我有类别:
class Metaclass(type):
def __new__(self, name, bases, attrs):
for k, v in attrs.iteritems():
if isinstance(v, types.FunctionType):
attrs[k] = self.decorator(v)
return super(MetaClass, self).__new__(self, name, bases, attrs)
@classmethod
def decorator(cls, func):
def wrapper(*args, **kwargs):
print(func.__name__)
但我不能func.__class__.__name__
打印元类。我希望获得func.__name__
课程的名称。
显然它不可能是python 2.7(?)
我可以将classname作为参数传递吗?调用这个装饰器的方法?
即如果testmethod
调用metaclass装饰器并将Testclass
作为参数传递。所以,如果我知道func.__name__
来自特定类,那就做特定的事情。
答案 0 :(得分:1)
您忘了从装饰器返回包装器。它应该是:
class Metaclass(type):
def __new__(self, name, bases, attrs):
for k, v in attrs.iteritems():
if isinstance(v, types.FunctionType):
attrs[k] = self.decorator(v)
return super(Metaclass, self).__new__(self, name, bases, attrs)
@classmethod
def decorator(cls, func):
def wrapper(*args, **kwargs):
print(func.__name__)
return wrapper
您现在可以使用元类:
>>> class TestClass(object):
__metaclass__ = Metaclass
def foo(self):
return "bar"
>>> t = TestClass()
>>> x = t.foo()
foo
>>> print x
None
该方法已被包装器正确替换。
在您之前(现已删除)的问题中,您要求 print"此方法来自Testclass" 。您只需将类名传递给装饰器:
class Metaclass(type):
def __new__(self, name, bases, attrs):
for k, v in attrs.iteritems():
if isinstance(v, types.FunctionType):
attrs[k] = self.decorator(v, name)
return super(Metaclass, self).__new__(self, name, bases, attrs)
@classmethod
def decorator(cls, func, name):
def wrapper(self, *args, **kwargs):
print "%s defined in %s called from %s instance" % (func.__name__,
name, self.__class__.__name__)
return func(self, *args, **kwargs)
return wrapper
您现在获得:
>>> class TestClass(object):
__metaclass__ = Metaclass
def foo(self):
return "bar"
>>> t = TestClass()
>>> x = t.foo()
foo defined in TestClass called from TestClass instance
>>> print x
bar