I am making development, and my POST request come like config/alert and I am trying to catch it with action decorater below.BUT Django does not catch it It wants config/alert/ but It is not possible to change incoming url add the SLASH.
#action(methods=['post'], detail=False, url_path='alert')
def get_post(self, request, pk=None, *args, **kwargs):
How can I modify the action to listen without slash in POST request?
I guess you have to check the APPEND_SLASH setting in your settings.py file.
You can find more information about option here: https://docs.djangoproject.com/en/dev/ref/settings/#append-slash
I found the solution adding extra APIView and putting ? end of the re_path in urls.py
re_path(r'alerta_forward/alert/?', post_data)
With using ? it became option slash, then I called an APIView.
Related
My Django web app allows users to connect via both HTTP and HTTPS. I'm curious: is there a way to detect the url scheme in settings.py? If so, how?
This can't be done in settings.py file as it contains only constants and variables. It doesn't process requests. But this can be done in a middleware.
Here's a middleware in it's simplest form to check URL scheme:
class DetectUrlScheme(object):
""" Middleware for detecting URL scheme """
def process_request(self, request):
if request.is_secure():
# HTTPS
# do something ...
else:
# HTTP
# do something ...
return
Although, if you've got only a couple of views, writing a middleware seems an overhead. So, you can check URL scheme in your views, too.
def myview(request):
if request.is_secure():
# HTTPS
# do something ...
else:
# HTTP
# do something ...
I'd like to add a 'Last seen' url list to a project, so that last 5 articles requested by users can be displayed in the list to all users.
I've read the middleware docs but could not figure out how to use it in my case.
What I need is a simple working example of a middleware that captures the requests so that they can be saved and reused.
Hmm, don't know if I would do it with middleware, or right a decorator. But as your question is about Middleware, here my example:
class ViewLoggerMiddleware(object):
def process_response(self, request, response):
# We only want to save successful responses
if response.status_code not in [200, 302]:
return response
ViewLogger.objects.create(user_id=request.user.id,
view_url=request.get_full_path(), timestamp=timezone.now())
Showing Top 5 would be something like;
ViewLogger.objects.filter(user_id=request.user.id).order_by("-timestamp")[:5]
Note: Code is not tested, I'm not sure if status_code is a real attribute of response. Also, you could change your list of valid status codes.
I am using django with rest framework and I try to test POST on existing object but I keep getting 405. my ViewSet looks like this:
class Agents(viewsets.ModelViewSet):
serializer_class = serializer.AgentSerializer
model = serializer_class.Meta.model
....
and in the urls:
router = routers.SimpleRouter()
router.register(r'rest/agents', api_views.Agents, "Agent")
...
urlpatterns += router.urls
I call the post request from within APITestCase class (rest testing), my post request looks like this:
response = self.client.post(url, {'available': True, 'online':True}, format='json')
and printing url shows "/chat/rest/agents/1910_1567/", while 1910_1567 is the valid id of an existing agent (I create the agent during the setup and use its id).
I've seen other questions about rest post getting 405, but all the solutions there were url-related and in my case the url is correct. I even ran the setup outside the test and accessed the url via browser to get the object and the object indeed exist. but when I try to post to it - 405.
any ideas?
thanks!
Most probably your url is somehow matching with some different url's regex, and the request is being dispatched to some other view which disallows post request. Could you mention others urls in your urls.py? Infact you can verify this by adding pdb(debugger) in dispatch method of your view. If you made it to the pdb, then you can assume I was wrong.
If that is not the case, then you can evaluate the issue from dispatch method with debugger. Just in case you have any doubt about how to do that -
class Agents(viewsets.ModelViewSet):
serializer_class = serializer.AgentSerializer
model = serializer_class.Meta.model
def dispatch(self, *args, **kwargs):
import ipdb; ipdb.set_trace()
return super(Agents, self).dispatch(*args, **kwargs)
Solution Found:-
You are passing id in url, so this request would be routed to detail view and detail view only allow GET, PUT, DELETE operations on resource since resource already exists. So to create resource, don't provide the id. Otherwise use PUT request and provide support for creation in PUT.
POST are supposed to be for creation purpose. If you want to update with a ViewSet you'll need to PUT or PATCH (partial update) instead.
Edit:
For more about this, here's some explanation on the HTTP methods used for REST API:
http://restful-api-design.readthedocs.org/en/latest/methods.html
This is also described in the DRF documentation at:
http://www.django-rest-framework.org/api-guide/routers/#simplerouter
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 trying to allow registration (using this django-registration register view) to one of my applications from a modal dialog.
Since this form is in a modal box, I'd like to get an json reponse on success (instead of the default redirection)
How can I use this view (django-registration register) to manage the registration and send back a json response on success ?
I know how to make ajax/json responses, the question is how to use the django-registration view without the redirection behavior or wrap it into an other view to manage the response.
First you need to change the urls.py to wrap the existing view with another functionality. To do that you have to create a new backend package in backends folder and change urls.py there while keeping everything else intact, or you could just go ahead and modify the existing urls.py in the backend package.
I have not tested this, but it should work.
Point url to the new view:
# urls.py
url(r'^register/$', register_wrap,
{'backend': 'registration.backends.default.DefaultBackend'},
name='registration_register'),
# your new view that wraps the existing one
def register_wrap(request, *args, **kwargs):
# call the standard view here
response = register(request, *args, **kwargs)
# check if response is a redirect
if response.status_code == 302:
# this was redirection, send json response instead
else:
# just return as it is
return response
If you are going to need this for more views you can just create a decorator using this.
Why I would do is to check if request.is_ajax() in your normal after-successfull-registration-redirect view and return json response there.
You ask how you can use the existing view to manage the registration and send back a json response on success. Since the HttpResponseRedirect is pretty much hard coded in the view, you can't use the view as it is. Instead, either fork it, or write your own view and change the urls.py so that r'^register/$' directs to your new view.
As far as the json response is concerned, on success you can do something like this:
from django.utils import simplejson as json
def register_ajax(request):
...
return HttpResponse(json.dumps(dict(success=True, **dict_containing_data)))
Hope this helps