我一直在开发一个使用i18n / locale作为根/登录页面中的url前缀的应用程序。我尝试早期部署我的应用程序以进行测试,DEBUG = True,一切都运行良好。如果尝试访问domain.com,则会重定向到domain.com/en /
现在,由于我禁用了DEBUG = False,它不会重定向到/ en /,而是会显示404错误。
我看到了这个问题,这与我的情景很接近
Django 1.4 LocaleMiddleware not working with Apache, but works with runserver
但是,对于我的情况,我已经有404页面设置及其处理程序和404.html文件。
以下是我的代码
settings.py
LANGUAGE_CODE = 'en-us'
ADMIN_LANGUAGE_CODE = LANGUAGE_CODE
gettext = lambda s: s
LANGUAGES = (
('ar', gettext('Arabic')),
('en', gettext('English')),
)
USE_I18N = True
MIDDLEWARE_CLASSES = (
'django.contrib.sessions.middleware.SessionMiddleware',
'django.middleware.locale.LocaleMiddleware',
'django.middleware.common.CommonMiddleware',
'django.middleware.csrf.CsrfViewMiddleware',
'django.contrib.auth.middleware.AuthenticationMiddleware',
'django.contrib.messages.middleware.MessageMiddleware',
'apps.accounts.middleware.AccountSocialAuthExceptionMiddleware',
'apps.core.middleware.DefaultSiteMiddleware',
# 'debug_toolbar.middleware.DebugToolbarMiddleware',
# Uncomment the next line for simple clickjacking protection:
# 'django.middleware.clickjacking.XFrameOptionsMiddleware',
)
TEMPLATE_CONTEXT_PROCESSORS = (
"django.contrib.auth.context_processors.auth",
"django.core.context_processors.debug",
"django.core.context_processors.request",
"django.core.context_processors.i18n",
"django.core.context_processors.media",
"django.core.context_processors.static",
"django.core.context_processors.tz",
"django.contrib.messages.context_processors.messages",
# custom context processors
"snowflake.apps.core.context_processors.project_name",
"snowflake.apps.core.context_processors.sites",
)
这是我的网址
from django.conf import settings
from django.conf.urls import patterns, include, url
from django.conf.urls.i18n import i18n_patterns
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
urlpatterns = patterns('',
url(r'^admin/', include('apps.mod.urls')),
url(r'^redirect/login/$', 'apps.core.views.redirect_to_accounts', name='login-redirect'),
)
urlpatterns += i18n_patterns('',
url(r'^$', 'apps.publication.views.index', name='home'),
url(r'^publication(s)?/', include('apps.publication.urls')),
url(r'^comment(s)?/', include('apps.comment.urls')),
url(r'^rating(s)?/', include('apps.rating.urls')),
url(r'^tag(s)?/', include('apps.taggable.urls')),
url(r'^search/', include('apps.search.urls')),
url(r'^guard(s)?/', include('apps.guard.urls')),
)
handler404 = 'apps.core.views.handler404'
P.S am am django 1.5
答案 0 :(得分:10)
您的观点apps.core.view.handler404
会返回HttpResponse
个对象...但其HTTP状态代码为200.
response = render_to_response("...")
response.status_code = 404
return response
应该解决你的问题(以前有过)。