How can I localize my urls in Django?
url(u'^{0}/$'.format(_('register')), RegisterView.as_view(), name='register'),
I tried the above but it only seems to work when I stop and start the server. I guess that the urls are translated when the application start.
So if there a way to solve this?
It's a bit more complicated than just throwing _() in urls.py. You have spotted the reason yourself: The URLs are evaluated once, when Django starts, and not for every request. Therefore you will have to
a) put every possible translation in urls.py, or
b) implement routing yourself
A. All in urls.py
url('hello', hello, name="hello_en"),
url('hallo', hello, name="hello_de"),
url('buenos_dias', hello, name="hello_es"),
Obviously not a nice solution, but it works for small projects.
B. Implementing routing
That has it's own drawback, especially when it comes to use reverse(). However, it works in principle:
urls.py:
#...
url('(?<path>.+)', dispatcher),
#...
views.py:
def dispatcher(request, path):
if path == "hallo":
lang = "de"
elif path == "buenos_dias":
lang = "de"
else:
lang = "en"
Of course, you can make the lookup more intelligent, but then you have to make preassumptions:
# if request.session['language'] can be trusted:
def dispatcher(request, path):
list_of_views = ['contact', 'about', 'foo']
v = None
for view in list_of_views:
if _(view) == path:
v = view
break
if v is None:
# return 404 or the home page
Internationalization of URLs has been introduced in django 1.4
see https://docs.djangoproject.com/en/dev/topics/i18n/translation/#url-internationalization
this is exactly what you are looking for
Related
I have a django-cms site, that uses i18n_patterns in urls.py. This works, urls are built like /lang/here-starts-the-normal/etc/.
Now, I would like to have urls like this: /prefix-lang/here-starts.... As there will be a couple of country specific domains, this wille be like /ch-de/here-... for Switzerland/.ch domain, /us-en/here-starts.... for the states, and some more. So, when the url would be /ch-de/..., the LANGUAGE would still be de. Hope this is clear?
As the content is filled with existing LANGUAGES=(('de', 'DE'), ('en', 'EN'), ...), I cannot change LANGUAGES for every domain - no content would be found in the cms, modeltranslation, only to mention those two.
How can I prefix the language slug in i18n_patterns? Is it possible at all?
I think a way without hacking Django too much would be to use URL rewrite facility provided by the webserver you run, for example, for mod_wsgi you can use mod_rewrite, similar facility exists also for uWSGI.
You may need to also post-process the output from Django to make sure that any links are also correctly re-written to follow the new schema. Not the cleanest approach but seems doable.
Working example, though the country/language order is reversed (en-ch instead of ch-en), to have it as django expects it when trying to find a language (ie, setting language to "en-ch", it will find "en", if available).
This solution involves a modified LocaleMiddleware, i18n_patterns, LocaleRegexResolver. It supports no country, or a 2 char country code, setup with settings.SITE_COUNTRY. It works by changing urls to a lang-country mode, but the found language code in middleware will still be language only, 2 chars, and work perfectly with existing LANGUAGES, that contain 2 chars language codes.
custom_i18n_patterns.py - this just uses our new resolver, see below
from django.conf import settings
from ceco.resolvers import CountryLocaleRegexURLResolver
def country_i18n_patterns(*urls, **kwargs):
"""
Adds the language code prefix to every URL pattern within this
function. This may only be used in the root URLconf, not in an included
URLconf.
"""
if not settings.USE_I18N:
return list(urls)
prefix_default_language = kwargs.pop('prefix_default_language', True)
assert not kwargs, 'Unexpected kwargs for i18n_patterns(): %s' % kwargs
return [CountryLocaleRegexURLResolver(list(urls), prefix_default_language=prefix_default_language)]
resolvers.py
import re
from django.conf import settings
from django.urls import LocaleRegexURLResolver
from modeltranslation.utils import get_language
class CountryLocaleRegexURLResolver(LocaleRegexURLResolver):
"""
A URL resolver that always matches the active language code as URL prefix.
extended, to support custom country postfixes as well.
"""
#property
def regex(self):
language_code = get_language() or settings.LANGUAGE_CODE
if language_code not in self._regex_dict:
if language_code == settings.LANGUAGE_CODE and not self.prefix_default_language:
regex_string = ''
else:
# start country changes
country_postfix = ''
if getattr(settings, 'SITE_COUNTRY', None):
country_postfix = '-{}'.format(settings.SITE_COUNTRY)
regex_string = '^%s%s/' % (language_code, country_postfix)
# end country changes
self._regex_dict[language_code] = re.compile(regex_string, re.UNICODE)
return self._regex_dict[language_code]
middleware.py - only very few lines changed, but had to replace the complete process_response.
from django.middleware.locale import LocaleMiddleware
from django.conf import settings
from django.conf.urls.i18n import is_language_prefix_patterns_used
from django.http import HttpResponseRedirect
from django.urls import get_script_prefix, is_valid_path
from django.utils import translation
from django.utils.cache import patch_vary_headers
class CountryLocaleMiddleware(LocaleMiddleware):
"""
This is a very simple middleware that parses a request
and decides what translation object to install in the current
thread context. This allows pages to be dynamically
translated to the language the user desires (if the language
is available, of course).
"""
response_redirect_class = HttpResponseRedirect
def process_response(self, request, response):
language = translation.get_language()
language_from_path = translation.get_language_from_path(request.path_info)
urlconf = getattr(request, 'urlconf', settings.ROOT_URLCONF)
i18n_patterns_used, prefixed_default_language = is_language_prefix_patterns_used(urlconf)
if (response.status_code == 404 and not language_from_path and
i18n_patterns_used and prefixed_default_language):
# Maybe the language code is missing in the URL? Try adding the
# language prefix and redirecting to that URL.
# start country changes
language_country = language
if getattr(settings, 'SITE_COUNTRY', None):
language_country = '{}-{}'.format(language, settings.SITE_COUNTRY)
language_path = '/%s%s' % (language_country, request.path_info)
# end country changes!
path_valid = is_valid_path(language_path, urlconf)
path_needs_slash = (
not path_valid and (
settings.APPEND_SLASH and not language_path.endswith('/') and
is_valid_path('%s/' % language_path, urlconf)
)
)
if path_valid or path_needs_slash:
script_prefix = get_script_prefix()
# Insert language after the script prefix and before the
# rest of the URL
language_url = request.get_full_path(force_append_slash=path_needs_slash).replace(
script_prefix,
'%s%s/' % (script_prefix, language_country),
1
)
return self.response_redirect_class(language_url)
if not (i18n_patterns_used and language_from_path):
patch_vary_headers(response, ('Accept-Language',))
if 'Content-Language' not in response:
response['Content-Language'] = language
return response
I'm using path converter in my django app like so:
# urls.py
from . import views
from django.urls import path
urlpatterns = [
path('articles/<str:collection>', views.ArticleView),
]
# views.py
#login_required
def ArticleView(request, collection):
print(collection)
if collection == "None":
articles_query = ArticleModel.objects.all()
...
This works fine in development for a url suck as : http://localhost:8000/articles/My Collection which gets encoded to http://localhost:8000/articles/My%20Collection, and is decoded properly in the ArticleView. However, in development, I have to edit the view like so to get it to work:
# views.py
import urllib.parse
#login_required
def ArticleView(request, collection):
collection = urllib.parse.unquote(collection)
print(collection)
if collection == "None":
articles_query = ArticleModel.objects.all()
...
Otherwise, the print(collection) shows My%20Collection and the whole logic in the rest of the view fails.
requirements.txt
asgiref==3.2.10
Django==3.1.1
django-crispy-forms==1.9.2
django-floppyforms==1.9.0
django-widget-tweaks==1.4.8
lxml==4.5.2
Pillow==7.2.0
python-pptx==0.6.18
pytz==2020.1
sqlparse==0.3.1
XlsxWriter==1.3.3
pymysql
What am I doing wrong here?
Thanks in advance!
The URL is being urlencoded which encodes spaces as %20. There are a number of other encodings. As you've discovered you need to decode that parameter in order to compare it to what you'd expect. As you've likely realized, if you have a value that actually wants The%20News and not The News, you have no recourse. To handle this people will create a slug field. Django has a model field for this in the framework.
This is typically a URL-friendly, unique value for the record.
Assuming you add a slug = models.SlugField() to ArticleModel, your urls and view can change into:
urlpatterns = [
# Define a path without a slug to identify the show all code path and avoid the None sentinel value.
path('articles', views.ArticleView, name='article-list'),
path('articles/<slug:slug>' views.ArticleView, name='article-slug-list'),
]
#login_required
def ArticleView(request, slug=None):
articles_query = ArticleModel.objects.all()
if slug:
articles_query = articles_query.filter(slug=slug)
It seems like the original URL querying function has been removed from Django 3.1. Does anyone know how to do it with a new package?
The url.py:
urlpatterns = [
re_path(r'^portfolio/(?P<title>[\w-]+)/$' , BlogApp_View.displayPortfolio, name='displayPortfolio'),
path('portfolio/', BlogApp_View.selectPortfolio, name='selectPortfolio'),]
The view.py
def displayPortfolio(request):
title = request.GET.get('title')
portfolio = Article.objects.filter(articleType__name__contains = "Portfolio", title=title)
print(title)
DICT = {}
return render(request, 'Article/', DICT)
The problem is now if I visit http://127.0.0.1:8000/Blog/portfolio/?title=A_UAV_Positioning_Approach_Using_LoRa/, it will skip the re_path shows in url.py.
Instead, it goes to the path one.
I have tried str:title method but that is actually not what I want. I prefer using the question mark pattern to finish the query.
The part after the questionmark is the querystring [wiki] and is not part of the path. This thus means that regardless what patterns you write, you can not distinguish on this, since the path patterns, regardless whether it is a path or re_path, are never matched against a URL with a query string.
You thus should write a single view, and inspect the request.GET query dict (which is a dictionary-like representation of the query string and see if it contains a value for title.
Your urlpatterns thus look like:
urlpatterns = [
path('portfolio/', BlogApp_View.selectPortfolio, name='selectPortfolio'),
]
and in the view, you can see if it contains a title:
def selectPortfolio(request):
if 'title' in request.GET:
# contains a ?title=…
title = request.GET.get('title')
portfolio = Article.objects.filter(
articleType__name__contains='Portfolio',
title=title
)
data = {'portfolio': portfolio}
return render(request, 'some_template.html', data)
else:
# contains no ?title=…
# …
return …
I have build a view and a form in Django1.5. If the POST request is successful (based on some values I set) then I need the page to redirect to another URL which is created simultaneously.
Otherwise, if the POST was not successful I need to stay on the same page. Right now I have solved the problem as following but I am quite sure this is not the best way to do it:
This is a part of my view:
def layer_create(request, template='layers/layer_create.html'):
if request.method == 'GET':
....
elif request.method == 'POST':
out = {}
...
new_table = 'something that comes from the form'
if form.is_valid():
...
try:
...
out['success'] = True
except:
...
out['success'] = False
finally:
if out['success']:
status_code = 200
# THIS IS THE PART WHICH I THINK I CAN IMPROVE
template = '/something/workspace:' + new_table + '/metadata'
else: # if form not valid
out['success'] = False
return render_to_response(template, RequestContext(request, {'form': form}))
This part of the code:
template = '/something/workspace:' + new_table + '/metadata'
seems very ugly to me. But as I am quite new in Django I am not sure how to approach this matter.
A side note first about Django 1.5 - you're highly advised to upgrade to a supported version like 1.8.
Redirecting
For redirecting you can use the redirect shortcut. (Or HttpResponseRedirect)
from django.shortcuts import redirect
# out of a view context
return redirect('/url/to/redirect/to/')
Building URLs
Indeed - as you did mention, your attempt with template = '/something/workspace:' + new_table + '/metadata' is not the cleanest way :)
Django provides a really nice way with the URL dispatcher.
A complete solution here would go too far (or definitely would require more detailed information about your project structure) - I would recommend you to dive into the Django URL dispatcher.
In short you would do something like:
# app/urls.py
urlpatterns = [
#...
url(r'^workspace/(?P<id>[0-9]+)/metadata/$', views.workspace_detail, name='workspace-detail-metadata'),
#...
]
Then you are able to reverse your URL patterns:
from django.core.urlresolvers import reverse
url = reverse('workspace-detail-metadata', kwargs={'id': 123})
# would result in:
# .../workspace/123/metadata/
After all, I have used the "reverse" method as follows:
layer = 'geonode:' + new_table
return HttpResponseRedirect(
reverse(
'layer_metadata',
args=(
layer,
)))
Where my urls.py file includes:
url(r'^(?P<layername>[^/]*)/metadata$', 'layer_metadata', name="layer_metadata"),
As described here this is the most appropriate way to do it.
I run a Pinax-site for collaborative purposes. I added 'account.middleware.AuthenticatedMiddleware' to 'MIDDLEWARE_CLASSES' in order to not allow anonymous access to anything on the site.
But now I need public APIs to be enabled. Is there any solutions besides adding 'login_required'-decorator at all the views that still need to be private?
edit
Gregor Müllegger answer doesn't work. settings.AUTHENTICATED_EXEMPT_URLS seems to get overwritten somewhere in the code
class AuthenticatedMiddleware(object):
def __init__(self, login_url=None, redirect_field_name=REDIRECT_FIELD_NAME):
if login_url is None:
login_url = settings.LOGIN_URL
self.redirect_field_name = redirect_field_name
self.login_url = login_url
self.exemptions = [
r"^%s" % settings.MEDIA_URL,
r"^%s" % settings.STATIC_URL,
r"^%s$" % login_url,
]
print "settings.AUTHENTICATED_EXEMPT_URLS ",settings.AUTHENTICATED_EXEMPT_URLS
if ( settings.AUTHENTICATED_EXEMPT_URLS):
self.exemptions += settings.AUTHENTICATED_EXEMPT_URLS
print "settings.AUTHENTICATED_EXEMPT_URLS ",settings.AUTHENTICATED_EXEMPT_URLS
doesn't print my settings but this:
settings.AUTHENTICATED_EXEMPT_URLS ['^/account/signup/$', '^/account/password_reset', '^/account/confirm_email', '^/openid']
I will try to fix it.
Have a look at the source code of AuthenticatedMiddleware.
It reveals that there is a setting called AUTHENTICATED_EXEMPT_URLS. It can contain regular expressions that are left public. Set it to something like this in your settings.py:
AUTHENTICATED_EXEMPT_URLS = (r"^api/",)
This will make any URLs below /api/ available without being logged in.t