强制动态加载的类来扩展或实现接口
任何人都知道在PHP中是否有可能强制类扩展或实现接口而子类不必声明它？
例：
interface Requirements
{
public function __construct();
public function kittens();
}
class DingleBerry
{
public function __construct()
{
// yadda yadda yadda
}
}
// Example of my initial hope
// of what you could do
$kittens = new DingleBerry implements Requirements;
interface Requirements
{
public function __construct();
public function kittens();
}
class DingleBerry
{
public function __construct()
{
// DO BAD STUFF (i.e. eat your soul)
}
}
// Example of what I CANNOT
// do.
$kittens = new DingleBerry;
if( !($kittens instanceof Requirements) )
{
// eat pizza.
}
没有找到相关结果
已邀请:
2 个回复
喷乡顾沥沪
// checks if class Bar implements Foo if (in_array('Foo', class_implements('Bar'))) { $foo = new Bar; } else { throw new Exception('Interface not implemented'); }委婪绷冗诉
<?php $reflection = new ReflectionClass('DingleBerry'); $reflection->implementsInterface('Requirements'); ?>