Checking which `guard` is loggedin

问题

I have a multiauth laravel 5.2 app, with the fallowing guards defined on config/auth.php:

...
'admin' => [
    'driver' => 'session',
    'provider' => 'admin',
],
'user' => [
    'driver' => 'session',
    'provider' => 'user',
],
...

So, admin and user.

The problem resides in the view layer, since this two loggedin guards share some views, ex:

Hello {{Auth::guard('admin')->user()->name}}

In this case the guard is hardcoded into the view to be always admin (it gives error when loggedin guard is user), but, to avoid have to do another equal view just for this little change, I would like have it dinamic, something like:

Hello {{Auth::guard(<LOGGEDIN GUARD>)->user()->name}}

PS: I know that this could be achieved getting the corresponding url segment, ex: www.site.com/pt/user/dasboard which in the case it would be segment 2, but this way the app would lose scalability, since in the future the corresponding segment may not be the same (2 in the example above)

回答1:

One way to do this is to extend the Laravel authentication class in the IoC container to include, for instance, a name() method that check which guard is used for the current session, and calls user() on that Guard instance.

Another way is to simply use an if-statement in your Blade template:

@if(Auth::guard('admin')->check())
    Hello {{Auth::guard('admin')->user()->name}}
@elseif(Auth::guard('user')->check())
    Hello {{Auth::guard('user')->user()->name}}
@endif

However, this is a little dirty. You can clean this up a bit by using a partial, or by passing the view a variable containing the guard name, either directly from your Controller, or via a ViewComposer, and then doing:

Hello {{Auth::guard($guardName)->user()->name}}

in your View.

Extending Laravel's authentication is your best option, imo.

回答2:

Since Laravel 5.5, this is easy to do with the @auth template directive.

@auth("user")
    You're a user!
@endauth

@auth("admin")
    You're an administrator!
@endauth

@guest
    You're not logged in!
@endguest

Reference: https://laravel.com/docs/5.6/blade#if-statements

回答3:

I recommend to use global helper function like

function activeGuard(){

foreach(array_keys(config('auth.guards')) as $guard){

    if(auth()->guard($guard)->check()) return $guard;

}
return null;
}

回答4:

This will get the guard name that is used for current logged in user

Auth::getDefaultDriver()

When you log in, by default it will get you the:

'web'

Dependable through which guard you've been logged in it will get you that guard name.

This is not applicable for APIs!!! Because API in laravel by default donesn't use session.

回答5:

Depends of Harat answer, I built a Class names CustomAuth, and it give me easy access to Auth facade methods: user() and id().

<?php

namespace App\Utils;

use Illuminate\Support\Facades\Auth;

class CustomAuth{
    static public function user(){
        return Auth::guard(static::activeGuard())->user() ?: null;
    }

    static public function id(){
        return static::user()->MID ?: null;
    }

    static private function activeGuard(){

        foreach(array_keys(config('auth.guards')) as $guard){

            if(auth()->guard($guard)->check()) return $guard;

        }
        return null;
    }
}

来源：https://stackoverflow.com/questions/38999725/checking-which-guard-is-loggedin