Auth::guard('user')->user() 在 Laravel 5.3 中为空

Question

我正在尝试像这样将当前用户对象传递给所有模板：

class Controller extends BaseController
{
    public function __construct()
    {
        view()->share('usr', Auth::guard('user'));
    }
}

每个控制器都由控制器扩展。但是，如果我尝试转储 Auth::guard('user')->user() Laravel 返回 null，尽管我已登录。此外，当我将此变量传递给模板时，{{ $usr->user() }} 返回当前用户。我做错了什么？

我的配置/auth.php

'defaults' => [
    'guard' => 'user',
    'passwords' => 'users',
],
'guards' => [
        'user' => [
            'driver' => 'session',
            'provider' => 'user',
        ],
    ],
'providers' => [
        'user' => [
            'driver' => 'eloquent',
            'model' => App\User::class,
        ],
    ],

内核.php

         protected $middleware = [
     \Illuminate\Foundation\Http\Middleware\CheckForMaintenanceMode::class,
 ];

/**
 * The application's route middleware groups.
 *
 * @var array
 */
protected $middlewareGroups = [
    'web' => [
        \App\Http\Middleware\EncryptCookies::class,
        \Illuminate\Cookie\Middleware\AddQueuedCookiesToResponse::class,
        \Illuminate\Session\Middleware\StartSession::class,
        \Illuminate\View\Middleware\ShareErrorsFromSession::class,
        //\App\Http\Middleware\VerifyCsrfToken::class,
        \Illuminate\Routing\Middleware\SubstituteBindings::class,
    ],

    'api' => [
        'throttle:60,1',
        'bindings',
    ],
];

/**
 * The application's route middleware.
 *
 * These middleware may be assigned to groups or used individually.
 *
 * @var array
 */
protected $routeMiddleware = [
    'auth' => \Illuminate\Auth\Middleware\Authenticate::class,
    'auth.basic' => \Illuminate\Auth\Middleware\AuthenticateWithBasicAuth::class,
    'bindings' => \Illuminate\Routing\Middleware\SubstituteBindings::class,
    'can' => \Illuminate\Auth\Middleware\Authorize::class,
    'guest' => \App\Http\Middleware\RedirectIfAuthenticated::class,
    'throttle' => \Illuminate\Routing\Middleware\ThrottleRequests::class,
];

我自己的登录功能：

public function authorizes(Request $request)
{
        $this->validate($request, [
        'login' => 'required|max:50',
        'password' => 'required|max:50'
    ]);

        $credentials = $request->only(['login', 'password' ]);
        $remember = $request->get('remember', false) == 1 ? true : false;

    if ($this->guard->attempt( $credentials, $remember)) {
        $user = $this->guard->user();
        $user->last_login = date('Y-m-d H:i:s');
        $user->save();
        return redirect()->route( 'homepage' )->withSuccess(trans('app.login.success'));
    }
    return redirect()->back()->withErrors(trans('app.wrong.credentials'));
}

Answer 1

尝试这样做：

view()->share('usr', Auth::user());

或者这个：

view()->share('usr', auth()->user());

Answer 2

首先检查用户是否登录，然后将当前用户共享为：

public function __construct()
{
    if (auth()->check()) {
        $this->currentUser = auth()->user();
        View::share('currentUser', $this->currentUser);
    } else {
      // you can redirect the user to login page.
    }
}

Answer 3

在你的情况下，有两件事需要考虑

1. 要获得实际的用户模型，你应该这样做Auth::guard('user')->user()
2. Auth::user() 在调用view()->share() 时实际上还没有初始化，见https://github.com/laravel/framework/issues/6130

因此，您宁愿使用视图作曲家。在AppServiceProvider 的boot 方法中添加：

\View::composer('*', function ($view) {
    $view->with('usr', \Auth::guard('user')->user());
});

Answer 4

在 Laravel 5.3 中，您应该像这样更改控制器构造函数以使其正常工作（假设您至少使用 Laravel 5.3.4）：

public function __construct()
{
    $this->middleware(function ($request, $next) {
        view()->share('usr', Auth::guard('user'));

        return $next($request);
    });
}

您可以在Upgrade guide 中看到此更改：

在以前的 Laravel 版本中，您可以访问会话变量或 控制器构造函数中经过身份验证的用户。这是 从未打算成为框架的明确功能。在 Laravel 中 5.3，由于中间件尚未运行，您无法访问控制器构造函数中的会话或经过身份验证的用户。

作为替代方案，您可以直接定义基于闭包的中间件 在控制器的构造函数中。在使用此功能之前，请确保 你的应用程序正在运行 Laravel 5.3.4