pickuppointnum == 1){
    //Pickup Point information for Pickuppoint #1
  }
  elseif ($pickuppointnum ==2) {
    //Pickup Point information for Pickuppoint #2
  }
}

在PHP中创建一个接口，它可以指定一个类型来确定要使用的类

2 个答案: