Question

这是我的第一个代码＆amp;有错误：

可捕获的致命错误：传递给Shop :: __ construct（）的参数1必须是Generator的实例，没有给定，在第7行的C：\ xampp \ htdocs \ oop \ config.php中调用，并在C：\中定义第8行的xampp \ htdocs \ oop \ shop.class.php

第8行：

    public function __construct(Generator $generator)

我的代码：

class Shop
{
    private $generator;
    private $vates;

    public function __construct(Generator $generator)
    {
        $this->Generator = $generator;
        $this->vates = 'Connected with 250 vates!';
    }

    public function connect()
    {
        if ($this->Generator->isDown())
        {
            echo 'Sorry, the generator is down!';
        }
        else
        {
            echo 'Sucessfully', $this->vates;
        }
    }
}

CONFIG.PHP：

include("system.class.php");
include("shop.class.php");

$generator = new Generator;
$shop = new Shop;

System.class

class Generator
{
    private $power = false;

    public function powerUp() 
    {
        $this->power = true;
        echo 'You powered up the generator';
    }
    public function shutDown()
    {
        $this->power = false;
        echo 'The generator slowly shutting down...';
    }

    public function isDown()
    {
        return $this->power;
    }
}

我做错了什么？谢谢;）

Answer 1

试试这个：

include("system.class.php");
include("shop.class.php");

$generator = new Generator;
$shop = new Shop($generator);

商店类的构造签名是type hinting参数，这意味着您只能传入Generator对象。

Answer 2

您没有将任何参数传递给构造函数：

$generator = new Generator();
$shop = new Shop($generator);

传递给Shop :: __ construct（）的PHP OOP参数1必须是。的实例

2 个答案: