On Jul 21, 5:56 am, huydo <[email protected]> wrote: > Hi, > > Just trying to do a direct from the __call__ method in my base.py file > and this is the following exception which I am getting. > > redirect(url(controller='redirect', action='index')) > File 'c:\\apps\\xbits\\ktrack\\python\\lib\\site-packages\\pylons-1.0- > py2.5.egg\\pylons\\controllers\\util.py', line 208 in redirect > raise exc(location=url).exception > HTTPFound: 302 Found > Content-Type: text/html; charset=UTF-8 > Content-Length: 0 > Location: /redirect/index > > Adding the following line to my middleware.py file (just after the > #CUSTOM MIDDLE HERE) fixed it. > > app = httpexceptions.make_middleware(app) > > Is this a bug ? or am I doing something wrong ?
I'd say it's more that you're doing something wrong. ;) redirect works by raising an exception (the one you're seeing). Normally, this would be caught by the base WSGIController class (in _inspect_call) and converted into an HTTP response, but because __call__ is a setup function and not part of the "action context", the exception isn't caught, presumably by design. In other words, you should only use redirect from within the "action context"--either in __before__, a particular action method, or __after__. If you really want to redirect from __call__, you can create an instance of HTTPFound and return that from __call__. -- You received this message because you are subscribed to the Google Groups "pylons-discuss" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/pylons-discuss?hl=en.
