相关疑难解决方法(0)

为什么decorator不能装饰静态方法或类方法呢？

from decorator import decorator

@decorator
def print_function_name(function, *args):
    print '%s was called.' % function.func_name
    return function(*args)

class My_class(object):
    @print_function_name
    @classmethod
    def get_dir(cls):
        return dir(cls)

    @print_function_name
    @staticmethod
    def get_a():
        return 'a'

双方get_dir并get_a导致AttributeError: <'classmethod' or 'staticmethod'>, object has no attribute '__name__'.

为什么decorator依赖属性__name__而不是属性func_name？(Afaik所有函数,包括classmethods和staticmethods,都有func_name属性.)

编辑:我正在使用Python 2.6.

我正在寻找一种减少样板装饰器的方法.我们有很多使用@decorate的类.例如:

class MyClass(Base):
     @decorate
     def fun1(self):
         pass
     @decorate
     def fun2(self):
         pass
     def fun3(self):
         pass

我想这样做,默认情况下装饰者在那里,除非有人另有说明.

我使用此代码进行自动装配

from functools import wraps

def myDecorator(func):
    @wraps(func)
    def decorator(self, *args, **kwargs):
        try:
            print 'enter'
            ret = func(self, *args, **kwargs)
            print 'leave'
        except:
            print 'exception'
            ret = None

        return ret

    return decorator

class TestDecorateAllMeta(type):
    def __new__(cls, name, bases, local):
        for attr in local:
            value = local[attr]
            if callable(value):
                local[attr] = myDecorator(value)
        return type.__new__(cls, name, bases, local)

class TestClass(object):
    __metaclass__ = TestDecorateAllMeta

    def test_print2(self, val):
        print …