Jwt-auth: Laravel Middleware - Json Reponse

[ThomasRalee]

Hi,

I was wondering if anyone knows how I can get JSON Response for "Invalid Tokens" or "Expired Token" instead of the full html response. I am not sure if this is the right place to ask this, please redirect me.

Appreciate your help.

[smujaddid]

Look at the wiki https://github.com/tymondesigns/jwt-auth/wiki/Authentication section "Global exception handling".

Hope this helps.

[bul-ikana]

Tried that, but it seems that exceptions are instances of Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException, not from jwt-auth exception classes. Is there any other way to check what is the cause of the exception and send a related message in json?

[mikkezavala]

I had it working like:

<?php

namespace App\Http\Middleware;

use Tymon\JWTAuth\Exceptions\JWTException;
use Tymon\JWTAuth\Exceptions\TokenExpiredException;
use Tymon\JWTAuth\Middleware\GetUserFromToken;

class VerifyJWT extends GetUserFromToken
{
    /**
     * Handle an incoming request.
     *
     * @param  \Illuminate\Http\Request $request
     * @param  \Closure $next
     * @return mixed
     */
    public function handle($request, \Closure $next)
    {
        $user = null;
        if ($request->expectsJson()) {
            if (!$token = $this->auth->setRequest($request)->getToken()) {
                $this->tokenFailureHandler();
            }
            try {
                $user = $this->auth->authenticate($token);
            } catch (TokenExpiredException $e) {
                $this->tokenFailureHandler();
            } catch (JWTException $e) {
                $this->tokenFailureHandler();
            }

            if (!$user) {
                return $this->respond('tymon.jwt.user_not_found', 'user_not_found', 404);
            }

            $this->events->fire('tymon.jwt.valid', $user);

            return $next($request);
        }

        // If there is any custom Implementation on non Ajax
        return parent::handle($request, $next);
    }

and my App\Http\Kernel

<?php

namespace App\Http;

use App\Http\Middleware\VerifyJWT;
// .... other imports.
use Illuminate\Foundation\Http\Kernel as HttpKernel;


class Kernel extends HttpKernel
{
//... Your Kernel impl....
    protected $routeMiddleware = [
        'jwt.auth'    => VerifyJWT::class,
        'jwt.refresh' => RefreshToken::class,
    ];
}

for my $this->tokenFailureHandler() I just handled as json

private function tokenFailureHandler()
{
    return response()->json(['my_custom_key' => 'value']);
}

[SimeonDominiq]

I'm using the dev-develop version. I have tried the various solutions provided still shows

Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException: Token Signature could not be verified.

I want to return a json error instead of the HTML error

Thanks

[SimeonDominiq]

I had it working like this

`

namespace App\Http\Middleware;

use Closure;
use JWTAuth;
use Exception;
use Tymon\JWTAuth\Http\Middleware\BaseMiddleware;
use App\Http\Traits\ApiTrait;

class authJWT extends BaseMiddleware
{
use ApiTrait;

/**
 * Handle an incoming request.
 *
 * @param  \Illuminate\Http\Request  $request
 * @param  \Closure  $next
 * @return mixed
 */
public function handle($request, Closure $next)
{
    try {
        $user = JWTAuth::parseToken()->authenticate();
    } catch (Exception $e) {
        if ($e instanceof \Tymon\JWTAuth\Exceptions\TokenInvalidException){
            return $this->respondWithError("Token is Invalid");
        }else if ($e instanceof \Tymon\JWTAuth\Exceptions\TokenExpiredException){
            return $this->respondWithError(['error'=>'Token is Expired']);
        }else{
            return $this->respondWithError(['error'=>'Something is wrong']);
        }
    }
    return $next($request);
}

}`

With this extending the BaseMiddleware Class, you can return the error in json format without tampering with the vendor code.

I could see that this thread is closed but i had to re-open it as i was facing issues while using the dev-develop version

@tymondesigns Hope this will be fixed

[nsulistiyawan]

I'm facing the same problem when using Laravel 5.6 and "tymon/jwt-auth": "1.0.*"
Here's my approach to fix that.

1. Open app\Exceptions\Handler.php
2. Edit render method

/**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $exception
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $exception)
    {
        // detect instance
        if ($exception instanceof UnauthorizedHttpException) {
            // detect previous instance
            if ($exception->getPrevious() instanceof TokenExpiredException) {
                return response()->json(['status' => 'token_expired'], $exception->getStatusCode());
            }
            else if ($exception->getPrevious() instanceof TokenInvalidException) {
                return response()->json(['status' => 'token_invalid'], $exception->getStatusCode());
            }
            else if ($exception->getPrevious() instanceof TokenBlacklistedException) {
                return response()->json(['status' => 'token_blacklisted'], $exception->getStatusCode());
            }
        }
        return parent::render($request, $exception);
    }
````

also dont forget to import all of required class.

namespace App\Exceptions;

use Exception;
use Illuminate\Auth\AuthenticationException;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
use Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException;
use Tymon\JWTAuth\Exceptions\TokenBlacklistedException;
use Tymon\JWTAuth\Exceptions\TokenExpiredException;
use Tymon\JWTAuth\Exceptions\TokenInvalidException;

class Handler extends ExceptionHandler
{
```
Hope this could help someone :)

[AndreiMireichyk]

При отправке запроса необходимы два заголовка, для ответа сервера в формате json
Content-Type : application/json
Accept: application/json

[Flayshon]

Faced this issue in Laravel 6.2 and tried using @nsulistiyawan code, but using

$exception->getPrevious()

always returns null. Probably because something changed in Laravel's Router implementation. So I modified the code to use UnauthorizedHttpException messages and it got the job done.

public function render($request, Exception $exception)
    {
        if ($exception instanceof UnauthorizedHttpException) {
            return response()->json(['error' => $exception->getMessage()], $exception->getStatusCode());
        }

        return parent::render($request, $exception);
    }

But it would be nice to modify the message according to the exception type. Does anyone know how to solve this now?

[mpandar]

Faced this issue in Laravel 6.2 and tried using @nsulistiyawan code, but using

$exception->getPrevious()

always returns null. Probably because something changed in Laravel's Router implementation. So I modified the code to use UnauthorizedHttpException messages and it got the job done.

public function render($request, Exception $exception)
    {
        if ($exception instanceof UnauthorizedHttpException) {
            return response()->json(['error' => $exception->getMessage()], $exception->getStatusCode());
        }

        return parent::render($request, $exception);
    }

But it would be nice to modify the message according to the exception type. Does anyone know how to solve this now?

I have the same problem, $exception->getPrevious() get null in laravel 7, Does anyone know how to solve this now?

[mpandar]

I'm facing the same problem when using Laravel 5.6 and "tymon/jwt-auth": "1.0.*"
Here's my approach to fix that.

1. Open app\Exceptions\Handler.php
2. Edit render method

/**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request  $request
     * @param  \Exception  $exception
     * @return \Illuminate\Http\Response
     */
    public function render($request, Exception $exception)
    {
        // detect instance
        if ($exception instanceof UnauthorizedHttpException) {
            // detect previous instance
            if ($exception->getPrevious() instanceof TokenExpiredException) {
                return response()->json(['status' => 'token_expired'], $exception->getStatusCode());
            }
            else if ($exception->getPrevious() instanceof TokenInvalidException) {
                return response()->json(['status' => 'token_invalid'], $exception->getStatusCode());
            }
            else if ($exception->getPrevious() instanceof TokenBlacklistedException) {
                return response()->json(['status' => 'token_blacklisted'], $exception->getStatusCode());
            }
        }
        return parent::render($request, $exception);
    }

also dont forget to import all of required class.

<?php

namespace App\Exceptions;

use Exception;
use Illuminate\Auth\AuthenticationException;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
use Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException;
use Tymon\JWTAuth\Exceptions\TokenBlacklistedException;
use Tymon\JWTAuth\Exceptions\TokenExpiredException;
use Tymon\JWTAuth\Exceptions\TokenInvalidException;

class Handler extends ExceptionHandler
{

Hope this could help someone :)

Does this work in laravel 7? $exception->getPrevious() get null