Question

是否可以从php

中的派生对象获取基础对象

像这样的东西

class base {}
class derived extends base{
public function getBase()
{
         return (base)$this;
}

上面的代码，抛出错误

Answer 1

您可以使用parent::来解析父方法，属性或常量。

Answer 2

如果您正在尝试获取基类的名称，请执行以下操作：

class base {
  public function getClassName() {
    return "base";
  }
}
class derived extends base{
  public function getBaseClassName() {
    return parent::getClassName();
  }
  public function getClassName() {
    return "derived";
  }
}

$d = new derived();
echo $d->getBaseClassName();

编辑：当您使用继承扩展某个类（例如：derived extends base）时，您说derived 是一种 base，derived 的所有实例也是base的实例。在大多数OO语言中，base和derived这两个实例不是单独的实体，不能单独处理。（在这方面，C ++是规则的例外）。

如果您需要单独处理实例，那么继承是该作业的错误工具。通过包含使用扩展，而不是继承。这看起来如下所示：

class base {

  public someBaseFunction() { 
    // ...
  }
}

class derived {

  /**
   * each instance of `derived` *contains* an in instance of `base`
   * that instance will be stored in this protected property
   */
  protected $base;

  /**
   * constructor
   */
  function __construct() {

    // remember to call base's constructor
    // passing along whatever parameters (if any) are needed
    $this->base = new base();

    // ... now do your constructor logic, if any ...

  }

  /**
   * Here's the method that fetches the contained
   * instance of `base`
   */
  public function getBase() {
    return $this->base;
  }

  /**
   * If needed, `derived` can implement public elements
   * from `base`'s interface. The logic can either delegate
   * directly to the contained instance of base, or it can
   * do something specific to `derived`, thus "overriding" 
   * `base`'s implementation.
   */
  public function someBaseFunction() {
    return $this->base->someBaseFunction();
  }

  /**
   * of course, `derived` can implement its own public
   * interface as well...
   */
  public function someOtherFunction() {
  }
}

PHP中的对象切片

2 个答案: