Question

我有用户模型，用户可以是1或2。根据要创建的用户类型，我想将配置文件与模型相关联。如果输入1，则为Person，输入2为公司。

我尝试过在models.py中编写代码，并遵循以下教程：https://simpleisbetterthancomplex.com/tutorial/2016/07/28/how-to-create-django-signals.html

signals / apps / entitiesmodels.py

class CompanyModel(AuditedModel):
    name = models.CharField(max_length=64, db_index=True, verbose_name='Name', null=True, blank=True)

class PersonModel(AuditedModel):   
    name = models.CharField(max_length=64, db_index=True, verbose_name='Name', null=True, blank=True)

class Tester(PersonModel,PersistentModel):
    # Link with user
    user = models.OneToOneField(settings.AUTH_USER_MODEL, on_delete=models.PROTECT, blank=True, related_name='%(class)s_user')

class Company(CompanyModel,PersistentModel):
    user = models.OneToOneField(settings.AUTH_USER_MODEL, on_delete=models.PROTECT, blank=True, related_name='%(class)s_user')

signals / apps / entities / signals.py

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

from . import models as entities_models       

@receiver(post_save, sender=settings.AUTH_USER_MODEL)
def user_create_profile(sender, instance, created, **kwargs):
    if created:
        if instance.user_type == 1:
            entities_models.Tester.objects.create(user=instance)
        elif instance.user_type == 2:
            entities_models.Company.objects.create(user=instance)
        else:
            pass

signals / apps / entities / app.py

from django.apps import AppConfig
class EntitiesConfig(AppConfig):
    name ='entities'

    def ready(self):
        import entities.signals

signals / apps / entities / api_v1 / views.py

from signals.apps.entities import models           
from . import serializers                          
from signals.libs.views import APIViewSet          

class PersonViewSet(APIViewSet):                                                             
    queryset = models.Person.objects.all()         
    serializer_class = serializers.PersonSerializer

signals / apps / entities / api_v1 / urls.py

from rest_framework.routers import DefaultRouter
from signals.apps.entities.api_v1 import views

# Create a router and register our viewsets with it.
app_name='entities'
router = DefaultRouter()
router.register(r'persons', views.PersonViewSet, base_name="entities-persons")
urlpatterns = router.urls

settings.py

LOCAL_APPS = (                                   
    'signals.apps.authentication',               
    'signals.apps.entities.apps.EntitiesConfig', 
)

运行服务器时，错误是：

 File "/home/gonzalo/Playground/signals3/signals/signals/apps/entities/api_v1/urls.py", line 2, in <module>
   from signals.apps.entities.api_v1 import views
 File "/home/gonzalo/Playground/signals3/signals/signals/apps/entities/api_v1/views.py", line 1, in <module>
   from signals.apps.entities import models
 File "/home/gonzalo/Playground/signals3/signals/signals/apps/entities/models.py", line 47, in <module>
   class Person(PersonModel):
 File "/home/gonzalo/.virtualenvs/signals-test/lib/python3.6/site-packages/django/db/models/base.py", line 108, in __new__
   "INSTALLED_APPS." % (module, name)
untimeError: Model class signals.apps.entities.models.Person doesn't declare an explicit app_label and isn't in an application in INSTALLED_APPS

如果有人想检查一下，我在github上有示例代码：https://github.com/gonzaloamadio/django-signals3

Answer 1

感谢您的回答，当使用这种在设置上引用应用程序的方式时，则应使用如下所示的导入方式：

signals / apps / entities / api_v1 / views.py

from signals.apps.entities import models

和urls.py

from entities.api_v1 import views

django在继承的配置文件上发出信号

1 个答案: