So, kind of a in a weird situation here. I've got a Django project using TastyPie to power its API and some views/templates that will be used to power plugins for various sites. Rather than building this plugins as standard Django templates, I've been asked to use our API to process requests to the plugins, which means I'm calling a view on my server from another view, and for some reason that isn't working with any of my views. For reference:
#views.py
from django.http import HttpResponse
from django.shortcuts import render_to_response
from django.template.context import Context, RequestContext
import json, urllib, urllib2
SITE_NAME = "http://localhost:8000/"
API_PATH = "api/v1/"
def aggregate(request):
template = 'lemonwise/plugins/aggregate.html'
sku = request.GET.get('sku')
url = ''.join([SITE_NAME, API_PATH, 'product/?sku=', sku])
product = json.loads(urllib.urlopen(url).read())['objects'][0]
return render_to_response(template, product)
def reviews(request):
template = 'lemonwise/plugins/reviews.html'
sku = request.GET.get('sku')
url = ''.join([SITE_NAME, API_PATH, 'product/?format=json&sku=', sku])
#Comment the next line out and the url is passed correctly
response = urllib2.build_opener().open(url).read()
return HttpResponse(url)
#page = opener.open(url).read()
#return HttpResponse(url)
#product = json.loads(urllib2.build_opener().open(url).read())['objects'][0]
#return HttpResponse(url)
#reviews = [json.loads(urllib.urlopen(SITE_NAME + uri)) for uri in product['reviews']]
#return render_to_response(template, {'reviews': reviews})
def survey(request):
template = 'lemonwise/plugins/survey.html'
sku = request.GET.get('sku')
url = ''.join([SITE_NAME, API_PATH, 'product/?sku=', sku])
product = json.loads(urllib2.build_opener().open(url).read())['objects'][0]
return render_to_response(template, product)
def mosthelpfulpositive(request):
template = 'lemonwise/plugins/mosthelpfulpositive.html'
sku = request.GET.get('sku')
url = ''.join([SITE_NAME, API_PATH, 'product/?sku=', sku])
product = json.loads(urllib2.build_opener().open(url).read())['objects'][0]
uri = product['most_helpful_positive']
most_helpful_positive = json.loads(urllib.urlopen(SITE_NAME + uri))
return render_to_response(template, most_helpful_positive)
def mosthelpfulnegative(request):
template = 'lemonwise/plugins/mosthelpfulnegative.html'
sku = request.GET.get('sku')
url = ''.join([SITE_NAME, API_PATH, 'product/?sku=', sku])
product = json.loads(urllib2.build_opener().open(url).read())['objects'][0]
uri = product['most_helpful_negative']
most_helpful_negative = json.loads(urllib.urlopen(SITE_NAME + uri))
return render_to_response(template, most_helpful_negative)
And the corresponding urls.py (in a different app):
#urls.py
from django.conf import settings
from django.conf.urls.defaults import patterns, include, url
from django.contrib import admin
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
from tastypie.api import Api
from lemonwise.reviews.api import *
admin.autodiscover()
v1_api = Api(api_name='v1')
v1_api.register(UserResource())
v1_api.register(MerchantResource())
v1_api.register(ProductFamilyResource())
v1_api.register(ProductResource())
v1_api.register(BooleanAttributeResource())
v1_api.register(SlideAttributeResource())
v1_api.register(ProductAttributeResource())
v1_api.register(SubmissionResource())
v1_api.register(ReviewResource())
v1_api.register(ReviewProductAttributeResource())
v1_api.register(CommentResource())
v1_api.register(BestUseResource())
v1_api.register(HelpfulVoteResource())
#Acess the api via http://127.0.0.1:8000/api/v1/user/?format=json
urlpatterns = patterns('',
url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
url(r'^admin/', include(admin.site.urls)),
# Lemonwise apps
url(r'^reviews/', include('lemonwise.reviews.urls')),
#API
url(r'^api/', include(v1_api.urls)),
)
if settings.DEBUG:
urlpatterns += staticfiles_urlpatterns()
Any idea on how to fix this? I can't find any writings on the subject.
EDIT: More specifically, what's happening is that when I load any of these views they hang trying to read the api page. ALso, my server shows no indications of any requests being processed. (Though I can load the api pages directly.)
Now your comment has clarified the situation, I can take a guess at what the problem is. It is that you are using the built-in development server, which is single-threaded. So while it's processing the original request, it's not able to process the internal request for another URL - so it hangs indefinitely.
The solution, as Mao points out, is to think of a better architecture. If you can't do that, you may have some luck with using something like gunicorn instead of the built-in server.
Particular to Tastypie you can use your Resources in views, if you need to, accessing the objects directly and calling the get methods and building the bundle.
You can find out how: http://django-tastypie.readthedocs.org/en/latest/cookbook.html
Go to the section titled - Using Your Resource In Regular Views
Agreed, your development server likely only has one instance. Launch another one on a different port. ie/
SITE_NAME = "http://localhost:8001/"
Okay, so I already said that having views making HTTPRequests against other views is kind of silly, but I do see what your problem currently is. You are calling json.loads(urllib.urlopen(SITE_NAME + uri)), but urllib.urlopen returns a file-like object, not a string, so it should actually be json.load(urllib.urlopen(SITE_NAME + uri)).
I just encountered the same problem i.e. Calling an API from the same server it is hosted. As Daniel suggested, I was able to solve this by using multi threads in gunicorn:
gunicorn --certfile=crt --keyfile=key --bind 0.0.0.0:8000 app.wsgi --threads 5
Related
Hello! Please tell me how to organize a redirect correctly.
There is an old version of the site and a new one. In the old version (another CMS, not Django) the objects have their own URL, in the new revised scheme, and the objects have a different URL.
In each object on the new site, there is a completed field with the old URL. In model.py it looks like this:
old_url = models.CharField('Old URL', blank=True, max_length=100)
I specifically moved the old url to a separate field. Or was it not necessary to do this?
Question. How to set up a redirect correctly, so that after going to the object using the old URL, the site visitor will be redirected to the new URL of this object?
IMHO, I don't think writting old_url for each and every object is pretty inefficient. Instead you can implement a custom 404 view, and handle the redirection there.
I think you can create some regex or plain url maps to new url and redirect accordingly.
import re
from django.http import HttpResponseNotFound
OLD_URL_MAP = { 'old_url_regex': 'new_url_path'}
def handler404(self, request):
for old_re, new_url in OLD_URL_MAP.items():
if re.match(old_re, request.path):
return redirect(new_url, request.resolver_match.kwargs)
return HttpResponseNotFound('not found')
# inside urls.py
handler404 = 'myapp.views.handler404'
Here I have used a map hard coded in python, you can create a model for that as well.
Update
A costly solution is to use middleware. You can try like this:
import re
from django.urls import resolve
OLD_URL_MAP = { 'old_url_regex': 'new_url_path'}
class RerouteMiddleware:
def __init__(self, get_response):
self.get_response = get_response
def __call__(self, request):
try:
resolve(request.path_info) # trying to find if the current url exists in your django project. if not, it will throw exception.
except:
for old_re, new_url in OLD_URL_MAP.items(): # iterating through urls
if re.match(old_re, request.path):
return redirect(new_url, request.resolver_match.kwargs)
response = self.get_response(request)
return response
And add that middleware at the bottom of MIDDLEWARE settings.
FYI, its a regex based solution, assuming those urls are dynamic. Instead you can use plain text urls, but its up to you.
Use redirect() from django.shortcuts [the same library from where you import render]. Also, assuming, that the old_url contains only the relative url.
from django.shortcuts import render, redirect
def someView(request):
q = ~some queryset returning the current object~
current_url = request.get_full_path().strip("http://www.example.com/")
if q.old_url == current_url:
redirect(q.new_url)
else:
pass
Remember, redirect() returns an HttpResponse.
I've created an app "Blog". In my app I've got several models including "BlogIndex(Page)". When I run local server I find myself at "home_page.html". What I want is to start my local server at "blog_index.html". I know that I can set a root page at settings>site>localhost to make my "blog_index.html" a root page, but I can't do this because in my app I've got some other models that live at the same level as "BlogIndex(Page)" and they are children of the root which is "HomePage" so it would brake my code. So my question is: can I make a redirect from "HomePage(Page)" to my "BlogIndex" so that when i start my server I would be automatically redirected from "HomePage" to "BlogIndex"? How can I do it? How much it will affect the performance of the site and it's optimization?
I know that there is settings>redirect but it works only for inactive pages, but i need "HomePage" to be active.
Thank you.
Perhaps a better approach would be to display your blog posts (and any other models you want) on your homepage. Just override get_context(). See here: Wagtail Views: extra context
Update:
You can redirect by overriding the serve() method. For example, in your model, you would do something like:
# home/models.py
...
from django.http import HttpResponseRedirect
from django.urls import reverse
class HomePage(Page):
body = RichTextField(blank=True)
content_panels = Page.content_panels + [
FieldPanel('body', classname="full"),
]
def serve(self, request):
# Redirect to blog index page
return HttpResponseRedirect('/blog/')
# only do this if you're using urls.py and namespaces
# return HttpResponseRedirect(reverse('blog:index'))
More info: http://docs.wagtail.io/en/latest/reference/pages/model_recipes.html?highlight=serve()#overriding-the-serve-method
I am using django-two-factor-auth for a webapp. I cannot access the admin page.
I know I am entering the correct credentials. When I input incorrect credentials, I get an appropriate error message.
When I input the correct credentials, the page simply reloads with this URL:
http://localhost:8080/account/login/?next=/inveskore/
These are my settings related to two_factor:
LOGIN_URL = 'two_factor:login'
LOGIN_REDIRECT_URL = '/inveskore'
TWO_FACTOR_SMS_GATEWAY = 'two_factor.gateways.twilio.gateway.Twilio'
This is the associated URL path:
path('admin/', admin.site.urls),
According to this, it results from the admin user not having 2FA set.
So, how do you set 2FA for the admin user if you can't access the site?
EDIT:
I took down the 2FA login requirements for the site and then added a phone device. No luck.
I recently ran into this scenario and created this solution based on a comment there:
https://github.com/Bouke/django-two-factor-auth/issues/219#issuecomment-494382380
I subclassed AdminSiteOTPRequired and then specified it as the admin class to use
from django.conf import settings
from django.contrib.auth import REDIRECT_FIELD_NAME
from django.contrib.auth.views import redirect_to_login
from django.http import HttpResponseRedirect
from django.shortcuts import resolve_url
from django.urls import reverse
from django.utils.http import is_safe_url
from two_factor.admin import AdminSiteOTPRequired, AdminSiteOTPRequiredMixin
class AdminSiteOTPRequiredMixinRedirSetup(AdminSiteOTPRequired):
def login(self, request, extra_context=None):
redirect_to = request.POST.get(
REDIRECT_FIELD_NAME, request.GET.get(REDIRECT_FIELD_NAME)
)
# For users not yet verified the AdminSiteOTPRequired.has_permission
# will fail. So use the standard admin has_permission check:
# (is_active and is_staff) and then check for verification.
# Go to index if they pass, otherwise make them setup OTP device.
if request.method == "GET" and super(
AdminSiteOTPRequiredMixin, self
).has_permission(request):
# Already logged-in and verified by OTP
if request.user.is_verified():
# User has permission
index_path = reverse("admin:index", current_app=self.name)
else:
# User has permission but no OTP set:
index_path = reverse("two_factor:setup", current_app=self.name)
return HttpResponseRedirect(index_path)
if not redirect_to or not is_safe_url(
url=redirect_to, allowed_hosts=[request.get_host()]
):
redirect_to = resolve_url(settings.LOGIN_REDIRECT_URL)
return redirect_to_login(redirect_to)
Then in urls.py:
from django.contrib import admin
admin.site.__class__ = AdminSiteOTPRequiredMixinRedirSetup
I cannot comment yet but the answer above by #saschwarz worked like a charm with Django 2.2 and django-two-factor-auth. The only thing missing from his code is the additional import:
from django.http import HttpResponseRedirect
After that I was able to use this line:
admin.site.__class__ = AdminSiteOTPRequiredMixinRedirSetup
to quickly enforce two factor for admin users.
This is really missing from the docs for django-two-factor-auth... I looked at the documentation for quite a while and couldn't find a clear way to generate a qr code. It's only really demonstrated in the demo app, and not in a very modular way.
The solution from saschewarz didn't solve the problem in my app.
So I found another one where standard admin login is shown.
django-two-factor-auth by default patch your admin urls with function 'patch_admin' from two_factor\admin.py so you always will see your standard site login for admin login.
To solve this you can comment out 2 functions in two_factor\admin.py
def patch_admin()
and
def unpatch_admin()
And comment out in two_factor\apps.py
def ready(self)
And to use two factor authentication in admin site add this code in main urls.py (where is path to admin):
from django_otp.admin import OTPAdminSite
admin.site.__class__ = OTPAdminSite
I am new to Python and Django. I did a experiment on enforcing request method (e.g. for certain url you can only use GET). Here is my code.
tests.py
from django.test import TestCase, Client
client = Client()
class MyTests(TestCase):
def test_request_method:
""" Sending wrong request methods should result in 405 error """
self.assertEqual(client.post('/mytest', follow = True).status_code, 405)
urls.py
from django.conf.urls import url
from . import views
urlpatterns = [
url(r'^$', views.index, name = 'index'),
url(r'^mytest/', views.mytest, name = 'mytest'),
]
views.py
from django.http import HttpResponse
def mytest(request):
if request.method == 'GET':
return HttpResponse("Not implemented", status = 500)
else:
return HttpResponse("Only GET method allowed", status = 405)
But the test always returns status 500.
I saw here that this may be related to using follow=True in the client.post() call. However if I use follow=False I will get status 301 instead.
Any ideas? Thank you!
Does it perhaps redirect /mytest to /mytest/? The documentation suggests that by default, a trailing slash is added by doing a redirect if no URL pattern matches without the slash, and to quote:
Note that the redirect may cause any data submitted in a POST request to be lost.
A request caused by commonly used redirect status codes is always a GET request. You could either make the request to /mytest/ or remove the trailing slash from your URL pattern.
I'm pretty new to django and i'm working on a website that needs a dynamic URL for a database table, and it all works fine, but i would like to know how to remove the "?id=" from the url, so rather than
localhost:8000/dynamicurl/?id=XXXXX
The url becomes
localhost:8000/dynamicurl/XXXXX
Instead
I did a fine amount of searching in the documentation and didn't find a lot, though it's pretty likely i missed something.
EDIT:
thanks everyone for helping, the simplest answer was to remove the object i was using to fetch the ID and just replace it for ID in evert instance,
so my url became
url(r'^dynamicurl/(?P[0-9]+)/$', views.dynamicurl)
and my view became
def dynamicurl(request, id):
i'm like very very new to django FYI
you can capture a variable in url definition in the apps urls.py file. It would look something like this:
url(r'^dynamicurl/(?P<id>[0-9]+)?$', dynamicurl, name = 'dynamicurl'),
then in your view function you receive that parameter:
def dynamicurl(request, id):
If you are talking about how to change your url inside the urls, I suggest you to use code that already answered above: https://stackoverflow.com/a/41988051/6396981
But, if you talking about how to redirect localhost:8000/dynamicurl/?id=XXXXX to localhost:8000/dynamicurl/XXXXX, hope this usefull..
1. views.py
from django.http import HttpResponse
from django.views.generic.base import RedirectView
from django.core.urlresolvers import reverse
class RedirectView(RedirectView):
permanent = False
def get_redirect_url(self):
get_id = self.request.GET.get('id')
if get_id is not None:
return reverse('redirected_page', kwargs={'id': id})
return reverse('origin_page')
def redirected_view(request, id):
# your final view goes here...
return HttpResponse("You're looking for id: %d." % id)
2. urls.py
from django.conf.urls import url
from yourapp.views import views (RedirectView, redirected_view)
urlpatterns = [
# first view the pool to doing redirection
url(r'^pool/$', RedirectView.as_view(), name='origin_page'),
# the final url
url(r'^pool/(?P<id>[\d]+)/$', redirected_view, name='redirected_page'),
]