34

我需要使用模型进行保存，但我需要在保存之前断开一些信号接收器。

我是说，

我有一个模型：

class MyModel(models.Model):
    ...

def pre_save_model(sender, instance, **kwargs):
    ...

pre_save.connect(pre_save_model, sender=MyModel)

在代码的另一个地方我需要类似的东西：

a = MyModel()
...
disconnect_signals_for_model(a)
a.save()
...
reconnect_signals_for_model(a)

因为我需要在这种情况下保存模型而不执行函数 pre_save_model。

4

5 回答 5

37

对于干净且可重用的解决方案，您可以使用上下文管理器：

class temp_disconnect_signal():
    """ Temporarily disconnect a model from a signal """
    def __init__(self, signal, receiver, sender, dispatch_uid=None):
        self.signal = signal
        self.receiver = receiver
        self.sender = sender
        self.dispatch_uid = dispatch_uid

    def __enter__(self):
        self.signal.disconnect(
            receiver=self.receiver,
            sender=self.sender,
            dispatch_uid=self.dispatch_uid,
            weak=False
        )

    def __exit__(self, type, value, traceback):
        self.signal.connect(
            receiver=self.receiver,
            sender=self.sender,
            dispatch_uid=self.dispatch_uid,
            weak=False
        )

现在，您可以执行以下操作：

from django.db.models import signals

from your_app.signals import some_receiver_func
from your_app.models import SomeModel

...
kwargs = {
    'signal': signals.post_save,
    'receiver': some_receiver_func,
    'sender': SomeModel, 
    'dispatch_uid': "optional_uid"
}
with temp_disconnect_signal(**kwargs):
    SomeModel.objects.create(
        name='Woohoo',
        slug='look_mom_no_signals',
    )

注意：如果您的信号处理程序使用 a dispatch_uid，则必须使用dispatch_uidarg。

于 2014-10-07T21:09:18.860 回答
27

您可以像Haystack在 RealTimeSearchIndex 中那样连接和断开信号，这似乎更标准：

from django.db.models import signals
signals.pre_save.disconnect(pre_save_model, sender=MyModel)
a.save()
signals.pre_save.connect(pre_save_model, sender=MyModel)
于 2012-06-28T09:23:02.383 回答
10

我没有测试以下代码，但它应该可以工作：

from django.db.models.signals import pre_save


def save_without_the_signals(instance, *args, **kwargs):
    receivers = pre_save.receivers
    pre_save.receivers = []
    new_instance = instance.save(*args, **kwargs)
    pre_save.receivers = receivers
    return new_instance

它将使来自所有发送者的信号静音，但不仅仅是instance.__class__.

此版本仅禁用给定模型的信号：

from django.db.models.signals import pre_save
from django.dispatch.dispatcher import _make_id


def save_without_the_signals(instance, *args, **kwargs):
    receivers = []
    sender_id = _make_id(instance.__class__)
    for index in xrange(len(self.receivers)):
        if pre_save.receivers[index][0][1] == sender_id:
            receivers.append(pre_save.receivers.pop(index))
    new_instance = instance.save(*args, **kwargs)
    pre_save.receivers.extend(receivers)
    return new_instance
于 2010-02-05T19:29:06.833 回答
8

如果您只想断开并重新连接一个自定义信号，您可以使用以下代码：

def disconnect_signal(signal, receiver, sender):
    disconnect = getattr(signal, 'disconnect')
    disconnect(receiver, sender)

def reconnect_signal(signal, receiver, sender):
    connect = getattr(signal, 'connect')
    connect(receiver, sender=sender)

通过这种方式，您可以这样做：

disconnect_signal(pre_save, pre_save_model, MyModel)
a.save()
reconnect_signal(pre_save, pre_save_model, MyModel)
于 2010-02-08T15:24:34.400 回答
0

我需要防止在单元测试期间触发某些信号，所以我根据 qris 的响应制作了一个装饰器：

from django.db.models import signals

def prevent_signal(signal_name, signal_fn, sender):
    def wrap(fn):
        def wrapped_fn(*args, **kwargs):
            signal = getattr(signals, signal_name)
            signal.disconnect(signal_fn, sender)
            fn(*args, **kwargs)
            signal.connect(signal_fn, sender)
        return wrapped_fn
    return wrap

使用它很简单：

@prevent_signal('post_save', my_signal, SenderClass)
def test_something_without_signal(self):
    # the signal will not fire inside this test
于 2015-11-12T02:33:55.050 回答