处理laravel中的jwt auth错误

Question

我正在做一个RESTAPI项目<我在为一个问题挣扎。当我得到令牌过期错误时，生成的代码如下所示：

public function authenticate(Request $request){
    $this->checkForToken($request);

    try {
        if (! $this->auth->parseToken()->authenticate()) {
            throw new UnauthorizedHttpException('jwt-auth', 'User not found');
        }
    } catch (JWTException $e) {
        throw new UnauthorizedHttpException('jwt-auth', $e->getMessage(), $e, $e->getCode());
    }
}

这段代码写在这个文件中：

供应商/tymon/jwt-auth/src/Http/Middleware/BaseMiddleware.php

如何将其作为JSON类型返回？

Answer 1

在App\Exceptions\Handler类的render方法中捕获该异常，并返回格式为json的响应：

// Handler.php
// import the class of the exception you want to render a json response for at the top
use Symfony\Component\HttpKernel\Exception\UnauthorizedHttpException;
...

public function render($request, Exception $exception)
{   

  // if your api client has the correct content-type this expectsJson() 
  // should work. if not you may use $request->is('/api/*') to match the url.


  if($request->expectsJson()) 
  {

    if($exception instanceof UnauthorizedHttpException) {

      return response()->json('Unauthorized', 403);

    }

  }

  return parent::render($request, $e);

}