PHP類的鏈實例化

Question

我有三個php類。我可以實例他們這樣說：

$piza = new Mashrooms(new SeaFood(new PlainPiza())); 

然而，當我嘗試動態實例化它們以這樣的方式

$temp = Mashrooms(new SeaFood(new PlainPiza())); 
$piza = new $temp; 

失敗，並顯示以下錯誤：

Fatal error: Class 'SeaFood(new Mashrooms' not found.

您的幫助將不勝感激。

Answer 1

$temp是一個對象，而不是一個類，並且您不能在現有對象上使用new關鍵字。

$plain = 'PlainPiza'; 
$seafood = 'SeaFood'; 
$mashrooms = 'Mashrooms'; 

$piza = new $mashrooms(new $seafood(new $plain))); 

鑑於新的信息

The problem is I don't know how many classes I will instantiate

我覺得你的做法我是錯的。你有沒有想過有一個比薩類和添加您的頂部對象比薩餅對象？例如：

<?php 

class Pizza 
{ 
    private $_toppings; 
    private $_placements = array('left', 'right', 'whole'); 

    public function _construct() 
    { 
     foreach($this->_placements as $placement) 
     { 
      $this->_toppings[$placement] = array(); 
     } 
    } 

    public function add_topping(Base_Topping $topping, $placement) 
    { 
     if(in_array($placement, $this->_placements)) 
     { 
      array_push($this->_toppings[$placement], $topping); 
     } 
    } 
} 

abstract class Base_Topping 
{ 
    protected $_price = 0.00; 
    protected $_name = 'No Name'; 

    public function get_name() 
    { 
     return $this->_name; 
    } 

    public function get_price() 
    { 
     return $this->_price; 
    } 
} 

class Mushrooms extends Base_Topping 
{ 
    protected $_price = '1.00'; 
    protected $_name = 'Mushrooms'; 
} 

// assuming $_POST['toppings'] = array('Mushrooms' => 'whole', 'Pepperoni' => 0, 'Sausage' => 0, etc...) 
$pizza = new Pizza(); 
$toppings = array_filter($_POST); // will return anything with a non-false value 
foreach($toppings as $name => $coverage) 
{ 
    $topping = new $name(); 
    $pizza->add_topping($topping, $coverage); 
} 

?> 

Answer 2

這將工作：

$temp = "Mashrooms"; 
$pizza = new $temp(new SeaFood(new PlainPiza())); 

在這裏看到：http://3v4l.org/RXDLX