Question

我有三个php类。我可以这样实例化它们：

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

但是，当我尝试以这种方式动态实例化它们时：

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

失败并显示此错误：

致命错误：Class'SeaFood（找不到新的Mashrooms。

我们将不胜感激。

Answer 1

$temp是一个对象，而不是一个类，您不能在现有对象上使用new关键字。

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

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

鉴于新信息

问题是我不知道我将实例化多少个类

我认为你的方法我错了。您是否考虑过使用Pizza课程并将您的顶级对象添加到披萨对象中？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

这将有效：

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

见这里：http://3v4l.org/RXDLX

php类的链实例化

2 个答案: