django捕获任何URL? [英] django catching any url?

问题描述

当用户要求输入不需要的URL时，我试图找到一种显示django登录页面的方法吗?我应该使用哪种语法?截止到今天，我有

I am trying to find a way to display my django login page when the user ask for an unwanted url? Which syntax should I user ? As of today I have

from django.conf.urls import patterns, include, url

从django.contrib导入管理员

from django.contrib import admin

urlpatterns = pattern(''，#示例:

urlpatterns = patterns('', # Examples:

url(r'^login'           , 'database.views.index', name='login'),
url(r'^create-user/'    ,  'database.views.account_creation', name='create_user'),

url(r'^get-details/'    ,  'database.views.get_details', name='get-details'),
url(r'^upload-csv'  ,  'database.views.upload_csv', name='upload_csv'),
# url(r'^blog/', include('blog.urls')),

url(r'^admin/', include(admin.site.urls)),
#url(r'^'           , 'database.views.index', name='login'),

)

我希望如果用户要求一个疯狂的URL，它将被定向到登录URL(即view.index函数).有什么主意吗?

I would like that if a user ask for a crazy url, it would be directed to the login url (ie view.index function). Any idea ?

推荐答案

在不评论您是否应该执行此操作的情况下，Django会尝试按顺序匹配您的url模式.因此，如果您想要一个全面/全面的处理程序，请放在最后:

Without commenting on whether you should do this, Django will attempt to match your url patterns in order. So if you want a fall-through / catch-all handler, put this last:

url(r'^.*', 'database.views.index', name='unmatched')

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