laravel路由问题

<code>php</code><code>/**
     * Render the given HttpException.
     *
     * @param  \Symfony\Component\HttpKernel\Exception\HttpException  $e
     * @return \Symfony\Component\HttpFoundation\Response
     */
    protected function renderHttpException(HttpException $e)
    {
        if (view()->exists('errors.'.$e->getStatusCode()))
        {
            return response()->view('errors.'.$e->getStatusCode(), [], $e->getStatusCode());
        }
        else
        {
            return (new SymfonyDisplayer(config('app.debug')))->createResponse($e);
        }
    }
  比如你想自定义 404 错误页面的话，只要创建一个 resources/views/errors/404.blade.php 的视图文件  
404内容自定义了
</code>

<code><?php namespace App\Exceptions;

use Exception;
use Illuminate\Foundation\Exceptions\Handler as ExceptionHandler;
use Symfony\Component\HttpKernel\Exception\HttpException;
use Symfony\Component\HttpKernel\Exception\MethodNotAllowedHttpException;
use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;

class Handler extends ExceptionHandler
{

    /**
     * A list of the exception types that should not be reported.
     *
     * @var array
     */
    protected $dontReport = [
        //'Symfony\Component\HttpKernel\Exception\HttpException'
    ];

    /**
     * Report or log an exception.
     *
     * This is a great spot to send exceptions to Sentry, Bugsnag, etc.
     *
     * @param  \Exception $e
     *
     * @return void
     */
    public function report(Exception $e)
    {
        return parent::report($e);
    }

    /**
     * Render an exception into an HTTP response.
     *
     * @param  \Illuminate\Http\Request $request
     * @param  \Exception               $e
     *
     * @return \Illuminate\Http\Response
     */
    public function render($request,Exception $e)
    {
        if($e instanceof NotFoundHttpException)
        {
            return \Response::view('sys::missing',[],404);    //你需要的在这里，404
        }

        //其他一些异常 照着写就行了
        //...
    }
}

</code></code>