我知道我可以在运行时通过执行
$obj = (object)array( foo => bar );+
我可以用这种方式
echo $obj->foo; //bar
如果要使$obj从现有类继承，该怎么办？
What I wanna achive: I m forking paris project on github (https://github.com/balanza/paris). It s an active record class. I wonder I need to declare a class for every object, even if it s empty:
class User extends Model{}
我想我可能会使用动态对象来避免这些无聊的东西。
您可以始终执行eval（类User extends Model｛｝），但这不是一个好主意。实际上，你应该只在一个文件中创建类，然后操作码缓存就会正常工作，它可以被版本跟踪，等等。
tl；博士：定义模型类，这是正确的做法™.
如果真的没有其他方法，可以用我的Dynherit类创建一个“动态”继承。它创建了您想要的继承，但继承来自文件，并且只使用尚未加载的类。
下载v0.62:http://optimisationphp.komrod.com/source/sha/test/dynamic_inheritance_01/dynherit_v0.62.zip
GitHub链接：https://github.com/Komrod/Dynherit
严格地说，对于你的问题，我认为唯一可能的方法是使用Runkit，它允许您动态地重新声明类。不确定该解决方案的可移植性。
你可以使用原型设计模式来克隆一个对象，然后你可以在克隆的对象中添加你需要的一切
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