Django:如何告诉Django它应该在哪里寻找应用程序?

时间:2018-05-16 07:27:44

标签: django python-module django-settings pythonpath django-apps

在阅读了Django 1.11的两个Scoops之后,我改变了我的项目结构:

myname_project
  ├── config/
  │   ├── settings/
  │   │   ├── base.py
  │   │   ├── local.py
  │   │   ├── staging.py
  │   │   ├── prod.py
  │   ├── __init__.py
  │   ├── urls.py
  │   └── wsgi.py
  ├── docs/
  ├── myname/
  │   ├── accounts/  # App
  │   ├── blog/ # App
  │   ├── core/ # App
  │   ├── media/ # Development only!
  │   ├── static/
  │   └── templates/
  ├── .gitignore
  ├── Makefile
  ├── README.md
  ├── manage.py
  └── requirements/

我现在的问题是,Django再也找不到该应用了。以下是我的设置文件外观的要点:

import os
from pathlib import Path

from django.core.exceptions import ImproperlyConfigured


def get_env_variable(var_name):
    """Get the environment variable or return exception."""
    try:
        return os.environ[var_name]
    except KeyError:
        error_msg = 'Set the {} environment variable'.format(var_name)
        raise ImproperlyConfigured(error_msg)

# Build paths inside the project like this: BASE_DIR / 'media'
BASE_DIR = Path(__file__).resolve().parent.parent

PROJECT_ROOT = BASE_DIR.parent / 'myname'

INSTALLED_APPS = [
    'django.contrib.admin',
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.messages',
    'django.contrib.staticfiles',
    ...
    'accounts',
    'blog',
    'core',
]

...

ROOT_URLCONF = 'config.urls'

...

现在问题是,当我运行python manage.py makemigrations --settings=config.settings.local时,我收到错误:ModuleNotFoundError: No module named 'accounts'

所以Django似乎无法再找到我的应用了。我如何告诉Django根目录应该在哪里查找应用程序?

我尝试使用PROJECT_ROOT = BASE_DIR.parent / 'myname',但这没有帮助。而且我找不到如何在Django文档中设置(伪代码:) APP_DIR

1 个答案:

答案 0 :(得分:1)

sys.path.append(os.path.join(BASE_DIR, "myname"))行添加到settings.py

BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__))) 
sys.path.append(os.path.join(BASE_DIR, "myname"))

或使用pathlib:

BASE_DIR = Path(__file__).resolve().parent.parent
sys.path.append(str(BASE_DIR / 'myname'))