创建被调用类的新实例而不是父类

Question

当方法位于父类中时，如何返回被调用类的实例。

例如，在下面的示例中，如果我拨打B::foo();，如何返回B的实例？

abstract class A 
{ 
    public static function foo() 
    { 
     $instance = new A(); // I want this to return a new instance of child class. 
      ... Do things with instance ... 
     return $instance; 
    } 
} 

class B extends A 
{ 
} 

class C extends A 
{ 
} 

B::foo(); // Return an instance of B, not of the parent class. 
C::foo(); // Return an instance of C, not of the parent class. 

我知道我能做到这样的事情，但有一个更合适的方法：

abstract class A 
{ 
    abstract static function getInstance(); 

    public static function foo() 
    { 
     $instance = $this->getInstance(); // I want this to return a new instance of child class. 
      ... Do things with instance ... 
     return $instance; 
    } 
} 

class B extends A 
{ 
    public static function getInstance() { 
     return new B(); 
    } 
} 

class C extends A 
{ 
    public static function getInstance() { 
     return new C(); 
    } 
} 

Answer 1

$instance = new static; 

您正在寻找Late Static Binding。

Answer 2

http://www.php.net/manual/en/function.get-called-class.php

<?php 

class foo { 
    static public function test() { 
     var_dump(get_called_class()); 
    } 
} 

class bar extends foo { 
} 

foo::test(); 
bar::test(); 

?> 

结果

string(3) "foo" 
string(3) "bar" 

所以你的函数将会是：

public static function foo() 
{ 
    $className = get_called_class(); 
    $instance = new $className(); 
    return $instance; 
} 

Answer 3

所有你需要的是：

abstract class A { 
    public static function foo() { 
     $instance = new static(); 
     return $instance ; 
    } 
} 

或者

abstract class A { 
    public static function foo() { 
     $name = get_called_class() ; 
     $instance = new $name; 
     return $instance ; 
    } 
}