Question

我正在尝试使用Django的独立脚本，继承自https://github.com/syntarsus/minimal-django

当我离开调用开发服务器到命令行调用时，事情进展顺利：

python minimal.py runserver

服务器很高兴启动并运行：

Performing system checks...

System check identified no issues (0 silenced).
September 24, 2017 - 00:58:36
Django version 1.11.5, using settings None
Starting development server at http://127.0.0.1:8000/
Quit the server with CTRL-BREAK.

我想修改脚本以在Django抽象中降低级别并使用以下命令调用服务器：

from django.core import management    
management.call_command('runserver')

这会引发异常

AppRegistryNotReady: Apps aren't loaded yet.

正如在其他SO问题中所建议的那样，我尝试用以下方法覆盖例外：

import django
django.setup()

......但显然它不再起作用，所以我迷路了。

Answer 1

这对我有用，只需执行python minimal.py即可调用它，它将运行本地服务器：

<强> minimal.py

import sys

import django
from django.conf import settings
from django.conf.urls import url
from django.core.management import call_command
from django.http import HttpResponse

settings.configure(
    DEBUG=True,
    SECRET_KEY='A-random-secret-key!',
    ROOT_URLCONF=sys.modules[__name__],
)
django.setup()


def index(request):
    return HttpResponse('<h1>A minimal Django response!</h1>')

urlpatterns = [
    url(r'^$', index),
]

if __name__ == '__main__':
    call_command('runserver')

试图在独立的Django中覆盖AppRegistryNotReady

1 个答案: