模型方法通过装饰器的Django信号？

I'm trying to do something like these proposed signal decorators. In addition to having a decorator that connects the decorated method to a signal (with the signal's sender as an argument to the decorator), I would like to use the decorator on class methods.

我正在尝试做这些提议的信号装饰器。除了有一个装饰器将装饰方法连接到一个信号（信号的发送者作为装饰器的参数），我想在类方法上使用装饰器。

I'd like to use the decorator like so:

我想像这样使用装饰器：

class ModelA(Model):

    @connect.post_save(ModelB)
    @classmethod
    def observe_model_b_saved(cls, sender, instance, created, **kwargs):
        # do some stuff
        pass

The decorator is:

装饰者是：

from django.db.models import signals
def post_save(sender):
    def decorator(view):
        signals.post_save.connect(sender=sender, receiver=view)
        return view
    return decorator

The error I get when I do this is:

我这样做的错误是：

File "/Library/Python/2.6/site-packages//lib/python2.6/site-packages/django/dispatch/dispatcher.py", line 78, in connect
AssertionError: Signal receivers must be callable.

I guess the problem is that @classmethod returns a class method object which is not callable. I don't really understand how classmethod works under the hood, but I surmise from this reference page that the class method object is not translated into a callable until it is accessed from the class, e.g., ModelA.observe_model_b_saved. Is there any way that I can both (1) define my method as a class or instance method on a model, and (2) connect it to a signal using a decorator directly on the method definition? Thanks!

我想问题是@classmethod返回一个不可调用的类方法对象。我真的不明白classmethod是如何工作的，但我从这个参考页面推测，在从类中访问类方法对象之前，类方法对象不会被转换为可调用方法，例如ModelA.observe_model_b_saved。有什么方法可以兼顾（1）将我的方法定义为模型上的类或实例方法，以及（2）直接在方法定义上使用装饰器将其连接到信号？谢谢！

3 个解决方案

#1

It's not clear from your example code, so I'd be asking if the signal listener actually has to be a @classmethod? I.e. Will a regular method do (and then use self.__class__ if you still need to access the class itself)? Does it need to be a method at all (can you just use a function)?

从你的示例代码中不清楚，所以我会问信号监听器是否真的必须是@classmethod？即一个常规方法会做（然后使用self .__ class__，如果你仍然需要访问类本身）？是否需要一个方法（你能使用一个函数）吗？

Another option might be to use a second method to listen to the signal and delegate the call to the @classmethod:

另一种选择可能是使用第二种方法来监听信号并将调用委托给@classmethod：

class ModelA(Model): 

    @classmethod 
    def do_observe_model_b_saved(cls, sender, instance, created, **kwargs): 
        # do some stuff 
        pass 

    @connect.post_save(ModelB) 
    def observe_model_b_saved(self, sender, instance, created, **kwargs): 
        self.do_observe_model_b_saved(sender, instance, created, **kwargs)

#2

Could you make it a @staticmethod instead? That way, you can just swap the order of the decorators.

你能把它改成@static方法吗？这样，您就可以交换装饰器的顺序。

class ModelA(Model):

    @staticmethod
    @connect.post_save(ModelB)
    def observe_model_b_saved(sender, instance, created, **kwargs):
        # do some stuff
        pass

You'd have to refer to the class by full name instead of getting passed the cls argument, but this would allow you to keep a similar code organization.

您必须通过全名引用该类，而不是通过cls参数，但这将允许您保留类似的代码组织。

#3

Based off of Matt's answer, the @staticmethod trick worked for me. You can use a string to reference a model non-concretely.

基于Matt的回答，@ staticmethod技巧对我有用。您可以使用字符串非具体地引用模型。

class Foo(Model):

    @staticmethod
    @receiver(models.signals.post_save, sender='someappname.Foo')
    def post_save(sender, instance, created, **kwargs):
            print 'IN POST SAVE', sender, instance.id, created

#1