我正在应用程序的 __init__.py
文件中导入信号文件。问题发生在我运行 manage.py runserver 时,我发现 __init__.py
被导入了两次,并且我的回调运行了两次。
但是,当我运行 manage.py shell 时,__init__.py
按预期工作(一次)
(Django 1.8.3)
app/__init__.py
:
print 'Hello!'
import signals
应用程序/模型.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()
应用程序/信号.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 的解决方案:
app/__init__.py
:
print 'Hello!'
import app.signals
最佳答案
在 AppConfig.ready
方法中导入您的 signals.py
以防止回调被注册两次。以下是执行此操作的推荐方法:
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/__init__.py
default_app_config = 'app.apps.MyAppConfig'
app/apps.py
from django.apps import AppConfig
class MyAppConfig(AppConfig):
name="app"
def ready(self):
from . import signals
apps/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
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 dispatch_uid
到 receiver
装饰器。
关于python - Django 信号 : How to initialize bindings?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31973707/