Question

我正在做一个休息api项目。
我正在努力解决一个问题。当我收到令牌到期错误时，生成的代码将是这样的：

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());
    }
}

此代码写在此文件中：

供应商/ tymon / JWT-AUTH / SRC / HTTP /中间件/ BaseMiddleware.php

如何将其作为JSON类型返回？

Answer 1

在 App \ Exceptions \ Handler 类'渲染方法中捕获该异常，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);

}

在laravel中处理jwt auth错误

1 个答案: