On Jul 22, 2:17 am, Wyatt Baldwin <[email protected]> wrote:
> 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__.

Hi Wyatt,

Thanks for the pointer. It does appear redirecting in __call__ does
not work.
I have moved all my logic which was in __call__ to __before__ and
redirect works in there.

It would be nice to be able to return a response in __before__ as well
rather then only be allowed to raise an exception.

Cheers and thanks again.

Huy

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