laravel 如何从laravel5中的中间件抛出禁止的异常？
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How throw forbidden exception from middleware in laravel5?
提问by gsk
I am writing a middleware in laravel 5. I want to throw a forbidden exception with code 403 from middleware. My middleware function is given below:
我正在 Laravel 5 中编写一个中间件。我想从中间件中抛出一个代码为 403 的禁止异常。我的中间件功能如下：
use Exception;
public function handle($request, Closure $next)
{
if (!Auth::check()) {
throw new Exception("Access denied", 403);
}
return $next($request);
}
I am calling my middleware from controller and I am getting error message with code 500 but not 403. How can I resolve this?
我正在从控制器调用我的中间件，但收到代码为 500 但不是 403 的错误消息。我该如何解决这个问题？
回答by lukasgeiter
You can simply use the abort()helper. (Or App::abort())
您可以简单地使用abort()帮助程序。（或App::abort()）
public function handle($request, Closure $next) {
if (!Auth::check()) {
abort(403, 'Access denied');
}
return $next($request);
}
You can handle these exceptions inside App\Exceptions\Handlerby overriding render()For example:
您可以App\Exceptions\Handler通过覆盖render()在内部处理这些异常，例如：
public function render($request, Exception $e)
{
if($e instanceof HttpException && $e->getStatusCode() == 403){
return new JsonResponse($e->getMessage(), 403);
}
return parent::render($request, $e);
}
