Re: How to define a url that does not belong to any app?

What you can do is populate a path in your DIRS list in TEMPLATES in your settings file to the directory where your template is located.  Then, you can use TemplateView from django.views.generic to pass in a HTML file to be served up by Django for that URL.

Something like this:

urls.py

from django.conf.urls import url
from django.views.generic import TemplateView

urlpatterns = [
    url(r'^', include('grid.urls')),
    url(r'^learn-more$', TemplateView.as_view(template_name='index.html'))
    url(r'^silk/', include('silk.urls', namespace='silk'))
]

url_patterns = [

]

settings.py

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [os.path.join(BASE_DIR, '../templates')],
        'APP_DIRS': True,
        'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug',
                'django.template.context_processors.request',
                'django.contrib.auth.context_processors.auth',
                'django.contrib.messages.context_processors.messages',
            ],
        },
    },
]

What you're doing here is telling django to look in this location for templates.  Then, you're defining a URL that takes in the template name as a parameter to return when the URL his hit.

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