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python - 龙卷风 RequestHandler 中的 Python AsyncHttpClient 抛出异常

问题描述

我将在 RequestHandler 中通过龙卷风 AsyncHttpClient 调用端点，但它会引发运行时异常This event loop is already running

class RegistrationHandler(tornado.web.RequestHandler):

  def post(self, *args, **kwargs):
      call_async_register("some params")


def call_async_register(parameters):
    def call():
        http_client = AsyncHTTPClient()
        future = Future()
        http_request = HTTPRequest(url, request_type.name, headers={'X-Peering': '1'}, body=body)

        def handle_future(f: Future):
            future.set_result(f.result())

        fetched_future = http_client.fetch(http_request)

        fetched_future.add_done_callback(handle_future)

        return future
    try:
        instance = io_loop.IOLoop.current()
        response = instance.run_sync(call)
        return response.body.decode()
    except Exception as err:
        self.logger.exception("Account Request Failed: {}".format(err))
        return None

标签： pythonpython-3.xasynchronoustornado

解决方案

这是问题所在：

instance = io_loop.IOLoop.current()
response = instance.run_sync(call)

run_sync本身尝试启动 ioloop。但从您的代码中可以明显看出，instance它已经在运行。所以你得到了错误。

如果您想将call()方法返回的值发送回用户，请将您的方法转换为协程（使用async/await语法）。

例子：

class RegistrationHandler(tornado.web.RequestHandler):

    async def post(self, *args, **kwargs):
        response = await call_async_register("some params")

        self.write(response)


async def call_async_register(parameters):
    http_client = AsyncHTTPClient()
    http_request = HTTPRequest(url, request_type.name, headers={'X-Peering': '1'}, body=body)

    try:
        response = await http_client.fetch(http_request)
        return response.body.decode()
    except Exception as err:
        self.logger.exception("Account Request Failed: {}".format(err))
        return None

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