There is a complex app (not possible to just paste the code). Going to try to explain.
Django
There is a urls.py file from the native Django app. The urlpatterns defines and register its urls. The ^foo/ defines a group of related urls and the foonamepsace.
urlpatterns = patterns('',
...
url(r'^foo/', include('foo.urls', namespace='foonamespace')),
...
Now there is a method generate_report which does some logic inside and then uses render_to_string to return the HTML:
def generate_report(..):
...
return render_to_string('foo/foo_report.html', args)
Everything works inside the app, the url get reversed successfully.
Django Rest Framework (DRF)
Now there is a DRF implementation and one of its resources is supposed to return a report in a binary format.
class PDFReportViewSet(APIView):
renderer_classes = (BinaryFileRenderer, )
def get(..):
...
pdf = generate_report() # <-- fails with NoReverseMatch
...
return response
Problem
The ViewSet calls the generate_report, however one gets an error when trying to parse the HTML:
NoReverseMatch: foonamespace' is not a registered namespace
Question
Any clues why DRF cannot reverse the namespcae/url from the the core of Django app? How to make sure DRF can reverse a namespace from the core urls.py urlpattern?
Added
After investigation, inside the foo_report.html any usage of the url, for example {% url 'foonamespace:123' %} or {% url 'barnamespace:123' %} produces the error - only if ran from the DRF (running the same page using native Django works fine).
URLS
foo.urls.py
from django.conf.urls import patterns, url
from foo.views import (FooListView, FooDetailView...)
urlpatterns = patterns('',
url(r'^$', FooListView.as_view(), name='foo_list'),
url(r'^(?P<pk>\d+)/$', FooDetailView.as_view(), name='foo_details'),
Important note. The app is served at some.domain.com/, while the REST is served from some.domain.com/rest. So may be this way /rest just don't include anything because it is a parent of the root (which includes the foo.urls.py)
I was managed to resolve my issue with the help from #dirkgroten. It was difficult to see the problem without looking at the source code.
Solution
Updated the routers.py file:
urlpatterns = router.urls
urlpatterns += patterns('',
url(r'^foo/', include('foo.urls', namespace='foonamespace')),
)
Explanation
Basically, the app was serve from the root url / while the rest was served from /rest. The DRF router simply didn't include any of the root routes. Adding them manually like it is shown in solution resolved the problem and made foonamespace visible for all DRF elements.
I have a Django backend that returns json data. I'm able to get data back on my localhost but got a 404 on production server. I'm running nginx in front of gunicorn server. Any ideas why I'm getting a 404? Shouldn't this be able to work to retrieve json data, or do I need to use django rest framework and implement viewsets to make this work?
Not Found
The requested URL /about was not found on this server.
urls.py
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^about', about.get_info),
]
about.py
from django.http import JsonResponse
def get_info(req):
return JsonResponse({"test": "hello"})
The problem is inside url.py. The way the rules are defined currently, it would only allow you to open about/ and admin/, i.e. with the / at the end. To fix this, you can define the URLs as following:
urlpatterns = [
url(r'^admin/$', admin.site.urls),
url(r'^about/$', about.get_info),
]
Now you should be able to use both admin/ and admin to access the page.
Very new to django. I'm using version 1.5.2 and I just did a fresh install. I'm using the django development server; I'll be moving to Apache down the road, but I want to understand the django's particular flavor of MVC methodology before doing taking that step.
So I start up the django server with `python manage.py runserver 0.0.0.0:8000' through the terminal in my project directory (django_books). I get this error:
ViewDoesNotExist at /
Could not import django_books.views.home. Parent module django_books.views does not exist.
So my view doesn't exist. My view.py file is empty because the tutorial I was following did not include one. I'm not sure if this is the problem. If it is, how do I create this file (what goes in it)?
Directory Structure:
django_books
beer (from the tutorial lol)
migrations
__init__.py
models.py
views.py
random_book
(same as beer above)
django_books (this is my actual django project, beer and random_book are apps)
__init__.py
settings.py
urls.py
wsgi.py
media
.gitignore
manage.py
requirements.txt (output from pip freeze command)
urls.py
from django.conf.urls import patterns, include, url
# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()
urlpatterns = patterns('',
# Examples:
url(r'^$', 'django_books.views.home', name='home'),
# url(r'^django_books/', include('django_books.foo.urls')),
# Uncomment the admin/doc line below to enable admin documentation:
# url(r'^admin/doc/', include('django.contrib.admindocs.urls')),
# Uncomment the next line to enable the admin:
url(r'^admin/', include(admin.site.urls)),
)
If you keep your urls.py the way it is, that means you need to create views.py within /django_books/django_books/
Within that file, create a new function called home.
Alternately, if you have any functions inside of /django_books/beer/, you could reference them from urls.py.
All urls.py does is expose a python path to a function and route a URL there. So you can see that you don't have a module or file called views within django_books/django_books, which is why you get the failure.
View is basically a python function that receives HTTP Request and returns HTTP Response.
Quote from docs:
A view function, or view for short, is simply a Python function that
takes a Web request and returns a Web response. This response can be
the HTML contents of a Web page, or a redirect, or a 404 error, or an
XML document, or an image . . . or anything, really. The view itself
contains whatever arbitrary logic is necessary to return that
response. This code can live anywhere you want, as long as it’s on
your Python path. There’s no other requirement–no “magic,” so to
speak. For the sake of putting the code somewhere, the convention is
to put views in a file called views.py, placed in your project or
application directory.
This line url(r'^$', 'django_books.views.home', name='home'), in urls.py points the index / of your site to the home view - you should create it.
Create a python function called home in views.py:
from django.http import HttpResponse
import datetime
def home(request):
now = datetime.datetime.now()
html = "<html><body>It is now %s.</body></html>" % now
return HttpResponse(html)
Restart your development server and visit http://127.0.0.1:8000.
FYI, read the tutorial more carefully, part 3 is about dealing with urls and views.
Following the tutorial found here exactly, I cannot create a custom 500 or 404 error page. If I do type in a bad url, the page gives me the default error page. Is there anything I should be checking for that would prevent a custom page from showing up?
File directories:
mysite/
mysite/
__init__.py
__init__.pyc
settings.py
settings.pyc
urls.py
urls.pyc
wsgi.py
wsgi.pyc
polls/
templates/
admin/
base_site.html
404.html
500.html
polls/
detail.html
index.html
__init__.py
__init__.pyc
admin.py
admin.pyc
models.py
models.pyc
tests.py
urls.py
urls.pyc
view.py
views.pyc
templates/
manage.py
within mysite/settings.py I have these enabled:
DEBUG = False
TEMPLATE_DEBUG = DEBUG
#....
TEMPLATE_DIRS = (
'C:/Users/Me/Django/mysite/templates',
)
within mysite/polls/urls.py:
from django.conf.urls import patterns, url
from polls import views
urlpatterns = patterns('',
url(r'^$', views.index, name='index'),
url(r'^(?P<poll_id>\d+)/$', views.detail, name='detail'),
url(r'^(?P<poll_id>\d+)/results/$', views.results, name='results'),
url(r'^(?P<poll_id>\d+)/vote/$', views.vote, name='vote'),
)
I can post any other code necessary, but what should I be changing to get a custom 500 error page if I use a bad url?
Under your main views.py add your own custom implementation of the following two views, and just set up the templates 404.html and 500.html with what you want to display.
With this solution, no custom code needs to be added to urls.py
Here's the code:
from django.shortcuts import render_to_response
from django.template import RequestContext
def handler404(request, *args, **argv):
response = render_to_response('404.html', {},
context_instance=RequestContext(request))
response.status_code = 404
return response
def handler500(request, *args, **argv):
response = render_to_response('500.html', {},
context_instance=RequestContext(request))
response.status_code = 500
return response
Update
handler404 and handler500 are exported Django string configuration variables found in django/conf/urls/__init__.py. That is why the above config works.
To get the above config to work, you should define the following variables in your urls.py file and point the exported Django variables to the string Python path of where these Django functional views are defined, like so:
# project/urls.py
handler404 = 'my_app.views.handler404'
handler500 = 'my_app.views.handler500'
Update for Django 2.0
Signatures for handler views were changed in Django 2.0:
https://docs.djangoproject.com/en/2.0/ref/views/#error-views
If you use views as above, handler404 will fail with message:
"handler404() got an unexpected keyword argument 'exception'"
In such case modify your views like this:
def handler404(request, exception, template_name="404.html"):
response = render_to_response(template_name)
response.status_code = 404
return response
Official answer:
Here is the link to the official documentation on how to set up custom error views:
https://docs.djangoproject.com/en/stable/topics/http/views/#customizing-error-views
It says to add lines like these in your URLconf (setting them anywhere else will have no effect):
handler404 = 'mysite.views.my_custom_page_not_found_view'
handler500 = 'mysite.views.my_custom_error_view'
handler403 = 'mysite.views.my_custom_permission_denied_view'
handler400 = 'mysite.views.my_custom_bad_request_view'
You can also customise the CSRF error view by modifying the setting CSRF_FAILURE_VIEW.
Default error handlers:
It's worth reading the documentation of the default error handlers, page_not_found, server_error, permission_denied and bad_request. By default, they use these templates if they can find them, respectively: 404.html, 500.html, 403.html, and 400.html.
So if all you want to do is make pretty error pages, just create those files in a TEMPLATE_DIRS directory, you don't need to edit URLConf at all. Read the documentation to see which context variables are available.
In Django 1.10 and later, the default CSRF error view uses the template 403_csrf.html.
Gotcha:
Don't forget that DEBUG must be set to False for these to work, otherwise, the normal debug handlers will be used.
Add these lines in urls.py
urls.py
from django.conf.urls import (
handler400, handler403, handler404, handler500
)
handler400 = 'my_app.views.bad_request'
handler403 = 'my_app.views.permission_denied'
handler404 = 'my_app.views.page_not_found'
handler500 = 'my_app.views.server_error'
# ...
and implement our custom views in views.py.
views.py
from django.shortcuts import (
render_to_response
)
from django.template import RequestContext
# HTTP Error 400
def bad_request(request):
response = render_to_response(
'400.html',
context_instance=RequestContext(request)
)
response.status_code = 400
return response
# ...
Django 3.0+ 4.0+
here is link how to customize error views
here is link how to render a view
in the urls.py (the main one, in project folder), put:
handler404 = 'my_app_name.views.custom_page_not_found_view'
handler500 = 'my_app_name.views.custom_error_view'
handler403 = 'my_app_name.views.custom_permission_denied_view'
handler400 = 'my_app_name.views.custom_bad_request_view'
and in the mentioned app (my_app_name) put in the views.py:
def custom_page_not_found_view(request, exception):
return render(request, "errors/404.html", {})
def custom_error_view(request, exception=None):
return render(request, "errors/500.html", {})
def custom_permission_denied_view(request, exception=None):
return render(request, "errors/403.html", {})
def custom_bad_request_view(request, exception=None):
return render(request, "errors/400.html", {})
NOTE: errors/404.html is the path if you place your files into the projects (not the apps) template foldertemplates/errors/404.html so please place the files where you want and write the right path.
NOTE 2: After page reload, if you still see the old template, change in settings.py DEBUG=True, save it, and then again to False and save again (that will restart the server and collect the new files).
From the page you referenced:
When you raise Http404 from within a view, Django will load a special view devoted to handling 404 errors. It finds it by looking for the variable handler404 in your root URLconf (and only in your root URLconf; setting handler404 anywhere else will have no effect), which is a string in Python dotted syntax – the same format the normal URLconf callbacks use. A 404 view itself has nothing special: It’s just a normal view.
So I believe you need to add something like this to your urls.py:
handler404 = 'views.my_404_view'
and similar for handler500.
If all you need is to show custom pages which have some fancy error messages for your site when DEBUG = False, then add two templates named 404.html and 500.html in your templates directory and it will automatically pick up this custom pages when a 404 or 500 is raised.
In Django 3.x, the accepted answer won't work because render_to_response has been removed completely as well as some more changes have been made since the version the accepted answer worked for.
Some other answers are also there but I'm presenting a little cleaner answer:
In your main urls.py file:
handler404 = 'yourapp.views.handler404'
handler500 = 'yourapp.views.handler500'
In yourapp/views.py file:
def handler404(request, exception):
context = {}
response = render(request, "pages/errors/404.html", context=context)
response.status_code = 404
return response
def handler500(request):
context = {}
response = render(request, "pages/errors/500.html", context=context)
response.status_code = 500
return response
Ensure that you have imported render() in yourapp/views.py file:
from django.shortcuts import render
Side note: render_to_response() was deprecated in Django 2.x and it has been completely removed in verision 3.x.
No additional view is required. https://docs.djangoproject.com/en/3.0/ref/views/
Just put the error files in the root of templates directory
404.html
400.html
403.html
500.html
And it should use your error page when debug is False
settings.py:
DEBUG = False
TEMPLATE_DEBUG = DEBUG
ALLOWED_HOSTS = ['localhost'] #provide your host name
and just add your 404.html and 500.html pages in templates folder.
remove 404.html and 500.html from templates in polls app.
In Django 2.* you can use this construction in views.py
def handler404(request, exception):
return render(request, 'errors/404.html', locals())
In settings.py
DEBUG = False
if DEBUG is False:
ALLOWED_HOSTS = [
'127.0.0.1:8000',
'*',
]
if DEBUG is True:
ALLOWED_HOSTS = []
In urls.py
# https://docs.djangoproject.com/en/2.0/topics/http/views/#customizing-error-views
handler404 = 'YOUR_APP_NAME.views.handler404'
Usually i creating default_app and handle site-wide errors, context processors in it.
Make an error, on the error page find out from where django is loading templates. I mean the path stack. In base template_dir add these html pages 500.html , 404.html. When these errors occur the respective template files will be automatically loaded.
You can add pages for other error codes too, like 400 and 403.
As one single line (for 404 generic page):
from django.shortcuts import render_to_response
from django.template import RequestContext
return render_to_response('error/404.html', {'exception': ex},
context_instance=RequestContext(request), status=404)
# views.py
def handler404(request, exception):
context = RequestContext(request)
err_code = 404
response = render_to_response('404.html', {"code":err_code}, context)
response.status_code = 404
return response
# <project_folder>.urls.py
handler404 = 'todo.views.handler404'
This works on django 2.0
Be sure to include your custom 404.html inside the app templates folder.
Try moving your error templates to .../Django/mysite/templates/.
I am note sure about this one, but I think these need to be "global" to the website.
In Django root urls.py file, add the below lines
from django.conf.urls import (handler400, handler403, handler404, handler500)
handler400 = 'app.views.bad_request'
handler403 = 'app.views.permission_denied'
handler404 = 'app.views.page_not_found'
handler500 = 'app.views.server_error'
In your app's views.py file, create the respective functions.
def server_error(request, exception=None):
# return render(request, '500.html')
return redirect('/')
Finally, in your settings.py file, set DEBUG = False
I had an additional
TEMPLATE_DIRS
within my settings.py and that was causing the problem.
This answer was posted as an edit to the question Django, creating a custom 500/404 error page by the OP reZach under CC BY-SA 3.0.
In urls.py, enter this code:
from django.conf.urls import (handler400, handler403, handler404, handler500)
handler404 = 'my_app.views.page_not_found_view'
then add this code in your views.py
from django.shortcuts import render,get_object_or_404
def page_not_found_view(request, exception):
return render(request, '404.html', status=404)
Dont forget to set DEBUG = False and also set ALLOWED_HOSTS = [127.0.0.1] while you are testing in your laptop.
You don't need to do anything fancy, just create a 404.html file in your templates. Go to settings.py and set:
DEBUG = False
ALLOWED_HOSTS = ["*"]
It will automatically overwrite the default.
Django > 2.2
from django.shortcuts import render_to_response, render
from django.template import RequestContext
def handler500(request, *args, **argv):
context = {}
print(request.body, '==========')
response = render(request, '500.jinja', context=context)
response.status_code = 500
return response
in urls.py
handler500 = 'apps.core.views.handler500'