Capture subdomains with wildcard to a view (Django) [duplicate] - python

Tl; dr: Is there a way to override the default behaviour of reverse?
In my django project I have alot of urls such as
url(r'^\w+/company/', include("company.urls", namespace="company")),
Which allows for urls such as
.../companyA/company/
.../companyB/company/
So that I can then use a custom middleware to modify the request to include some specific details based upon what company is using my site
This all works fine except for when django is trying to decipher the full path with reverse and {% url .. %}...
It seems to be returning /x/company/ as a default match for the regex. since the django.utils.regex_helper method next_char has an escape mapping for \w to map to x
The url tag I have been able to override to replace the /x/ with the correct company name and I am wondering if there is a similar thing I can do to override reverse in the same way, or anything else that I can do to resolve this problem?
Previously, I was using
url(r'^(?P<company_name>\w+)/company/', include("company.urls", namespace="company"))
But this meant I had to include a parameter in every view
def view(request, company_name):
...
As well as include it in all my other calls to the view (i.e with the {% url %}) which I am trying to avoid.

For ease of use, Django packages as compiled a page full of every possible existing django package that can accomplish this. However below is my own simple implementation
I modified my nginx proxy config to use the following
server_name ~(?<short_url>\w+)\.domainurl\.com$;
... stuff related to static files here
location / {
proxy_set_header X-CustomUrl $short_url;
.... other proxy settings
}
What this does is create a variable inside a request header that can then be used within Django. This variable I then used within a custom middleware to extend a request with a reference to the model which allows its use anywhere.
class CompanyMiddleware(object):
def process_request(self, request):
if settings.DEBUG:
request.company = CompanyClass.objects.get(id=1)
return None
short_url = request.META.get("HTTP_X_CUSTOMURL")
try:
company = CompanyClass.objects.get(short_url=short_url)
except Model.DoesNotExist:
return HttpResponseBadRequest('Company not found')
request.company = company
return None
Examples:
www.companya.domainurl.com # short_url is companya
test.domainurl.com # short_url is test
To use this within a template, context processors must be added to the settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
"django.contrib.auth.context_processors.auth",
"django.core.context_processors.debug",
"django.core.context_processors.i18n",
"django.core.context_processors.media",
'django.core.context_processors.request' # This one in particular
)

Related

Is there a way to add context to Django reverse function?

Is it possible to pass context through the reverse function in Django in some way like reverse('foo', context={'bar':'baz'})? Or is there a better workaround?
As already described by Sir WillemVanOnsem in the above comment.
You can't provide context in reverse as it only produces a string: a path, eventually it goes to view.
reverse() can only take args and kwargs, see Reversing namespaced URLs for more detail.
Reverse generates an URL. The URL can parse or supply extra context. In urls.py,
path( 'action/<str:context>/', MyView.as_view(), name='foo' )
then
reverse('app:foo', kwargs={'context':'bar:baz+quux:help'} )
will generate the URL ending
.../appname/action/bar:baz+quux:help
and your view will parse context:
context = self.kwargs.get( context, '')
context_dir = {}
for kv in context.split('+'):
keyval = kv.split(':')
context_dir[ keyval[0].strip() ] = keyval[1].strip()
or something like this, ending with context_dir as {'bar':'baz', 'quux':'help'}
Alternatively you can append a querystring to the URL returned by reverse and retrieve that in the view you redirect to via request.GET
url = reverse('foo') + '?bar=baz&quux=help'
redirect, and then in that view request.GET.get('bar') will return "baz" etc.
Finally you can stuff an almost arbitrarily complex context into the user's session (which gets stored either as a cookie in his browser, or an object in your database). This is the most general but also the most complex. See the doc for using Django sessions

Django: Display website name in Template/Emails without using Sites framework

I want to render my website name in django templates. Django's own docs on Sites state:
Use it if your single Django installation powers more than one site
and you need to differentiate between those sites in some way.
My django app doesn't. I know I can still use it, but it seems like an overkill. I just want to pull a variable with my website's name (!= domain) in ANY template. I don't want to pass it in views either because that doesn't seem DRY enough.
Writing a custom processor seemed like a simple-enough option, but for some reason these variables aren't available in the .txt emails django-registration sends (while other variables definitely are, so I guess it's not impossible).
TIA
Edit: was asked to include code that doesn't work:
processors.py:
def get_website_name(request):
website_name = 'SomeWebsite'
return {'mysite_name': website_name}
Included successfully in context_processors in settings.py. It works nicely in "regular" templates, but not in emails.
Here's how I'm sending the emails, inside a change_email_view:
msg_plain = render_to_string('email_change_email.txt', context)
msg_html = render_to_string('email_change_email.html', context)
send_mail(
'Email change request',
msg_plain,
'my#email',
[profile.pending_email],
html_message=msg_html,
)
A further problem is that django-regitration further abstracts some of those views away: so when a user registers, wants to reset a password, etc...I don't even have access to the views.
Based on Django custom context_processors in render_to_string method you should pass the request to render_to_string.
msg_plain = render_to_string('email_change_email.txt', context, request=request)
msg_html = render_to_string('email_change_email.html', context, request=request)

Dynamically check if a slug exists

On a legacy app, I need to check if a URL exists, and if it does not, to redirect it to another location. The problem is that I need to check if that url is present in a set of values, in the urls file and I'm not clear on how best to do that.
For example, both projects and cities are sharing the same url pattern. e.g. /projects/london and /projects/my-project-name.
I want to first check if the slug matches a city, and if it does not to then return the project view (cities cannot match project names).
My urls are currently structured as follows:
url(r'^projects/(?P<project-name>[-\w]+)', get_project, name='project-view'),
url(r'^projects/.*', get_city, name='city-view'),
I know this is very messy, and a bad overall pattern but unfortunately it's not something that can be changed at the moment. So my goal is to figure out if I can first check if the project-name could be a city, and if it is, to redirect onto that view without falling into a redirect loop.
I wondered if I could do something like this:
url(r'^projects/(?P<city>london|paris|new-york)/', get_city, name='city-view'),
where london|paris|new-york are generated dynamically
You can dynamically generate a url with all of the city names, but the url will be cached once django accesses it the first time, so in order to modify the url regex, you'd have to restart the django process. If that's fine for your purposes, you can generate the url like this:
url(r'^projects/(?P<city>{})/$'.format(city_slugs.join('|')),
get_city, name='city-view')
But, it would probably be better to create a view routing method that implements the logic to send requests to their appropriate view:
# urls.py
# ...
url(r'^projects/(?P<slug>[-\w]+)/$',
project_city_router, name='project-city-router'),
# ...
# views.py
def is_a_city(slug):
# If they're in the database, something like:
# return City.objects.filter(slug=slug).exists()
return slug in ['london', 'paris', 'new-york', '...']
def project_city_router(request, slug=None):
if not slug:
# /projects/
return render(request, 'my/template.html', {'foo': 'bar'})
elif is_a_city(slug):
# /projects/<city>/
return get_city(request, city=slug)
else:
# /projects/<project-name/
return get_project(request, project_name=slug)
With this router, if the slug argument is a project or city, it returns the result of the get_project or get_city view itself.
This also allows for your list of cities to be checked dynamically against a database, or file.

How can I redirect in django middleware? global name 'view' is not defined

Django newbie here, need help on basic middleware to redirect to another view if a certain model field is empty.
I am creating a terms of agreement page that users must get redirected to right after they signup to the platform if their filed_terms field on their Profile model is empty.
I am using middleware for this. However I am unable to get this to work. This is my middleware class:
class TermsMiddleware(object):
def process_request(self, request):
if request.user.profile.filled_terms is None:
return redirect(reverse(terms))
This gives me the following error:
global name 'terms' is not defined
I also have the url matcher that works perfectly when I navigate to it manually:
url(r'^terms/', 'my_app.views.terms')
I have a terms.html template and a terms view in my views.py file that is working perfectly in all other respects. I have also added it to the settings middleware requirements to make sure it loads.
Do I have to import something from views or url dispatcher into my middleware file? If so what would that be? I have been at this for a while an cannot find anything helpful.
reverse function takes url name instead on the regex. So you need to add name on your url configuration. Here is the example.
url(r'^terms/', 'my_app.views.terms', name='terms')
Add this in your views.py
from django.core.urlresolvers import reverse
And you need to fix your reverse function into.
return redirect(reverse('terms'))
Python interpret your terms as a variable and you have no variable named terms while you need to put string on reverse.

How to do reverse URL search in Django namespaced reusable application

Consider that I include namespaced reusable application:
urlpatterns = patterns('',
# ella urls
url('^ella/', include('ella.core.urls', namespace="ella")),
)
Now, the Ella applications has urls like that:
urlpatterns = patterns( '',
url( r'^(?P<category>[a-z0-9-/]+)/$', category_detail, name="category_detail" ),
# object detail
url( r'^(?P<category>[a-z0-9-/]+)/(?P<year>\d{4})/(?P<month>\d{1,2})/(?P<day>\d{1,2})/(?P<content_type>[a-z0-9-]+)/(?P<slug>[a-z0-9-]+)/$',
object_detail, name="object_detail" )
)
Now, calling {% url ella:category_detail category="cat" %} works fine. However, when object tries to generate link to it's details, it calls
from django.core.urlresolvers import reverse
url = reverse('object_detail', kwargs={'required' : 'params'})
This is not working, unless rewritten as
from django.core.urlresolvers import reverse
url = reverse('ella:object_detail', kwargs={'required' : 'params'})
So, if I understand it correctly, including reusable application into namespace breaks all inner reverse()s inside given application.
Is it true? What have I missed? Is there any way around?
Since you have name-spaced url configuration, you need to mention namespace:view-name pattern in order to reverse it properly (especially from view).
But, if you want to avoid this, you may also pass namespace/appname as current_app parameter.
There are multiple ways to specify current_app when you are in template. But if you are in view, you need to hard-code as you did, or pass to current_app parameter
url = reverse('object_detail',
kwargs={'foo':'bar'},
current_app=app_name_or_name_space)
refer: http://docs.djangoproject.com/en/dev/topics/http/urls/#reverse
URL Namespaces can be specified in two ways.
Firstly, you can provide the application and instance namespace as arguments to include() when you construct your URL patterns. For example,:
(r'^help/', include('apps.help.urls', namespace='foo', app_name='bar')),
This is right from http://docs.djangoproject.com/en/dev/topics/http/urls/#defining-url-namespaces.
Try changing include('ella.core.urls', namespace="ella") to include('ella.core.urls', namespace="ella", app_name="ella"). I'm not 100% this will work, but its worth a shot.
At least in Django 1.8 you can write something like this:
url = reverse('%s:object_detail' % request.resolver_match.namespace, kwargs={'required' : 'params'})
request.resolver_match.namespace is a string containing the namespace of the currently running view.

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