Question

我有一个名为Plugins的PHP类，其中我有其他plguin样式的PHP类，它们从PLugins类扩展。

有没有办法让所有从Plugins类扩展的PHP CLass名称？也许在PHP中使用Reflection之类的东西

因此，在下面的示例中，我将能够获得值：

MyTestPlugin
AnotherTestPlugin

abstract class Plugins
{
    // class properties and methods
}


class MyTestPlugin extends PLugins
{
    // class properties and methods
}


class AnotherTestPlugin extends PLugins
{
    // class properties and methods
}

Answer 1

是的，你可以通过 Reflection ：

来完成

$children  = array();
foreach( get_declared_classes() as $class )
{
    $reflected = new ReflectionClass( $class );
    if( $reflected->isSubclassOf( 'Plugins' ) ) $children[] = $class;
}

Answer 2

<?php

abstract class Plugins
{
    // class properties and methods
}


class MyTestPlugin extends Plugins
{
    // class properties and methods
}


class AnotherTestPlugin extends Plugins
{
    // class properties and methods
}

$plugin1 = new MyTestPlugin();
$plugin2 = new AnotherTestPlugin();


$parentClassName = 'Plugins';

foreach(get_declared_classes() as $class){
    if(is_subclass_of($class, $parentClassName)){
        echo $class.' == is a child class of '.$parentClassName.'<br>';
    }

}

?>

获取扩展类的类的类类名？

2 个答案: