Django in single file with model and admin

mini_django.py

"""
Django in single file with model and admin. Based on:-

http://fahhem.com/blog/2011/10/django-models-without-apps-or-everything-django-truly-in-a-single-file/
"""
import sys
from os import path as osp

def rel_path(*p): return osp.normpath(osp.join(rel_path.path, *p))
rel_path.path = osp.abspath(osp.dirname(__file__))
this = osp.splitext(osp.basename(__file__))[0]
 
from django.conf import settings
SETTINGS = dict(
    SITE_ID=1,
    DATABASES = {},
    DEBUG=True,
    TEMPLATE_DEBUG=True,
    ROOT_URLCONF = this
)
SETTINGS['TEMPLATE_DIRS'] = (rel_path(),),
SETTINGS['DATABASES']={
    'default':{
        'ENGINE':'django.db.backends.sqlite3',
        'NAME':rel_path('db')
    }
}

SETTINGS['INSTALLED_APPS'] = (
    'django.contrib.auth',
    'django.contrib.contenttypes',
    'django.contrib.sessions',
    'django.contrib.sites',
    'django.contrib.messages',
    #'django.contrib.staticfiles',
    'django.contrib.admin', 
)

if not settings.configured:
    settings.configure(**SETTINGS)

from django.db import models
from django.contrib import admin
 
class SomeModel(models.Model):
    field_name = models.CharField(max_length=10)

    class Meta:
        app_label = this
    __module__ = this

try:
    admin.site.register(SomeModel)
except admin.sites.AlreadyRegistered:
    pass

admin.autodiscover()

from django.conf.urls.defaults import patterns, url, include
from django.http import HttpResponse
 
def view_name(request):
    out = []
    for obj in SomeModel.objects.all():
        out.append(obj.field_name)

    return HttpResponse(','.join(out))
 
urlpatterns = patterns('',
    (r'^$', view_name),
    url(r'^admin/', include(admin.site.urls)),
)

if __name__=='__main__':
    # override get_app to work with us
    get_app_orig = models.get_app
    def get_app(app_label,*a, **kw):
        if app_label==this:
            return sys.modules[__name__]
        return get_app_orig(app_label, *a, **kw)
    models.get_app = get_app
 
    models.loading.cache.app_store[type(this+'.models',(),{'__file__':__file__})] = this

    from django.core import management
    management.execute_from_command_line()