Laravel 經過Url肯定返回錯誤信息格式 阿星小棧

路徑：項目/app/Exceptions/Handler.php

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

        if(!empty($exception)){
            $url = URL::current();
            $urlData = parse_url($url);
            $midware = substr($urlData['path'],0,6);
            if($midware != '/admin'){
                echo '參數缺失，系統錯誤！';exit;
            }
            return response()->json(['message'=>'參數缺失，系統異常！', 'code'=>99999]);
        }
        return parent::render($request, $exception);
    }