Question

我遇到了PHP接口编码问题。我从“Head First Design Patterns”一书中获取了Java代码并将其转换为下面的代码。我正在使用MAMP / PHP 5.6.2和NetBeans 8.1。

我正在尝试在Menu类中实现扩展抽象类（MenuComponent）的接口“TestInterface”。 Menu类不会以“TestInterface”实现开头。当我注释掉“TestInterface＆＃34;在Menu类声明中作为下面的代码。虽然“TestInterface”已被注释掉，但即使在声明接口并将接口函数保持为Menu成员函数时，PHP也不会抛出任何错误。我已成功运行更简单的代码，同时使用与上述相同的平台同时扩展和实现。由于代码更简单，我相信我的代码中存在结构或语法错误。我希望有人可以帮我找到我做错的事。提前谢谢。

<?php

$run = new myclass;
$run->main();

class myclass {

   private $pancakeHouseMenu;
   private $allMenus;
   private $waitress;

   public function main(){

        echo "<br />hi main!<br />";

        $this->pancakeHouseMenu = new Menu("PANCAKE HOUSE MENU", "Breakfast");

        $this->allMenus = new Menu("ALL MENUS", "All menus combind");

        $this->allMenus->add($this->pancakeHouseMenu);

        $this->pancakeHouseMenu->add(new MenuItem(

            "Regular Pancake Breakfast",
            "Pancakes with eggs and sausage"));

        $this->waitress = new Waitress($this->allMenus);

        $this->waitress->printMenu(); 

    }

}

interface TestInterface {

    public function interfaceTest();


}

abstract class MenuComponent {

    public function add(MenuComponent $newMenuComponent) {

            throw new InvalidArgumentException("Exception thrown");

    }

    public function getName() {

            throw new InvalidArgumentException("Exception thrown");

    }

    public function getDescription() {

            throw new InvalidArgumentException("Exception thrown");

    }

    public function printOut() {

            throw new InvalidArgumentException("Exception thrown");

    }

}

class Waitress {

    private $allMenus;

    public function __construct(MenuComponent $allMenus) {

        $this->allMenus = $allMenus;
        $this->allMenus->add($allMenus);

    }

    public function printMenu() {

        $this->allMenus->printOut();

    }

}

class MenuItem extends MenuComponent {

    private $name;
    private $description;

    public function __construct($name, $description) {

        $this->name = $name;
        $this->description = $description;

    }

    public function getName() {

        return $this->name;

    }

    public function getDescription() {

        return $this->description;

    }

    public function printOut() {

        print(" " . $this->getName());   
        print("  -- " . $this->getDescription());

    }

}

class Menu extends MenuComponent /*** implements TestInterface ***/ {

    private $menuComponents = array();
    private $name;
    private $description;
   // private $testVar;

    public function __construct($name, $description) {

       $this->name = $name;
       $this->description = $description;
       $this->testVar = "Interface test succeeded";

    }

     public function interfaceTest(){

        return $this->testVar;

    }

    public function add(MenuComponent $newMenuComponent) {

        array_push($this->menuComponents, $newMenuComponent);

    }

    public function getName() {

       return $this->name;

    }

    public function getDescription() {

        return $this->description;

    }

    public function printOut() {

        print("<br />" . $this->getName());
        print(", " . $this->getDescription());
        print("<br />---------------------");
        print("<br />Testing interface var: ". $this->interfaceTest());

    }

}

?>

Answer 1

在您的代码中，您可以在类的声明之上创建一个对象。如果您的类不实现任何接口，这似乎没问题。由于您的类menu确实实现了接口TestInterface，因此PHP在声明类之前不接受您的对象实例化。

解决方案非常简单，将对象创建myclass置于对象声明之下：

<?php
class myclass {

    private $pancakeHouseMenu;
    private $allMenus;
    private $waitress;

    ...

    public function getDescription() {

          return $this->description;

     }

     public function printOut() {

          print("<br />" . $this->getName());
          print(", " . $this->getDescription());
          print("<br />---------------------");
          print("<br />Testing interface var: ". $this->interfaceTest());

     }

}

$run = new myclass;
$run->main();

?>

PHP接口编码

1 个答案: