>
> I have followed what both the book and sample files say can be done in
> routes.py:
>
> default_application = 'try'
> default_controller = 'default'
> default_function = 'index'
>
> This simply doesn't work. Entering www.mydomain.com as the target
> url in the browser routes to the wel
The answer appears to be: yes, you must restart web2py.
That means probably the sample in my first post would also work.
So, a question:
Why do the dictionaries need to be nested? In other words why is the
key "BASE" necessary?
My rant on the manual still stands. Sorry to be disagreeable. Lot
On Feb 11, 2012, at 12:51 PM, Lewis wrote:
> The answer appears to be: yes, you must restart web2py.
>
> That means probably the sample in my first post would also work.
>
> So, a question:
>
> Why do the dictionaries need to be nested? In other words why is the
> key "BASE" necessary?
Becau
The answer appears to be: yes, you must restart web2py.
That means probably the sample in my first post would also work.
So, a question:
Why do the dictionaries need to be nested? In other words why is the
key "BASE" necessary?
My rant on the manual still stands. Sorry to be disagreeable. L
On Feb 11, 2012, at 12:47 PM, Lewis wrote:
> This doesn't work either:
>
> routers = dict(
>BASE = dict(
>default_application = 'try',
>),
> )
>
> I placed the above as routes.py in the web2py folder (of /var as it
> happens). All applications run when explicitly referenced as
>
This doesn't work either:
routers = dict(
BASE = dict(
default_application = 'try',
),
)
I placed the above as routes.py in the web2py folder (of /var as it
happens). All applications run when explicitly referenced as
www.mydomain.com/try.
Do I need to restart web2py?
On Feb 1
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