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在python中，我可以创建一个在实例化时可以接收任意方法调用的类吗？我已阅读此内容，但无法将各个部分拼凑在一起

我想这与attribute lookup. 对于一个类Foo：

class Foo(object):
  def bar(self, a):
    print a

类属性可以通过 获得print Foo.__dict__，这给出

{'__dict__': <attribute '__dict__' of 'Foo' objects>, '__weakref__': <attribute '__weakref__' of 'Foo' objects>, '__module__': '__main__', 'bar': <function bar at 0x7facd91dac80>, '__doc__': None}

所以这段代码是有效的

foo = Foo()
foo.bar("xxx")

如果我打电话foo.someRandomMethod()，AttributeError: 'Foo' object has no attribute 'someRandomMethod'会导致。

我希望foo对象接收任何随机调用并默认为无操作，即。

def func():
    pass

我怎样才能做到这一点？我希望这种行为模拟一个对象进行测试。

4

1 回答 1

8

来自http://rosettacode.org/wiki/Respond_to_an_unknown_method_call#Python

class Example(object):
    def foo(self):
        print("this is foo")
    def bar(self):
        print("this is bar")
    def __getattr__(self, name):
        def method(*args):
            print("tried to handle unknown method " + name)
            if args:
                print("it had arguments: " + str(args))
        return method

example = Example()

example.foo()        # prints “this is foo”
example.bar()        # prints “this is bar”
example.grill()      # prints “tried to handle unknown method grill”
example.ding("dong") # prints “tried to handle unknown method ding”
                     # prints “it had arguments: ('dong',)”
于 2015-07-02T07:07:49.277 回答