Question

我开发了一个API，但令牌过期有问题，我尝试寻找方法来刷新API发送的令牌，我使用自定义中间件，当令牌过期时，将刷新的令牌添加到响应头。该应用仅需要搜索响应中是否包含此信息，如果是，则更新保存的令牌。我得到

{“代码”：103，“响应”：空}

我的中间件

<?php

namespace App\Http\Middleware;

use Carbon\Carbon;
use Closure;
use Illuminate\Http\JsonResponse;
use Illuminate\Http\Response;
use Illuminate\Support\Facades\Cache;
use Symfony\Component\HttpKernel\Exception\BadRequestHttpException;
use Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException;
use Tymon\JWTAuth\Exceptions\JWTException;
use Tymon\JWTAuth\Exceptions\TokenBlacklistedException;
use Tymon\JWTAuth\Exceptions\TokenExpiredException;
use Tymon\JWTAuth\Facades\JWTAuth;
use Tymon\JWTAuth\Http\Middleware\BaseMiddleware;

class JwtRefresh extends BaseMiddleware {

    public function handle($request, Closure $next)
    {
        try
        {
            if (! $user = JWTAuth::parseToken()->authenticate() )
            {
                return response()->json([
                    'code'   => 101, // means auth error in the api,
                   'response' => null // nothing to show
                 ]);
            }
        }
        catch (TokenExpiredException $e)
        {
            // If the token is expired, then it will be refreshed and added to the headers
            try
            {
                $refreshed = JWTAuth::refresh(JWTAuth::getToken());
                $user = JWTAuth::setToken($refreshed)->toUser();
                header('Authorization: Bearer ' . $refreshed);
            }
            catch (JWTException $e)
            {
                return response()->json([
                    'code'   => 103, // means not refreshable
                   'response' => null // nothing to show
                 ]);
            }
        }
        catch (JWTException $e)
        {
            return response()->json([
                'code'   => 101, // means auth error in the api,
                   'response' => null // nothing to show
            ]);
        }

        // Login the user instance for global usage
        Auth::login($user, false);

        return  $next($request);
    }
}

Answer 1

我认为您只需要这样做，

const url = this.channelArray.some(
    x => string.includes(x)
);

希望这会对您有所帮助。

JWT / LARAVEL 5.6刷新过期令牌

1 个答案: