php - 创建一个被调用的类的新实例而不是父类

Question

当方法在父类中时，如何返回被调用类的实例。

例如。在下面的示例中，如何返回Bif 我调用的实例B::foo();？

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.

我知道我可以这样做，但是有没有更简洁的方法：

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;

您正在寻找后期静态绑定。

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"

所以你的功能将是：

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 ;
    }
}