how to handle two different views witch match for the same url? (Both views need a database call to determine if the called element is available. Changing the url structure is not an option.)
url(r'^', include(wagtail_urls)),
This wagtail url is matching all url's and is raising an 404 if the according page/slug is not in the database.
But, I also have my own view witch behaves similar. How can I tell django to continue with the next url instead of raising a 404?
(I could place my own view before the wagtail view and modify it, but I don't know how to return to the next url?)
This is my solution at the end:
from wagtail.wagtailcore.views import serve
# ... in my view where I normally return a 404 => I set this:
return serve(self.request, self.request.path)
First of all, Sharing same url pattern in diffrent views is bad idea.
The bigger your project, the harder it will be to maintain it.
Nevertheless, there is a way if you want that way.
You can place your own view first in urls.py like your saying,
process some your own logic first and catch 404 exception when there is nothing to show in your view than simply call the "Wagtail" view with request original parameters(page, slug, etc..) with return statement.
Below is example.
This example is based on Django functional based view, but there is the way in class based view something like this.
def your_own_view(request, slugs):
try:
article = get_object_or_404(Article, slugs=slugs)
except Http404:
return some_wagtail_view(request, slugs)
...
return render(request, "article/view.html", context)
So I have this API URL on the back-end and I am wondering how to make a portion of it optional.
url(r'^api/documents/(?P<id>[0-9]+)$', GetDocumentsAPIView.as_view(), name='documents'),
So two things can happen coming from the front-end.
1) When the user logs in, it brings them to the /home which lists all of their projects. Clicking on the project brings them /documents/85 where the number is the number of the project and it lists all the documents for that project. This is sent to the API URL /api/documents/85. This much is working fine.
2) The other option is the user will just click on a Documents link in the navbar, which will just bring them to /documents. This should just go to the API URL /api/documents/ and eventually onto the serializer.py where their most recent project is determined and the documents for that are returned and rendered in the front-end
This I can't get working. Not sure if I can do it with just one url(), or if I need two. Was hoping the one I posted would look at the (?P<id>[0-9]+)$ as optional and if nothing was there would return the same GetDocumentsAPIView, but this is not the case.
EDIT: Including the view I am using for testing at this point. Just print() to see if anything is being routed to the view.
from rest_framework.permissions import IsAuthenticated
from rest_framework.response import Response
from rest_framework.views import APIView
class GetDocumentsAPIView(APIView):
permission_classes = [IsAuthenticated]
def get(self, request, *args, **kwargs):
print(kwargs.get('id'))
return Response('Test')
No you need to add a URL pattern for r'^api/documents/$', which can launch the same view by the way. Only inside the view you'll have to assume id is optional:
if kwargs.get('id'):
# return specific document
else:
# return list of documents
I would rename your first URL pattern to 'view_document' so you can use the name 'documents' on the other, more generic view.
Note: I think your URL structure is wrong. Reading your URL, according to REST, I would expect to see the document with id 85 when fetching documents/85. If you actually are listing the docs of a project, the URL should be projects/85/documents/ or if unspecified projects/documents.
i have a website developed using django 1.6.0 with python 2.7.5 . In my view.py file i have a method defined which i want to be executed only when request for that view is redirected request from some where. I want to restrict user from executing that view by typing the url.
suppose view.py:
def online_test(request):
return buy_test_final(request)
urls.py:
url(r'^test$',online_test),
i need to restrict access of online_test method from url.
For making that view accessible for only redirected requests, you can check if request.META has HTTP_REFERER or not.
def online_test(request):
if 'HTTP_REFERER' not in request.META:
raise Http404
return buy_test_final(request)
Edit
As Andrew Gorcester has pointed out, in a comment below, that HTTP headers can be manipulated manually. Not only that, someone can simply add a link on any of your website's page by using Chrome's Developer Tools. Like this: Test. If he clicks this link, request.META will have HTTP_REFERER, thereby executing that view.
Use the above piece of code carefully, if you must.
I am trying to set my Django app work with multiple domains (while serving slightly different content)
I wrote this middleware:
class MultiSiteMiddleware(object):
def process_request(self, request):
host = request.get_host()
host_part = host.split(':')[0].split('.com')[0].split('.')
host = host_part[len(host_part)-1] + '.com'
site = Site.objects.get(domain=host)
settings.SITE_ID = site.id
settings.CURRENT_HOST = host
Site.objects.clear_cache()
return
In views I use this:
def get_site(request):
current_site = get_current_site(request)
return current_site.name
def view(request, pk):
site = get_site(request)
if site == 'site1':
# serve content1
...
elif site == 'site2'
# serve content2
...
But now there are 404 errors (I sometimes find them in logs, don't see them while browsing my site manually) where they aren't supposed to be, like my site sometimes is serving content for wrong domains, can they happen because of some flaw in the above middleware and view code or I should look somewhere else?
I had a similar requirement and decided not to use the django sites framework. My middleware looks like
class MultiSiteMiddleware(object):
def process_request(self, request):
try:
domain = request.get_host().split(":")[0]
request.site = Site.objects.get(domain=domain)
except Site.DoesNotExist:
return http.HttpResponseNotFound()
then my views have access to request.site
If you're seeing 404's for sites that aren't yours in your logs it would seem like somebody has pointed their domain at your servers IP address, you could use apache/nginx to filter these out before they hit your app, but your middleware should catch them (though possibly by raising an uncaught 500 error instead of a 404)
Serve multiple domain from one website (django1.8 python3+)
The goal is, obviously, to serve multiple domains, from one Django instance. That means, we use the same models, the same database, the same logic, the same views, the same templates, but to serve different things.
Searching the interwebs, I came to the idea of using the sites framework. The sites framework was designed to do exactly that thing. In fact, the sites framework has been used to do exactly that. But I haven't been able to know on which version of Django it was, and actually, I came to the idea the sites framework was just a vestigial module. Basically, it is just a table, with SITE_ID and SITE_URL, that you can easily access with a function. But I've been unable to find how, from that, you can do a multi-domain website.
So my idea has been, how to modify the url resolver. The idea behind all these is easy : www.domain1.com/bar is resolved to /domain1/bar, and www.domain2.foo is resolved to /domain2/foo. This solves the problem because, if you want to serve multiple domains, you just have to serve multiple folders.
In Django, to achieve this, you have to modify two things :
* the way Django routes requests
* the way django write urls
The way Django routes requests
Django routes requests with middlewares. That's it. So we just have to write a middleware that re-route requests.
To make it easier, middlewares can have a process_request method that process requests (WOW), before requests are handled. So let's write a DomainNameMiddleware
#!python
#app/middleware.py
class DomaineNameMiddleware:
"""
change the request path to add the domain_name at the first
"""
def process_request(self, request):
#first, we split the domain name, and take the part before the extension
request_domain = request.META['HTTP_HOST'].split('.')[-2]
request.path_info = "/%s/%s" % (request_domain, request.path.split('/')[1:])
The way Django writes URL
When I'm talking about django writing url, i'm principally thinking of the {% url %} template tag, the get_absolute_url methods, the resolve and resolve_lazy functions, and those basic Django thing. If we rewrite the way Django handle url, we have to tell Django to write url that way.
But basically it's pretty easy, thanks to Django.
You can easily rewrite basic Django function by just rewriting them, typically in init.py files of modules you added as apps. So :
#!python
#anyapp/__init__.py
from django.core import urlresolvers
old_reverse = urlresolvers.reverse
def new_reverse(viewname, urlconf=None, args=None, kwargs=None, current_app=None):
"""
return an url with the first folder as a domain name in .com
"""
TLD = 'com'
old_reverse_url = old_reverse(viewname, urlconf, args, kwargs, current_app)
# admin will add itself everytime you reload an admin page, so we have to delete it
if current_app == 'admin':
return '/%s' % old_reverse_url[len('admin'):].replace('adminadmin', 'admin')
return '//%s.%s/%s' % (app, TLD, path)
How to use it ?
I use it with base urls.py as dispatchedr.
#!python
#urls.py
from django.conf.urls import include, url
from domain1 import urls as domain1_urls
from domain2 import urls as domain2_urls
urlpatterns = [
url(r'^domain1/', include(domain1_urls, namespace='domain1')),
url(r'^domain2/', include(domain2_urls, namespace='domain2)),
]
Django's Sites framework has built-in middleware to accomplish this.
Simply enable the Sites framework and add this to your MIDDLEWARE:
'django.contrib.sites.middleware.CurrentSiteMiddleware'
This automatically passes a request object to Site.objects.get_current() on every request. It solves your problem by giving you access to request.site on every request.
For reference, the code as of 1.11 is:
from django.utils.deprecation import MiddlewareMixin
from .shortcuts import get_current_site
class CurrentSiteMiddleware(MiddlewareMixin):
"""
Middleware that sets `site` attribute to request object.
"""
def process_request(self, request):
request.site = get_current_site(request)
I will suggest you to use django-multisite .It will fulfill your requirement.
Try using the "sites" framework in django to get the domain name. You already know this I guess.
Take a look here: https://docs.djangoproject.com/en/1.7/ref/contrib/sites/#getting-the-current-domain-for-full-urls
See this:
>>> Site.objects.get_current().domain
'example.com'
Without the https://www or http://www.. Probably your domains will end in .org or some country .pe .ru etc not just .com.
There might be a case when people don't point to your domain but to your IP address for some reason, maybe development of testing so you should always raise an exception with Site.DoesNotExist
I am working on a Django 1.4 project and writing one simple application using per-site cache as described here:
https://docs.djangoproject.com/en/dev/topics/cache/#the-per-site-cache
I have correctly setup a local Memcached server and confirmed the pages are being cached.
Then I set CACHE_MIDDLEWARE_ANONYMOUS_ONLY = True because I don't want to cache pages for authenticated users.
I'm testing with a simple view that returns a template with render_to_response and RequestContext to be able to access user information from the template and the caching works well so far, meaning it caches pages just for anonymous users.
And here's my problem. I created another view using a different template that doesn't access user information and noticed that the page was being cached even if the user was authenticated. After testing many things I found that authenticated users were getting a cached page if the template didn't print something from the user context variable. It's very simple to test: print the user on the template and the page won't be cached for an authenticated user, remove the user on the template, refresh the page while authenticated and check the HTTP headers and you will notice you're getting a cached page. You should clear the cache between changes to see the problem more clearly.
I tested a little more and found that I could get rid of the user in the template and print request.user right on the view (which prints to the development server console) and that also fixed the problem of showing a cached page to an authenticated user but that's an ugly hack.
A similar problem was reported here but never got an answer:
https://groups.google.com/d/topic/django-users/FyWmz9csy5g/discussion
I can probably write a conditional decorator to check if user.is_authenticated() and based on that use #never_cache on my view but it seems like that defeats the purpose of using per-site cache, doesn't it?
"""
A decorator to bypass per-site cache if the user is authenticated. Based on django.views.decorators.cache.never_cache.
See: http://stackoverflow.com/questions/12060036/why-django-1-4-per-site-cache-does-not-work-correctly-with-cache-middleware-anon
"""
from django.utils.decorators import available_attrs
from django.utils.cache import add_never_cache_headers
from functools import wraps
def conditional_cache(view_func):
"""
Checks the user and if it's authenticated pass it through never_cache.
This version uses functools.wraps for the wrapper function.
"""
#wraps(view_func, assigned=available_attrs(view_func))
def _wrapped_view_func(request, *args, **kwargs):
response = view_func(request, *args, **kwargs)
if request.user.is_authenticated():
add_never_cache_headers(response)
return response
return _wrapped_view_func
Any suggestions to avoid the need of an extra decorator will be appreciated.
Thanks!
Ok, I just confirmed my "problem" was caused by Django lazy loading the User object.
To confirm it, I just added something like this to my view:
test_var = "some text" + request.user
And I got an error message telling me I couldn't concatenate an str to a SimpleLazyObject. At this point the lazy loading logic hasn't got a real User object yet.
To bypass the lazy loading, hence return a non-cache view for authenticated users, I just needed to access some method or attribute to triggers an actual query on the User object. I ended up with this, which I think it's the simplest way:
bypass_lazyload = request.user.is_authenticated()
My conditional_cache decorator is no longer needed, although it was an interesting exercise.
I may not need to do this when I finish working with my views as I'll access some user methods and attributes on my templates anyway but it's good to know what was going on.
Regards.