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？我读过this却无法拼凑起来
我猜它与属性查找有关。对于类 Foo:
class Foo(object):
def bar(self, a):
print a
类属性可以通过print Foo.__dict__得到，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")
如果我调用 foo.someRandomMethod()，将导致 AttributeError: 'Foo' object has no attribute 'someRandomMethod'。
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？我希望此行为模拟一个对象以进行测试。
请您参考如下方法:
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',)”
