主要目的就是结合中间件，如果用户没有登录打开后台，实现跳转到指定页面。
1： php artisan make:middleware AdminLogin
这样Http下面的Middleware会生成AdmnLogin.php
2：里面写handle方法如下，这个结合自己的写，指定跳转到页面
3：在Http下面的Kernel.php里的protected $routeMiddleware加上路由中间件
个人分类： LaraverYou are here
Laravel 5.4 5.5 定义中间件，没有登录访问后台跳转到指定页面 未登录跳转 laravel 5.5 5.4 有大用 有大大用
星期三, 2018-05-16 09:15 — adminshiping1
来自 https://blog.csdn.net/myarche/article/details/79738417
protected function unauthenticated($request, AuthenticationException $exception)
{
if ($request->expectsJson()) {
return response()->json(['error' => 'Unauthenticated.'], 401);
}
return redirect()->guest(route('login'));
}但是Laravel 5.5这里已经移到了这个位置，vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php所以现在如何更改？我不想更改供应商目录中的内容，以防它被作曲家更新所覆盖。
protected function unauthenticated($request, AuthenticationException $exception)
{
return $request->expectsJson()
? response()->json(['message' => 'Unauthenticated.'], 401)
: redirect()->guest(route('login'));
}8个答案 正确答案
58
但是在Laravel 5.5中，此位置已移至该位置vendor / laravel / framework / src / Illuminate / Foundation / Exceptions / Handler.php，所以现在如何更改它？我不想更改供应商目录中的内容，以防它被作曲家更新所覆盖。
只是默认情况下不再存在该功能。
您可以像在5.4中一样重写它。只要确保包括
use Exception;
use Request;
use Illuminate\Auth\AuthenticationException;
use Response;在处理程序文件中。
例如我app/Exceptions/Handler.php看起来像这样：
<?php
namespace App\Exceptions;
use Exception;
use Request;
use Illuminate\Auth\AuthenticationException;
use Response;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
class Handler extends ExceptionHandler
{
(...) // The dfault file content
/**
* Convert an authentication exception into a response.
*
* @param \Illuminate\Http\Request $request
* @param \Illuminate\Auth\AuthenticationException $exception
* @return \Illuminate\Http\Response
*/
protected function unauthenticated($request, AuthenticationException $exception)
{
return $request->expectsJson()
? response()->json(['message' => 'Unauthenticated.'], 401)
: redirect()->guest(route('authentication.index'));
}
}- 1个
- 3
if (in_array('api', \Route::getCurrentRoute()->computedMiddleware)) { return response()->json(['error' => 'Unauthenticated.'], 401); }- paing 在20:18 1月27日'18 - 非常感谢，但是我必须交换线路响应并彼此重定向，因为我有一个api，其中有一个注销方法，所以在邮递员中，当我填写标题值（如授权）后注销用户时， =承载加上登录生成的访问令牌，第一次命中它注销了用户，但是如果我再次单击，它说没有消息，现在被json消息替换：未经身份 验证 – shashi verma 18年11月20日，5：14
20
这是我解决的方法。在渲染函数中，我捕获了异常类。如果是Authentication异常类，我会写我的代码以进行重定向（我会在以前版本的未经身份验证的函数中编写的代码）。
public function render($request, Exception $exception)
{
$class = get_class($exception);
switch($class) {
case 'Illuminate\Auth\AuthenticationException':
$guard = array_get($exception->guards(), 0);
switch ($guard) {
case 'admin':
$login = 'admin.login';
break;
default:
$login = 'login';
break;
}
return redirect()->route($login);
}
return parent::render($request, $exception);
}- 2谢谢它为我工作。我在youtube中使用了DevMarketer MultiAuth教程，但似乎在laravel 5.5中不起作用。这解决了我的问题。 – 小熊damayo 17年9月10日在7:02
5
但是在Laravel 5.5中，此位置已移至该位置vendor / laravel / framework / src / Illuminate / Foundation / Exceptions / Handler.php，所以现在如何更改它？我不想更改供应商目录中的内容，以防它被作曲家更新所覆盖。
我们只需要使用Illuminate \ Auth \ AuthenticationException即可；
然后像laravel 5.4一样工作
2
标准异常处理程序使用命名路由。
因此，您只需定义使用该名称的路线即可。
因此，在routes/web.php文件中，只需添加以下行：
Route::get('mylogin', 'MyLoginController@showLoginForm')->name('login');该name('login')位为该路由命名，因此未经身份验证的异常将使用此路由。
您无需费心制作自己的异常处理程序或修改标准异常处理程序。
可vendor/laravel/framework/src/Illuminate/Routing/Router.php在auth()函数中的文件中找到样板“ auth”代码使用的命名路由。（登录，注销，注册，password.request，password.email和password.reset）。使用Route::auth();路由文件中的行时，将添加这些路由。
0
只需在路由文件中添加用于登录的路由即可：
Route::get('/login', [
'uses' => 'UserController@getSignin',
'as' => 'login'
]);-1
除了之外overriding，您还可以直接对位于处的未经身份验证Handler.php的现有功能进行更改，以根据防护重定向到预期的路由。vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php
/**
* Convert an authentication exception into a response.
*
* @param \Illuminate\Http\Request $request
* @param \Illuminate\Auth\AuthenticationException $exception
* @return \Illuminate\Http\Response
*/
protected function unauthenticated($request, AuthenticationException $exception)
{
$guard = array_get($exception->guards(),0);
switch ($guard) {
case 'admin':
return $request->expectsJson()
? response()->json(['message' => $exception->getMessage()], 401)
: redirect()->guest(route('admin.login'));
break;
default:
return $request->expectsJson()
? response()->json(['message' => $exception->getMessage()], 401)
: redirect()->guest(route('login'));
break;
}
}-1
将您的app \ Exceptions \ Handler.php代码替换为以下内容。...
<?php
namespace App\Exceptions;
use Exception;
use Illuminate\Auth\AuthenticationException;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
class Handler extends ExceptionHandler
{
/**
* A list of the exception types that are not reported.
*
* @var array
*/
protected $dontReport = [
//
];
/**
* A list of the inputs that are never flashed for validation exceptions.
*
* @var array
*/
protected $dontFlash = [
'password',
'password_confirmation',
];
/**
* Report or log an exception.`enter code here`
*
* @param \Exception $exception
* @return void
*/
public function report(Exception $exception)
{
parent::report($exception);
}
/**
* Render an exception into an HTTP response.
*
* @param \Illuminate\Http\Request $request
* @param \Exception $exception
* @return \Illuminate\Http\Response
*/
public function render($request, Exception $exception)
{
return parent::render($request, $exception);
}
protected function unauthenticated($request, AuthenticationException $exception)
{
if ($request->expectsJson()) {
return response()->json(['message' => $exception->getMessage()], 401);
}
$guard = array_get($exception->guards(),0);
switch ($guard) {
case 'admin':
$login = 'admin.login';
break;
default:
$login = 'login';
break;
}
return redirect()->guest(route($login));
}
}-14
1.转到vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php文件。
2.搜索方法名称为unauthenticated。
3.然后将重定向网址从更改 redirect()->guest(route('login'))为redirect()->guest(route('api/login')) //whatever you want。
如果响应是API服务，则可以将响应作为JSON返回。
- 4
- 1个
- 如果您正在运行composer update命令，则没有任何影响。在laravel中，他们在访问API或Routes服务时具有一定级别的安全性。 – Nadimuthu Sarangapani 18年7月25日在18:04
34
In Laravel < 5.5 I could change this file app/Exceptions/Handler to change the unauthenticated user redirect url:
protected function unauthenticated($request, AuthenticationException $exception)
{
if ($request->expectsJson()) {
return response()->json(['error' => 'Unauthenticated.'], 401);
}
return redirect()->guest(route('login'));
}But in Laravel 5.5 this has been moved to this location vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php so how can I change it now? I don't want to change stuff in the vendor directory encase it gets overridden by composer updates.
protected function unauthenticated($request, AuthenticationException $exception)
{
return $request->expectsJson()
? response()->json(['message' => 'Unauthenticated.'], 401)
: redirect()->guest(route('login'));
}
- So you can catch the Authentication Error and then redirect before it gets caught by laravels. Nice thanks – Rob Jul 27 '17 at 4:57
- @robertmylne is it working ? because this solution is just for reporting !! – Maraboc Jul 27 '17 at 7:48
- @Maraboc it's not really doing what I had hoped it do. I was really hoping there was a place to override it. – Rob Jul 28 '17 at 1:44
58
But in Laravel 5.5 this has been moved to this location vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php so how can I change it now? I don't want to change stuff in the vendor directory encase it gets overridden by composer updates.
It's just the case that the function is not there by default anymore.
You can just override it as you did in 5.4. Just make sure to include
use Exception;
use Request;
use Illuminate\Auth\AuthenticationException;
use Response;in the Handler file.
For Example my app/Exceptions/Handler.php looks somewhat like this:
<?php
namespace App\Exceptions;
use Exception;
use Request;
use Illuminate\Auth\AuthenticationException;
use Response;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
class Handler extends ExceptionHandler
{
(...) // The dfault file content
/**
* Convert an authentication exception into a response.
*
* @param \Illuminate\Http\Request $request
* @param \Illuminate\Auth\AuthenticationException $exception
* @return \Illuminate\Http\Response
*/
protected function unauthenticated($request, AuthenticationException $exception)
{
return $request->expectsJson()
? response()->json(['message' => 'Unauthenticated.'], 401)
: redirect()->guest(route('authentication.index'));
}
}
- 1Best answer I've found, thanks! Just one thing: is there a way instead of using expectsJson(), checking if the request was via the api route? – Dave Nottage Nov 2 '17 at 22:53
- you are awesome men, i was thinking on disable the web middleware but this solution is best. – Andres Felipe Jan 12 '18 at 13:07
- 3
if (in_array('api', \Route::getCurrentRoute()->computedMiddleware)) { return response()->json(['error' => 'Unauthenticated.'], 401); }– paing Jan 27 '18 at 20:18 - unauthenticated function in kernal.php is not present in version 5.7. – UJ India Nov 15 '18 at 13:01
- thank you so much, but I had to exchange the line response and redirect with each other, as I am having an api, in which i have a logout method, so in postman when i logged out the user after filling the header value like authorization = bearer plus access token generated by login, on first hit it logged out the user but if i clicked again, it says no message , and now its been replaced by json message : unauthenticated – shashi verma Nov 20 '18 at 5:14
20
Here's how I solved it. In render function I caught exception class. And in case if it's Authentication exception class I wrote my code for redirect (the code I would write in unauthenticated function in previous version).
public function render($request, Exception $exception)
{
$class = get_class($exception);
switch($class) {
case 'Illuminate\Auth\AuthenticationException':
$guard = array_get($exception->guards(), 0);
switch ($guard) {
case 'admin':
$login = 'admin.login';
break;
default:
$login = 'login';
break;
}
return redirect()->route($login);
}
return parent::render($request, $exception);
}
- 2thanks it worked for me. i used DevMarketer MultiAuth tutorial in youtube but it seems it's not working in laravel 5.5. this solve my problem. – winnie damayo Sep 10 '17 at 7:02
- worked with me 2, and i am suing devMarketer...world is too small – Kareem Nour Emam Nov 11 '17 at 0:31
- Working on this same issue. When you got this to work, did you use the stock Laravel 5.5 function for
unauthenticated? Or did you also include the version of theunauthenticatedfunction from 5.4? – cmpreshn Jul 4 '18 at 1:44 - Have tried this code, but
$exception->guards()always returns nothing for me, so I have nothing to test against. If you have any additional resources you could recommend for a solution, it would be appreciated. Thanks for your answer. – cmpreshn Jul 4 '18 at 1:55
5
But in Laravel 5.5 this has been moved to this location vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php so how can I change it now? I don't want to change stuff in the vendor directory encase it gets overridden by composer updates.
We need to just include the use Illuminate\Auth\AuthenticationException;
and then it works as in laravel 5.4
2
The standard exception handler uses a named route.
So, you just define your route to use that name.
So, in your routes/web.php file, just add the line:
Route::get('mylogin', 'MyLoginController@showLoginForm')->name('login');The name('login') bit gives this route a name, so the unauthenticated exception will use this route.
You don't need to bother messing around making your own exception handler, or modifying the standard exception handler.
The named routes used by the boilerplate 'auth' code can be found in the vendor/laravel/framework/src/Illuminate/Routing/Router.php file, in the auth() function. (login, logout, register, password.request, password.email and password.reset). These routes are added when you use the Route::auth(); line in the route file.
- Thanks for a simple solution to the problem, I don't know why nobody else has upvoted this, this solved the issue for me, and the complex answers actually didn't even work at all. – Ashley Kleynhans Jun 26 '18 at 20:57
0
Just add a route for login in routes file:
Route::get('/login', [
'uses' => 'UserController@getSignin',
'as' => 'login'
]);
-1
Other than overriding, you could directly make changes in Handler.php to the existing function unauthenticated located at vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php to redirect to intended route based on guards.
/**
* Convert an authentication exception into a response.
*
* @param \Illuminate\Http\Request $request
* @param \Illuminate\Auth\AuthenticationException $exception
* @return \Illuminate\Http\Response
*/
protected function unauthenticated($request, AuthenticationException $exception)
{
$guard = array_get($exception->guards(),0);
switch ($guard) {
case 'admin':
return $request->expectsJson()
? response()->json(['message' => $exception->getMessage()], 401)
: redirect()->guest(route('admin.login'));
break;
default:
return $request->expectsJson()
? response()->json(['message' => $exception->getMessage()], 401)
: redirect()->guest(route('login'));
break;
}
}
- I'm working on a multiple authentication system, and I've been trying to get this to work, but when this exception is thrown,
$exception->guards()is coming back as empty for me. Do you know of any reason why the guards would not be available? I'm working in Laravel 5.5.40, and have been trying to implement multiple guards for authenticating different users. I'm working off of the video series on 'Multi-Auth' by devmarketer. – cmpreshn Jul 4 '18 at 1:40 - making changes in the vendor folder directly is highly bad practice! you should never ever do this – Shehzad Jan 7 at 6:46
-1
Replace Your app\Exceptions\Handler.php code with the following....
<?php
namespace App\Exceptions;
use Exception;
use Illuminate\Auth\AuthenticationException;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
class Handler extends ExceptionHandler
{
/**
* A list of the exception types that are not reported.
*
* @var array
*/
protected $dontReport = [
//
];
/**
* A list of the inputs that are never flashed for validation exceptions.
*
* @var array
*/
protected $dontFlash = [
'password',
'password_confirmation',
];
/**
* Report or log an exception.`enter code here`
*
* @param \Exception $exception
* @return void
*/
public function report(Exception $exception)
{
parent::report($exception);
}
/**
* Render an exception into an HTTP response.
*
* @param \Illuminate\Http\Request $request
* @param \Exception $exception
* @return \Illuminate\Http\Response
*/
public function render($request, Exception $exception)
{
return parent::render($request, $exception);
}
protected function unauthenticated($request, AuthenticationException $exception)
{
if ($request->expectsJson()) {
return response()->json(['message' => $exception->getMessage()], 401);
}
$guard = array_get($exception->guards(),0);
switch ($guard) {
case 'admin':
$login = 'admin.login';
break;
default:
$login = 'login';
break;
}
return redirect()->guest(route($login));
}
}-14
1.Go to vendor/laravel/framework/src/Illuminate/Foundation/Exceptions/Handler.php file.
2.Search the method name is unauthenticated.
3.And change the redirect url from redirect()->guest(route('login')) to redirect()->guest(route('api/login')) //whatever you want.
You could return the response as a JSON if it is API services.
- 4It is very bad practice to make changes in files from external packages! Don't do this. – thephper Jan 5 '18 at 14:38
- 1to ellaborate on @thephper's answer: if you make changes to external packages they will be lost if you ever update them. – SLow Loris Apr 16 '18 at 13:03
- If you want to lose all of your code by running a composer update command, use this!! – Erfun Jul 24 '18 at 17:34
- Nothing will impact if you are running the composer update command. In laravel, they are having certain level of security while accessing the API or Routes services. – Nadimuthu Sarangapani Jul 25 '18 at 18:04
19
By default if I am not logged and I try visit this in browser:
http://localhost:8000/homeIt redirect me to http://localhost:8000/auth/login
How can I change to redirect me to http://localhost:8000/login
41
I wanted to do the same thing in Laravel 5.5. Handling authentication has moved to Illuminate\Auth\Middleware\Authenticate which throws an Illuminate\Auth\AuthenticationException.
That exception is handled in Illuminate\Foundation\Exceptions\Hander.php, but you don't want to change the original vendor files, so you can overwrite it with your own project files by adding it to App\Exceptions\Handler.php.
To do this, add the following to the top of the Handler class in App\Exceptions\Handler.php:
use Illuminate\Auth\AuthenticationException;And then add the following method, editing as necessary:
/**
* Convert an authentication exception into an unauthenticated response.
*
* @param \Illuminate\Http\Request $request
* @param \Illuminate\Auth\AuthenticationException $exception
* @return \Illuminate\Http\Response
*/
protected function unauthenticated($request, AuthenticationException $exception)
{
if ($request->expectsJson()) {
return response()->json(['error' => 'Unauthenticated.'], 401);
}
return redirect()->guest('login'); //<----- Change this
}Just change return redirect()->guest('login'); to return redirect()->guest(route('auth.login')); or anything else.
I wanted to write this down because it took me more than 5 minutes to figure it out. Please drop me a line if you happened to find this in the docs because I couldn't.
- 7
- 2Do not change core files, this is a really bad idea and when composer runs an update all your changes will be over written – twigg Jan 10 '17 at 15:34
- 1
- 2@twigg that's no core file. This is the exeption handler located at App\Exceptions\Handler. Nothing will be overwritten – Moritz Ewert Mar 8 '17 at 13:51
- 1@Inigo the guest method saves the current url as the 'intended' destination while it redirects unauthenticated users. after logging in the user will be returned to the intended url – Moritz Ewert Oct 9 '17 at 12:59
31
Just to extend @ultimate's answer:
You need to modify
App\Http\Middleware\Authenticate::handle()method and changeauth/loginto/login.Than you need to add
$loginPathproperty to your\App\Http\Controllers\Auth\AuthControllerclass. Why? See Laravel source.
In result you'll have this in your middleware:
namespace App\Http\Middleware;
class Authenticate {
/**
* Handle an incoming request.
*
* @param \Illuminate\Http\Request $request
* @param \Closure $next
* @return mixed
*/
public function handle($request, Closure $next)
{
if ($this->auth->guest())
{
if ($request->ajax())
{
return response('Unauthorized.', 401);
}
else
{
return redirect()->guest('/login'); // <--- note this
}
}
return $next($request);
}
}And this in your AuthController:
namespace App\Http\Controllers\Auth;
class AuthController extends Controller
{
protected $loginPath = '/login'; // <--- note this
// ... other properties, constructor, traits, etc
}
11
This is Laravel 5.4 Solution:
There is a new unauthenticated() method in app/Exceptions/Handler.php which handles unauthenticated users and redirects to login path.
So change
return redirect()->guest('login');to
return redirect()->guest('auth/login');
- Hi @salil i have error Sorry, the page you are looking for could not be found. NotFoundHttpException in RouteCollection.php line 161: when loading the custom login url. May i know how to solve this issue? do i need to create a new view just for the custom login? – davidlee Apr 3 '17 at 10:19
- 1Christ on a bike, this has changed yet again in Laravel 5.5. The
unauthenticated()method is now removed from App/Exceptions/Handler.php. It's like with every version they want to obfuscate the inner workings of the authentication system further. Very frustrating for those of us wanting to modify- or even just understand- it. – Inigo Oct 16 '17 at 12:08
5
In Laravel 5.6, go to app/Exceptions folder and open the Handler.php, add a new method that overrides the unauthenticated method like so:
protected function unauthenticated($request, AuthenticationException $exception)
{
if($request->ajax())
{
return response([
"message" => "Unauthenticated.",
"data" => [],
],401);
}
return redirect()->to('/');
}This method is triggered when you access a protected route using the built-in "auth" middleware. Now you will have full control where to redirect or the response sent.
4
Authentication checks are made using middleware in Laravel 5.
And the middleware for auth is App\Http\Middleware\Authenticate.
So, you can change it in handle method of the middleware.
3
EDIT: On Laravel 5.1, simply add protected $redirectPath = '/url/you/want'; to AuthController would do the trick.
REFER : http://laravel.com/docs/5.1/authentication#included-authenticating
On Laravel 5.1, it is completely moved to another middleware named RedirectIfAuthenticated.php under App\Http\Middleware
public function handle($request, Closure $next)
{
if ($this->auth->check()) {
return redirect('/'); //change this part to anywhere you wish to be redirected to
}
return $next($request);
}Hope it helps.
1
could you please outputs php artisan route:list please
You are right you can set the following attributes:
protected $loginPath = 'xxx';
protected $redirectPath = 'xxx';
protected $redirectAfterLogout = 'xxx';Set this attribute to you AuthController.php
0
Since your other question was marked as duplicate..I will try to answer it here..
First you need to change your route like
<?php
Route::get(config('constants.cms_path') . '/login', [
'as' => 'login',
'uses' => 'Auth\AuthController@getLogin'
]);In your blade..make sure you use named route in the Login url link like
{{ route('login') }}In Middleware/Authenticate.php change the redirect guest to
return redirect()->guest(config('constants.cms_path') . '/login');0
To change the redirection after the login, you only have to go to app/Http/Controllers/Auth/LoginController.php and add that inside the class LoginController:
protected $redirectTo = '/redirect-url-here';Same for redirection after a new users register, but in that case, on AuthController.php
0
For Laravel 5.4 You can set protected $redirectTo = '/'; in LoginController.php FILE. Or in RegistersUsers.php file you can
protected function registered(Request $request, $user)
{
return redirect('tosomeRoute');
//Note: This code will run when
//The user has been registered
}
普通分类: