python django將模型拆分為單獨的文件

Question

我在學習python （3.5.2）和django （1.10.3）時絆腳石。 我有一個項目，那個項目有我的第一個應用程序。 這是會員專區。 我預計該應用程序將具有許多模型，並且我喜歡將模型拆分為自己的文件的想法。 我在這里看到了對此的解釋： https : //code.djangoproject.com/wiki/CookBookSplitModelsToFiles

    -- MyApplication
       |-- __init__.py
       |-- settings.py
       |-- urls.py
       |-- wsgi.py
    -- Members
       |-- __init__.py
       |-- admin.py
       |-- apps.py
       |-- migrations
       |   |-- __init__.py
       |-- models
       |   |-- __init__.py
       |   |-- address.py
       |-- tests.py
       |-- urls.py
       |-- views.py
    -- manage.py

Members/models/__init__.py看起來像

    from address import Address

Members/models/address.py看起來像

    from django.db import models

    class Address(models.Model):

        Member_ID = models.ForeignKey(
            settings.AUTH_USER_MODEL,
            on_delete=models.CASCADE
        )
        Order = models.PositiveIntegerField()

        ...

MyApplication/Settings.py具有

    INSTALLED_APPS = [
        'Members.apps.MembersConfig',
        'django.contrib.admin',
        'django.contrib.auth',
        'django.contrib.contenttypes',
        'django.contrib.sessions',
        'django.contrib.messages',
        'django.contrib.staticfiles',
        ...

當我嘗試嘗試設置數據庫遷移時

$ python manage.py makemigrations Members
Traceback (most recent call last):
  File "manage.py", line 22, in <module>
    execute_from_command_line(sys.argv)
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/site-packages/django/core/management/__init__.py", line 367, in execute_from_command_line
    utility.execute()
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/site-packages/django/core/management/__init__.py", line 341, in execute
    django.setup()
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/site-packages/django/__init__.py", line 27, in setup
    apps.populate(settings.INSTALLED_APPS)
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/site-packages/django/apps/registry.py", line 108, in populate
    app_config.import_models(all_models)
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/site-packages/django/apps/config.py", line 199, in import_models
    self.models_module = import_module(models_module_name)
  File "/home/myUser/.pyenv/versions/3.5.2/lib/python3.5/importlib/__init__.py", line 126, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
  File "<frozen importlib._bootstrap>", line 986, in _gcd_import
  File "<frozen importlib._bootstrap>", line 969, in _find_and_load
  File "<frozen importlib._bootstrap>", line 958, in _find_and_load_unlocked
  File "<frozen importlib._bootstrap>", line 673, in _load_unlocked
  File "<frozen importlib._bootstrap_external>", line 665, in exec_module
  File "<frozen importlib._bootstrap>", line 222, in _call_with_frames_removed
  File "/home/myUser/dev.aikidocia.org/MyApplication/Members/models/__init__.py", line 1, in <module>
    from address import Address
ImportError: No module named 'address'

我想念什么？

Answer 1

如@Evert建議，使用適當的相對導入。

from .address import Address