Question

这是我的代码：

文件Application.php：

class Application {

    private $upgrade;

    public function __construct($upgrade = null) {
        if (!is_null($upgrade) && ($upgrade instanceof ApplicationUpgrade)) {
            $this->setUpgrade($upgrade);

            if ($this->getUpgrade()->getVersion() > $this->getVersion()) {
                $this->getUpgrade()->upgrade();
            }
        }
    }

    public function getVersion() {
        return '1';
    }

    public function getUpgrade() {
        return $this->upgrade;
    }

    public function setUpgrade(ApplicationUpgrade $upgrade) {
        $this->upgrade = $upgrade;
    }

    public function sayHi($name) {
        echo 'Hi ' . $name . '!';
    }
}

文件ApplicationUpgrade.php：

class ApplicationUpgrade {

    public function getVersion() {
        return '1.1';
    }

    public function upgrade() {
        //code upgrade here...
    }

    public function sayHi($name) {
        echo 'Hi ' . $name . ', Method has been upgraded!';
    }
}

现在：

$app = new Application(new ApplicationUpgrade());
$app->sayHi('A');

我想在升级之前，说他回归＆＃39;嗨A！＆＃39;在升级之后，sayHi返回＆＃39;嗨A，方法已升级！＆＃39;。

Answer 1

你可以这样做，但我认为这有点不好，但无论如何你走了

<?php
class Application {
    public function sayHi($name) {
       if($this->upgrade !== null) {
          $this->upgrade->sayHi($name);
       }
       else {
          echo 'Hi ' . $name . '!';
       }
    }
}

从类外部覆盖方法或传递参数

是不可能的

另一个选项是扩展Application类并覆盖一些方法 http://php.net/manual/en/language.oop5.inheritance.php

你应该阅读文档，因为它是学习OOP基础知识的一个很好的起点

class Application {
  public function sayHi() {
    echo('hi version1');
  }
}

class ApplicationV2 extends Application {
  public function sayHi() {
    echo('hi version2');
  }
}

$app = new ApplicationV2();
$app->sayHi();

此外，我不会这样做来管理不同的版本，你应该google arround以获得更好的解决方案并做一些研究

如何使用PHP创建类的函数升级

1 个答案: