Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 8711113
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 13, 20262026-06-13T04:47:36+00:00 2026-06-13T04:47:36+00:00

I am creating REST based APIs for an app using Tastypie with Django. The

  • 0

I am creating REST based APIs for an app using Tastypie with Django. The problem is default API url in Tastypie contains version info in url patterns i.e.

http://lx:3001/api/v1/vservers/?username=someuser&api_key=someapikey

I want my url to be free from API version info like this:

http://lx:3001/api/vservers/?username=someuser&api_key=someapikey

urls.py

v1_api = Api()
v1_api.api_name = ''
v1_api.register(UserResource())
...
url(r'^api/', include(v1_api.urls)),

I am overwriting api_name with an empty string still

http://lx:3001/api/vservers/?username=someuser&api_key=someapikey does not work.

How can I get rid of the version info altogether?

Thanks..

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-06-13T04:47:37+00:00Added an answer on June 13, 2026 at 4:47 am

    Subclass Api and override urls to remove all the api_name-related bits:

    class MyApi(Api):
        @property
        def urls(self):
            """
            Provides URLconf details for the ``Api`` and all registered
            ``Resources`` beneath it.
            """
            pattern_list = [
                url(r"^%s$" % trailing_slash(), self.wrap_view('top_level'), name="api_top_level"),
            ]
    
            for name in sorted(self._registry.keys()):
                pattern_list.append((r"^/", include(self._registry[name].urls)))
    
            urlpatterns = self.override_urls() + patterns('',
                *pattern_list
            )
            return urlpatterns
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

I'm creating REST service using ASP.NET Web API. How can I return 401 status
I am creating a REST Web Service using Java and Jersey API. The basic
We are creating an API that suits the benefits of a REST based architecture.
I'm using the new django-rest-framework 2.0 and have been following the tutorial for creating
Background: I am creating a REST api, that will require users to use only
How does one go about creating a c# wrapper for rest based service. I
I have been exploring OAuth version 1.0 for the REST API I am currently
I'm creating a web based service that I want to expose as a REST
I'm writing a Drupal module to integrate with a custom Java-based REST API for
I'm creating REST application based on spring and apache tiles. I've added .css file

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.