>
> Not sure the below is a problem with routes or not.
> I have a feed() function. If I browse to feed.rss it works fine. But I
> would like /feed/ as the url (without.rss). How to achieve it?
>
You could do that with the pattern-based rewrite system, but then you
wouldn't be able to use the parameter-based system as you are now. As a
simple alternative, you could do:
def feed():
response.view = 'default/feed.rss' # or 'generic.rss' if you don't
have a 'default/feed.rss' view
That assumes you always want rss for the feed() function -- otherwise,
you'll need to add the appropriate logic. See info about response.view
here: http://web2py.com/books/default/chapter/29/4#response.
Anthony
