no, it is not. The first time the default/user is called - in the login process for example - there not a HTTP(303) (redirect) exception, and the user get the form. The second time, in this scenery, the call to auth() will cause a redirect (HTTP exception) and the user is effectively logged into the system.
The same goes for the logout, but there is not a form in the middle. El lun., 17 jun. 2019 a las 9:57, Vlad (<[email protected]>) escribió: > but isn't such a scenario catching a user in a process of logging in or > logging out, but before actual login/logout? > > On Monday, June 17, 2019 at 9:44:10 AM UTC-4, Yoel Benitez Fonseca wrote: >> >> I think you can do it in the default/user function. Having in account >> that the call to auth() probably raises a HTTP exceptiong in case of a >> success. This example is only a proof of concept: >> >> def user(): >> form = CAT() >> >> try: >> form = auth() >> except HTTP as e: >> if e.status == 303'login' in request.args: >> if ('login' in request.args) and auth.user: >> # probably a success, u can test if auth.user is o not None >> do_your_loging_event_here() >> elif 'logout' if request.args: >> do_your_logout_event_here() >> >> return dict(form=form) >> >> >> >> El lun., 17 jun. 2019 a las 9:19, Vlad (<[email protected]>) escribió: >> >>> What would be the simplest way to intercept login and logout? >>> (i.e. I'd like to perform some additional tasks when a user is logged >>> in, and when a user is logged out.) >>> web2py elegantly hides the functionality in default.user under >>> form=auth() and {{=form}} and somehow I can't think of a simple way to be >>> notified about successful login and logout actions... >>> >>> -- >>> Resources: >>> - http://web2py.com >>> - http://web2py.com/book (Documentation) >>> - http://github.com/web2py/web2py (Source code) >>> - https://code.google.com/p/web2py/issues/list (Report Issues) >>> --- >>> You received this message because you are subscribed to the Google >>> Groups "web2py-users" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to [email protected]. >>> To view this discussion on the web visit >>> https://groups.google.com/d/msgid/web2py/b9dbb7bb-4500-48b2-82fd-cd8ea644f1fe%40googlegroups.com >>> <https://groups.google.com/d/msgid/web2py/b9dbb7bb-4500-48b2-82fd-cd8ea644f1fe%40googlegroups.com?utm_medium=email&utm_source=footer> >>> . >>> For more options, visit https://groups.google.com/d/optout. >>> >> >> >> -- >> Msc. Yoel Benítez Fonseca >> >> >> >> >> >> >> -- > Resources: > - http://web2py.com > - http://web2py.com/book (Documentation) > - http://github.com/web2py/web2py (Source code) > - https://code.google.com/p/web2py/issues/list (Report Issues) > --- > You received this message because you are subscribed to the Google Groups > "web2py-users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > To view this discussion on the web visit > https://groups.google.com/d/msgid/web2py/ee8abe78-346c-44dd-b8a6-e9b243e4eb90%40googlegroups.com > <https://groups.google.com/d/msgid/web2py/ee8abe78-346c-44dd-b8a6-e9b243e4eb90%40googlegroups.com?utm_medium=email&utm_source=footer> > . > For more options, visit https://groups.google.com/d/optout. > -- Msc. Yoel Benítez Fonseca -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/web2py/CAFDb1rUYzrXURyuVquS1n3Mgk%2B4T-2aVd%3DkV9c4cTveOigAh_A%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
