Is there a way to get the route file to work without having to manually define all optional params as separate routes?
For example I have a logout route that normally is just "/logout" but I have an optional param to define a code in the url that will give the user a message like your were logged out due to inactivity was the user logging themselves out.
I would like just this route to take care of it:
PHP Code:
$routes->get('logout/(:any)', 'Login::logout/$1');
However the route does not work if you do not include the "/(:any)" in the actual url. The only way I have been able to make it work is using two routes:
PHP Code:
$routes->get('logout/(:any)', 'Login::logout/$1');
$routes->get('logout', 'Login::logout');
Since my controller has the parameter defined with a default why do I have to setup 2 routes? I am running into this situation on other pages too where I have multiple optional params and they all have defaults in the controller but I have to setup a route for every different variation of total number of params being passed in which is a little annoying and seems overkill. Maybe by default it wouldn't work this way but I don't see an option or anything in the routes to enable a specific url to work this way. Even if I had to do something like this would be okay:
PHP Code:
$routes->get('logout', 'Login::logout', ['optional_params'=>true,'param_count'=>5,'param_types'=>['(:any)']]);
Something like this would be nice you could setup to accept up to X params and provide specific param types for each one or if none or only 1 in the array set them all to the same like (:any)
Instead of
PHP Code:
$routes->get('logout', 'Login::logout')
$routes->get('logout/(:any)', 'Login::logout/$1')
$routes->get('logout/(:any)/(:any)', 'Login::logout/$1/$2')
$routes->get('logout/(:any)/(:any)/(:any)', 'Login::logout/$1/$2/$3')
$routes->get('logout/(:any)/(:any)/(:any)/(:any)', 'Login::logout/$1/$2/$3/$4')
$routes->get('logout/(:any)/(:any)/(:any)/(:any)/(:any)', 'Login::logout/$1/$2/$3/$4/$5')
