thank you for trying it. It is one of those features that not many people are using but should be solid.
Massimo On Apr 5, 9:44 pm, Alexei Vinidiktov <alexei.vinidik...@gmail.com> wrote: > Thanks a lot, Massimo! It worked! > > > > On Tue, Apr 6, 2010 at 10:36 AM, mdipierro <mdipie...@cs.depaul.edu> wrote: > > you need in db.py > > > auth.settings.allow_basic_login = True > > > it defaults to False > > > On Apr 5, 9:21 pm, Alexei Vinidiktov <alexei.vinidik...@gmail.com> > > wrote: > >> Hello, > > >> I'm trying to make a desktop client written in Python 2.5 (on Windows) > >> communicate with a web2py 1.76.5 based app using services. It works > >> fine with services that don't require authentication, but I was > >> wondering if I could make it authenticate the user with Basic > >> authentication. > > >> I've read in the Python docs that xmlrpclib supports Basic authentication: > > >> " > >> Both the HTTP and HTTPS transports support the URL syntax extension > >> for HTTP Basic Authentication:http://user:p...@host:port/path. The > >> user:pass portion will be base64-encoded as an HTTP `Authorization' > >> header, and sent to the remote server as part of the connection > >> process when invoking an XML-RPC method. > >> " > > >> In my modified Welcome web2py app I've added the setting > >> auth.settings.allow_basic_authentication = True in > >> welcome/models/db.py > > >> I've decorated the private_call action like so: > > >> @auth.requires_login() > >> def private_call(): > >> return private_service() > > >> And I've added a test action getstring: > > >> @private_service.xmlrpc > >> def getstring(): > >> if auth.is_logged_in(): > >> return "logged in!" > >> else: > >> return "not logged in!" > > >> In my desktop app I'm calling the service like this: > > >> self.server = > >> xmlrpclib.Server("http://myem...@gmail.com:mypassw...@127.0.0.1:8000/welcome/default/private_call/xmlrpc";) > >> print self.server.getstring() > > >> and I'm getting this traceback: > > >> Traceback (most recent call last): > >> File "C:\Users\Alexei\Documents\wxPython\Boa\xml_rpc_test\Frame1.py", > >> line 137 > >> , in OnButton4Button > >> print self.server3.getstring() > >> File "C:\Python25\lib\xmlrpclib.py", line 1147, in __call__ > >> return self.__send(self.__name, args) > >> File "C:\Python25\lib\xmlrpclib.py", line 1437, in __request > >> verbose=self.__verbose > >> File "C:\Users\Alexei\Documents\wxPython\Boa\xml_rpc_test\Frame1.py", > >> line 185 > >> , in request > >> headers > >> xmlrpclib.ProtocolError: <ProtocolError for myem...@gmail.com:mypassword@ > >> 127.0.0.1:8000/welcome/default/private_call/xmlrpc: 303 SEE OTHER> > > >> I'm obviously doing something wrong. Could you point me in the right > >> direction? > > >> Thanks. > > >> -- > >> Alexei Vinidiktov > > > -- > > You received this message because you are subscribed to the Google Groups > > "web2py-users" group. > > To post to this group, send email to web...@googlegroups.com. > > To unsubscribe from this group, send email to > > web2py+unsubscr...@googlegroups.com. > > For more options, visit this group > > athttp://groups.google.com/group/web2py?hl=en. > > -- > Alexei Vinidiktov -- You received this message because you are subscribed to the Google Groups "web2py-users" group. To post to this group, send email to web...@googlegroups.com. To unsubscribe from this group, send email to web2py+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/web2py?hl=en.
