如何使用带有信号的 Django 模型继承？

Question

我在 Django 中有几个模型继承级别：

class WorkAttachment(models.Model):
    """ Abstract class that holds all fields that are required in each attachment """
    work            = models.ForeignKey(Work)
    added           = models.DateTimeField(default=datetime.datetime.now)
    views           = models.IntegerField(default=0)

    class Meta:
        abstract = True


class WorkAttachmentFileBased(WorkAttachment):
    """ Another base class, but for file based attachments """
    description     = models.CharField(max_length=500, blank=True)
    size            = models.IntegerField(verbose_name=_('size in bytes'))

    class Meta:
        abstract = True


class WorkAttachmentPicture(WorkAttachmentFileBased):
    """ Picture attached to work """
    image           = models.ImageField(upload_to='works/images', width_field='width', height_field='height')
    width           = models.IntegerField()
    height          = models.IntegerField()

从WorkAttachmentFileBased 和WorkAttachment 继承了许多不同的模型。我想创建一个信号，它会在创建附件时更新父工作的attachment_count 字段。认为为父发送者 (WorkAttachment) 发出的信号也适用于所有继承的模型是合乎逻辑的，但事实并非如此。这是我的代码：

@receiver(post_save, sender=WorkAttachment, dispatch_uid="att_post_save")
def update_attachment_count_on_save(sender, instance, **kwargs):
    """ Update file count for work when attachment was saved."""
    instance.work.attachment_count += 1
    instance.work.save()

有没有办法让这个信号适用于从WorkAttachment继承的所有模型？

Python 2.7、Django 1.4 pre-alpha

附：我试过one of the solutions I found on the net，但它对我不起作用。

Answer 1

除了@clwainwright 答案之外，我还配置了他的答案，改为适用于 m2m_changed 信号。我必须将其发布为代码格式的答案才有意义：

@classmethod
    def __init_subclass__(cls, **kwargs):
        super().__init_subclass__(**kwargs)
        for m2m_field in cls._meta.many_to_many:
            if hasattr(cls, m2m_field.attname) and hasattr(getattr(cls, m2m_field.attname), 'through'):
                models.signals.m2m_changed.connect(m2m_changed_receiver, weak=False, sender=getattr(cls, m2m_field.attname).through)

它会进行一些检查以确保在未来的 Django 版本中发生任何变化时它不会中断。

Answer 2

我只是使用 python 的（相对）新的__init_subclass__ method：

from django.db import models

def perform_on_save(*args, **kw):
    print("Doing something important after saving.")

class ParentClass(models.Model):
    class Meta:
        abstract = True

    @classmethod
    def __init_subclass__(cls, **kwargs):
        super().__init_subclass__(**kwargs)
        models.signals.post_save.connect(perform_on_save, sender=cls)

class MySubclass(ParentClass):
    pass  # signal automatically gets connected.

这需要 django 2.1 和 python 3.6 或更高版本。请注意，使用 django 模型和相关元类时似乎需要 @classmethod 行，即使根据官方 python 文档不需要它。

Answer 3

您可以在不指定sender 的情况下注册连接处理程序。并过滤其中需要的模型。

from django.db.models.signals import post_save
from django.dispatch import receiver


@receiver(post_save)
def my_handler(sender, **kwargs):
    # Returns false if 'sender' is NOT a subclass of AbstractModel
    if not issubclass(sender, AbstractModel):
       return
    ...

参考：https://groups.google.com/d/msg/django-users/E_u9pHIkiI0/YgzA1p8XaSMJ

Answer 4

Michael Herrmann 的解决方案无疑是最符合 Django 的方法。 是的，它适用于所有子类，因为它们是在 ready() 调用中加载的。

我想贡献文档参考：

实际上，信号处理程序通常定义在与其相关的应用程序的信号子模块中。信号接收器连接在应用程序配置类的 ready() 方法中。如果您使用的是 receiver() 装饰器，只需在 ready() 中导入信号子模块。

https://docs.djangoproject.com/en/dev/topics/signals/#connecting-receiver-functions

并添加警告：

在测试期间可能会多次执行 ready() 方法，因此您可能希望防止信号重复，尤其是当您计划在测试中发送它们时。

https://docs.djangoproject.com/en/dev/topics/signals/#connecting-receiver-functions

因此，您可能希望在连接函数上使用 dispatch_uid 参数来防止重复信号。

post_save.connect(my_callback, dispatch_uid="my_unique_identifier")

在这种情况下，我会这样做：

for subclass in get_subclasses(WorkAttachment):
    post_save.connect(update_attachment_count_on_save, subclass, dispatch_uid=subclass.__name__)

https://docs.djangoproject.com/en/dev/topics/signals/#preventing-duplicate-signals

Answer 5

最简单的解决方案是不限制sender，而是在信号处理程序中检查各个实例是否是子类：

@receiver(post_save)
def update_attachment_count_on_save(sender, instance, **kwargs):
    if isinstance(instance, WorkAttachment):
        ...

但是，这可能会导致显着的性能开销，因为每次保存任何模型时，都会调用上述函数。

我想我找到了最符合 Django 的方法：最新版本的 Django 建议在名为 signals.py 的文件中连接信号处理程序。这是必要的接线代码：

your_app/__init__.py：

default_app_config = 'your_app.apps.YourAppConfig'

your_app/apps.py：

import django.apps

class YourAppConfig(django.apps.AppConfig):
    name = 'your_app'
    def ready(self):
        import your_app.signals

your_app/signals.py：

def get_subclasses(cls):
    result = [cls]
    classes_to_inspect = [cls]
    while classes_to_inspect:
        class_to_inspect = classes_to_inspect.pop()
        for subclass in class_to_inspect.__subclasses__():
            if subclass not in result:
                result.append(subclass)
                classes_to_inspect.append(subclass)
    return result

def update_attachment_count_on_save(sender, instance, **kwargs):
    instance.work.attachment_count += 1
    instance.work.save()

for subclass in get_subclasses(WorkAttachment):
    post_save.connect(update_attachment_count_on_save, subclass)

我认为这适用于所有子类，因为它们都将在调用 YourAppConfig.ready 时加载（因此导入 signals）。

Answer 6

此解决方案解决了并非所有模块都导入内存时的问题。

def inherited_receiver(signal, sender, **kwargs):
    """
    Decorator connect receivers and all receiver's subclasses to signals.

        @inherited_receiver(post_save, sender=MyModel)
        def signal_receiver(sender, **kwargs):
            ...

    """
    parent_cls = sender

    def wrapper(func):
        def childs_receiver(sender, **kw):
            """
            the receiver detect that func will execute for child 
            (and same parent) classes only.
            """
            child_cls = sender
            if issubclass(child_cls, parent_cls):
                func(sender=child_cls, **kw)

        signal.connect(childs_receiver, **kwargs)
        return childs_receiver
    return wrapper

Answer 7

post_save.connect(my_handler, ParentClass)
# connect all subclasses of base content item too
for subclass in ParentClass.__subclasses__():
    post_save.connect(my_handler, subclass)

祝你有美好的一天！

Answer 8

还可以使用内容类型来发现子类 - 假设您将基类和子类打包在同一个应用程序中。这样的事情会起作用：

from django.contrib.contenttypes.models import ContentType
content_types = ContentType.objects.filter(app_label="your_app")
for content_type in content_types:
    model = content_type.model_class()
    post_save.connect(update_attachment_count_on_save, sender=model)

Answer 9

你可以试试这样的：

model_classes = [WorkAttachment, WorkAttachmentFileBased, WorkAttachmentPicture, ...]

def update_attachment_count_on_save(sender, instance, **kwargs):
    instance.work.attachment_count += 1
    instance.work.save()

for model_class in model_classes:
    post_save.connect(update_attachment_count_on_save, 
                      sender=model_class, 
                      dispatch_uid="att_post_save_"+model_class.__name__)

（免责声明：我没有测试过以上）