Python Django信号：如何初始化绑定？

我正在导入应用程序的

\uuuu init\uuuu.py

文件中的信号文件。当我运行manage.py runserver时，发现

\uuuu init\uuuu.py

被导入了两次，我的回调运行了两次

但是，当我运行manage.py shell时，

\uuuuu init\uuuuu.py

按预期工作（一次）

（Django 1.8.3）

app/\uuuuu init\uuuuuuuuuuuuupy

：

print 'Hello!'
import signals

print 'Hello!'
import app.signals

app/models.py：

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

app/signals.py：

print '------> app/signals.py'

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask)
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

示例输出：

$ ./manage.py runserver
Hello!
------> app/signals.py
Hello!
------> app/signals.py
Performing system checks...

System check identified no issues (0 silenced).
August 12, 2015 - 19:50:38
Django version 1.8.3, using settings 'signal_loca.settings'
Starting development server at http://127.0.0.1:8000/
Quit the server with CONTROL-C.

Angel Velasquez提供的解决方案：

app/\uuuuu init\uuuuuuuuuuuuupy

：

print 'Hello!'
import signals

print 'Hello!'
import app.signals

---

在

AppConfig.ready

方法中导入

signals.py

，以防止回调被注册两次。以下是推荐的方法：

my_awesome_project/
|-- app/
|   |-- __init__.py   <-- set default_app_config here
|   |-- apps.py       <-- define your AppConfig subclass here
|   |-- models.py
|   |-- signals.py    <-- define your signal handling functions here
|-- other_app/
|-- my_awesome_project/
    |-- __init__.py
    |-- settings.py
    |-- wsgi.py

app/apps.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

apps/models.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

app/signals.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

编辑：

---

将

dispatch\u uid

添加到

receiver

decorator。

在

AppConfig.ready

方法中导入您的

signals.py

，以防止回调被注册两次。以下是推荐的方法：

my_awesome_project/
|-- app/
|   |-- __init__.py   <-- set default_app_config here
|   |-- apps.py       <-- define your AppConfig subclass here
|   |-- models.py
|   |-- signals.py    <-- define your signal handling functions here
|-- other_app/
|-- my_awesome_project/
    |-- __init__.py
    |-- settings.py
    |-- wsgi.py

app/apps.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

apps/models.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

app/signals.py

default_app_config = 'app.apps.MyAppConfig'

from django.apps import AppConfig

class MyAppConfig(AppConfig):
    name="app"

    def ready(self):
        from . import signals

from django.db import models

class RemoteTask(models.Model):
    title = models.CharField(max_length=50)

class RemoteTaskStatus(models.Model):
    remote_task = models.IntegerField()
    status = models.IntegerField()

from django.db.models.signals import post_save
from django.dispatch import receiver
from app.models import RemoteTask, RemoteTaskStatus

@receiver(post_save, sender=RemoteTask, dispatch_uid="status_new_signal_on_RemoteTask_post_save")
def status_new(sender, **kwargs):
    print '---------> Running callback'
    if kwargs.get('created', 'False') is True:
        instance = kwargs['instance']
        RemoteTaskStatus.objects.create(
            remote_task=instance,
            status=0
        )

编辑：

---

添加了

dispatch\u uid

到

receiver

decorator。

Hmm我听说你在运行测试？有没有可能是py.test？尝试删除.pyc文件或

\uuuu pycache\uuuu

文件夹。

嗯，我听说你在运行测试？有没有可能是py.test？请尝试删除.pyc文件或

\uuuuu pycache\uuuuuu

文件夹。

您可能对“防止重复信号”感兴趣：感谢的可能重复，但这些链接对解决方案没有帮助如果我尝试您的解决方案，则会出现以下错误：“django.core.exceptions.AppRegistryNotReady:应用程序尚未加载。”您是如何解决的？您可能对“防止重复信号”感兴趣：感谢的可能重复，但这些链接对解决方案没有帮助如果我尝试您的解决方案，则会出现以下错误：“django.core.exceptions.AppRegistryNotReady:应用程序尚未加载。”您是如何解决的？谢谢@chucksmash，但我发现了一些问题。@nachopro您在signals.py中的print语句不在if语句的范围内，它检查是否正在创建或更新

实例。如果在
if
块中移动
print'----------->Running callback'
你还会看到它两次吗？是的，Chucksash，当我运行测试时（失败是因为创建了一个RemoteTask，添加了两个RemoteTaskStatus），我得到了两个打印“Running callback”@nachopro在这种情况下，您可以将
dispatch\u uid
参数传递给
接收器
装饰器：（为什么是Django，为什么？！谢谢@chucksmash，但我对此有些问题。@nachopro signals.py中的print语句不在if语句的范围内，它检查
实例是否正在创建或更新。如果在
if
块中移动
print'----------->运行callback'
，您还会看到它两次吗？是的chucksmash，当我运行测试时（失败是因为创建了一个RemoteTask，添加了两个RemoteTaskStatus），我得到了两张打印“Running callback”@nachopro，在这种情况下，您可以将
dispatch\u uid
参数传递给
接收器
装饰器：（为什么是Django，为什么？！尝试：在
\uuuu init\uuuuuuuuuuuuupy中导入app.signals
尝试：在
中导入app.signals