On Sep 23, 5:24 pm, IanSR <ijsto...@crystal.harvard.edu> wrote: > Is there some trick to parse > > http://www.example.com/foo/42 > > such that "42" becomes an integer parameter to the view function, > instead of a string? > > Right now I have to do: > > from my.app.views import foo > > urlpatterns = patterns('', > (r'foo/(?P<bar_str>\d+)/?$' , foo), > ) > > my/app/views.py > > def foo(request, bar_str): > bar = int(bar_str) > > This is obviously not so convenient. It would be much nicer to simply > have the pattern auto-converted to an integer. I understand the > Python reasons for why it works this way, but if there were some > Django shortcut to help with this, that would be great. > > Ian
There's no built-in way. If you really need this, the best way might be to write a decorator that converts the arguments and passes them on to the wrapped function. I doubt it'll end up much less verbose than just calling int(), though. -- DR. --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Django users" group. To post to this group, send email to django-users@googlegroups.com To unsubscribe from this group, send email to django-users+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/django-users?hl=en -~----------~----~----~----~------~----~------~--~---
