Actually, I want to customize the code to get better feedback, and after
digging in I got this discussion
<https://groups.google.com/forum/#!searchin/web2py/not$20authorized$20message|sort:date/web2py/k-FnX_X6CcI/fqgUpRbwAwAJ>
I just don't know where to exactly place this code, or get something better:
auth.messages.ajax_failed_authentication = DIV(H4(T('Your session has
expired')),
T('Please '),
A(T('login'),
_href=auth.settings.login_url +
('?_next=' +
urllib.quote(request.env.http_web2py_component_location))
if
request.env.http_web2py_component_location else ''),
T(' again to view this content.'),
_class='not-authorized alert alert-block')
On Tue, May 5, 2020 at 11:09 PM Jim S <[email protected]> wrote:
> Can you show some code?
>
> -Jim
>
> On Tuesday, May 5, 2020 at 1:38:29 PM UTC-5, Maurice Waka wrote:
>>
>> I notice this message after a period of inactivity. See attached.
>> The problem is that on clicking on the login link on the app on
>> pythonanywhere, I get an error: invalid view (default/user/load).
>> How can I resolve this?
>> Regards
>>
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- http://github.com/web2py/web2py (Source code)
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