Question

我有两个Django应用程序，如下图所示。按原样，Django将创建两个数据库表，＆＃39; appa_modela＆＃39;和＆＃39; appb_modelb＆＃39;。 commA.py创建的ModelB对象将保存到＆＃39; appb_modeb＆＃39;表。但是，我希望appA创建的ModelB对象独立于appB创建的对象。我知道“Meta”＃39;子类，其中＆＃39; app_label＆＃39;和＆＃39; db_table＆＃39;已设定。如何才能确保“app_label＆＃39;被改为appA＆＃39;何时将ModelB导入appA模块？

的appA / models.py

from django.db import models
class ModelA(models.Model):
  list_of_modelB_pk = models.TextField(default='',
       validators=[validate_comma_separated_integer_list])

的appA /管理/命令/ commA.py

from django.core.management.base import BaseCommand
import json
from appA.models import ModelA
from appB.models import ModelB
class Command(BaseCommand):
  def handle(self, *args, **options):
    mb1 = ModelB()
    mb1.save()
    print mb1._meta.app_label # prints 'appB' why not 'appA'?
    mb2 = ModelB()
    mb2.save()
    ma = ModelA()
    ma.list_of_modelB_pk = json.dumps([mb1.pk,mb2.pk])
    ma.save()
    print ma._meta.app_label # prints 'appA'

程序appB / models.py

from django.db import models
class ModelB(models.Model):
  fieldB1 = models.IntegerField(default=0)

程序appB / manangement /命令/ commB.py

from django.core.management.base import BaseCommand
from appB.models import ModelB
class Command(BaseCommand):
  def handle(self, *args, **options):
    mb1 = ModelB()
    mb1.save()
    mb2 = ModelB()
    mb2.save()

使用另一个应用程序中的Django模型并在独立数据库表中创建条目

0 个答案: