Problem
The quixote.errors.AccessError exception results in a HTTP 403 Forbidden response. This response is desirable when using FormBasedAuthentication but should only be used after responding with a 401 Unauthorized when using HTTP's native authentication mechanisms.
Solution
Create a variation on AccessError that results in a 401 Unauthorized response instead of a 403 Forbidden response:
from quixote import get_response from quixote.errors import AccessError class UnauthorizedError(AccessError): """The request requires user authentication. This subclass of AccessError sends a 401 instead of a 403, hinting that the client should try again with authentication. """ status_code = 401 title = "Unauthorized" description = "You are not authorized to access this resource." def __init__(self, realm='Protected', public_msg=None, private_msg=None): self.realm = realm AccessError.__init__(self, public_msg, private_msg) def format(self): get_response().set_header('WWW-Authenticate', 'Basic realm="%s"' % self.realm) return AccessError.format(self)
When an UnauthorizedError is raised, a 401 Unauthorized response is sent with the required WWW-Authenticate headers. You should be able to use this anywhere AccessError may be used as follows:
def ensure_access(): r = get_request() ha = r.get_environ('HTTP_AUTHORIZATION',None) if ha: auth_type, auth_string = ha.split() login, passwd = b64decode(auth_string).split(':') ...process authentication... else: raise UnauthorizedError
Most browsers pop up a little username/password dialog containing the name of the Realm specified.
Discussion
Some info/code on handling Basic and Digest Authentication headers would be super useful too. -- RyanTomayko 2004-09-28 00:22:28
Links
RFC 2617 - HTTP Authentication: Basic and Digest Access Authentication
http://www.ietf.org/rfc/rfc2617.txtRFC 2616 - Hypertext Transfer Protocol -- HTTP/1.1
http://www.w3.org/Protocols/rfc2616/rfc2616.html