You need to customize. auth.settings.login_url = URL(r=request,f='usuraio', args='login')
The name of this function sounds bad in italian. LOL Massimo On Apr 10, 3:06 am, Álvaro Justen [Turicas] <alvarojus...@gmail.com> wrote: > Hello web2pyers and Massimo, > I'm testing Auth and all work fine when I use a function called "user" > to return auth(). > Well, I speak Portuguese and want to call it "usuario". My > controllers' code is like to: > > def usuario(): > return dict(meio=auth()) > > In my app I want to only permit CRUD if user is logged in, so I have: > > @auth.requires_login() > def dados(): > return dict(meio=crud()) > > But when I'm not logged in and try to > openhttp://localhost:8000/myapp/default/dados/web2py redirects me > tohttp://localhost:8000/myapp/default/user/login- that returns me to a > 'invalid function' page. > > In fact, I can solve this problem creating an entry in routes.py, but > I think this isn't the right way... > > Oh, with CRUD URLs I doesn't had problems - I've not tested so much, > although CRUD doesn't have to "guess" the function like auth. > > -- > Álvaro Justen > Peta5 - Telecomunicações e Software Livre > 21 3021-6001 / 9898-0141 > http://www.peta5.com.br/ --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "web2py Web Framework" group. To post to this group, send email to web2py@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 -~----------~----~----~----~------~----~------~--~---
