PHP：我可以在“父”类中定义一个新的类实例，并让新实例扩展父类吗？

Question

Im wanting to do something like:

class foo extends bar {
  private $norf = '';
  private $script;

  public function __construct(){
    $this->norf = 'blah blah';
  }

  public function qux(){
    $this->script = new someOtherClass();
    return $this->script->displayNorf();
  }
}

class someOtherClass {
    public function displayNorf(){
        return $this->norf;
    }
}

$test = new foo();
print($test->qux()); //blah blah

This example is really stupid but Im looking for a way to extend the instance of a class.

Answer 1

Yes, you can create instances of other classes inside your class. However, your code sample won't work simply because the class someOtherClass does not contain a property named $norf .

$this refers to the current instance of the class, so in someOtherClass->displayNorf() , since the class does not contain a property named $norf you'll receive an undefined property error.

In order for this to work, you'll need to change the foo class's property $norf from private to protected and extend the foo class through the someOtherClass class like so:

class foo extends bar {
  protected $norf = '';
  private $script;

  public function __construct(){
    $this->norf = 'blah blah';
  }

  public function qux(){
    $this->script = new someOtherClass();
    return $this->script->displayNorf();
  }
}

class someOtherClass extends foo {
    public function displayNorf(){
        return $this->norf;
    }
}

$test = new foo();
print($test->qux()); //blah blah

Answer 2

I think the answer is yes. If you run the code below, The output will be blah blah .

<code>
<?php
class foo {
  protected $norf = '';
  private $script;

  public function __construct(){
    $this->norf = 'blah blah';
  }

  public function qux(){
    $this->script = new someOtherClass();
    return $this->script->displayNorf();
  }
}

class someOtherClass extends foo {
    public function displayNorf(){
        return $this->norf;
    }
}
$test = new foo();
print($test->qux()); //blah blah
?>
</code>