带有 gevent-socketio 的 Turbogears：请求密钥错误

Question

我尝试在我的 TurboGears 2 网站中使用 gevent.socketio：

在我使用的ini文件中

[server:main]
use = egg:gevent-socketio#paster
transports = xhr-multipart, xhr-polling, websocket
host = 0.0.0.0
port = 8080

当我尝试在网络浏览器中访问控制器时：

@expose('wago.templates.test')
def index(self):
  socketio_manage(request.environ, {"/stat": StatNamespace}, request=request) 
  return dict()

我收到以下错误：

Traceback (most recent call last):
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/wsgiapp.py", line 105, in __call__
response = self.wrapped_dispatch(controller, environ, context)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/wsgiapp.py", line 278, in dispatch
return controller(environ, context)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/controllers/dispatcher.py", line 132, in __call__
response = self._perform_call(context)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/controllers/dispatcher.py", line 113, in _perform_call
r = self._call(func, params, remainder=remainder, context=context)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/controllers/decoratedcontroller.py", line 120, in _call
output = controller_caller(context_config, bound_controller_callable, remainder, params)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/decorators.py", line 42, in _decorated_controller_caller
return application_controller_caller(tg_config, controller, remainder, params)
File "/home/pi/tgenv/lib/python2.7/site-packages/tg/configuration/app_config.py", line 124, in call_controller
return controller(*remainder, **params)
File "/home/pi/tgenv/WAGO/wago/controllers/root.py", line 13, in index
socketio_manage(request.environ, {"/stat": StatNamespace}, request=request)
File "/home/pi/tgenv/lib/python2.7/site-packages/socketio/__init__.py", line 67, in socketio_manage
socket = environ['socketio']
KeyError: 'socketio'

我使用了几个关于金字塔的教程来向我介绍 gevent-socketio。 我使用 TurboGears2、gevent 和 gevent-socketio 的旧版本进行了尝试，我也尝试了 this 模块，但总是出现相同的错误。 我对套接字很陌生，所以也许我只是缺少一些明显的东西

Answer 1

gevent-socketio 仅识别来自特定 url (socket.io/1/) 的套接字请求 因为 TurboGears 使用 python 函数名称作为 url，所以我们无法使用“。”或常规方式上的“1”。一个简单的解决方案：

    @expose()
    def _default(self, *args):
      args = list(args)
      if "socketio" in request.environ:
        #do socketio stuff...
      else:
        abort(404)