Question

如何将self.key传递给装饰者？

    class CacheMix(object):

        def __init__(self, *args, **kwargs):
            super(CacheMix, self).__init__(*args, **kwargs)

        key_func = Constructor(
            memoize_for_request=True,
            params={'updated_at': self.key}
        )

        @cache_response(key_func=key_func)
        def list(self, *args, **kwargs):
            pass

    class ListView(CacheMix, generics.ListCreateAPIView):
        key = 'test_key'

我收到错误：

'self' is not defined

Answer 1

这是一个用类装饰器做的例子，我试图在评论中向你描述。我在你的问题中填写了一些未定义的引用，并使用了cache_response函数装饰器的超简化版本，但希望这将足够具体地传达这个想法，使你能够使它适应你的真实代码。 / p>

import inspect
import types

class Constructor(object):
    def __init__(self, memoize_for_request=True, params=None):
        self.memoize_for_request = memoize_for_request
        self.params = params
    def __call__(self):
        def key_func():
            print('key_func called with params:')
            for k, v in self.params.items():
                print('  {}: {!r}'.format(k, v))
        key_func()

def cache_response(key_func):
    def decorator(fn):
        def decorated(*args, **kwargs):
            key_func()
            fn(*args, **kwargs)
        return decorated
    return decorator

def example_class_decorator(cls):
    key_func = Constructor(  # define key_func here using cls.key
        memoize_for_request=True,
        params={'updated_at': cls.key} # use decorated class's attribute
    )
    # create and apply cache_response decorator to marked methods
    # (in Python 3 use types.FunctionType instead of types.UnboundMethodType)
    decorator = cache_response(key_func)
    for name, fn in inspect.getmembers(cls):
        if isinstance(fn, types.UnboundMethodType) and hasattr(fn, 'marked'):
            setattr(cls, name, decorator(fn))
    return cls

def decorate_me(fn):
    setattr(fn, 'marked', 1)
    return fn

class CacheMix(object):
    def __init__(self, *args, **kwargs):
        super(CacheMix, self).__init__(*args, **kwargs)

    @decorate_me
    def list(self, *args, **kwargs):
        classname = self.__class__.__name__
        print('list() method of {} object called'.format(classname))

@example_class_decorator
class ListView(CacheMix):
    key = 'test_key'

listview = ListView()
listview.list()

输出：

key_func called with params:
  updated_at: 'test_key'
list() method of ListView object called

Answer 2

我刚刚发现，如果您这样编写装饰器函数：

def decorator(the_func):
    @wraps(the_func)
    def wrapper(*args, **kwargs):
        the_func(*args, **kwargs)
    return wrapper

并修饰任何以self为参数的方法，self将出现在args中。因此，您可以执行以下操作：

from functools import wraps
class myClass:

    def __init__(self):
        self.myValue = "Hello"

    def decorator(the_func):     
        @wraps(the_func)
        def wrapper(*args, **kwargs):
            print(args[0].myValue)
            the_func(*args, **kwargs)
        return wrapper    

    @decorator
    def myFunction(self):
        print("World")

像平常一样打电话

foo = myClass()
foo.myFunction()

您应该获得

Hello
World

如何将自己传递给装饰者？

2 个答案: