CODEWITHSUNDEEP
djangourl
Romain Jouin
Romain Jouin

Reputation: 4838

django catching any url?

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

from django.contrib import admin

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'),

)

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 ?

Upvotes: 3

Views: 3838

Answers (1)

dylrei
dylrei

Reputation: 1738

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')

Upvotes: 7

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