【问题标题】:httplib.ResponseNotReady with api.aihttplib.ResponseNotReady 与 api.ai
【发布时间】:2026-01-20 20:55:02
【问题描述】:

我正在尝试使用 api.ai 制作自己的聊天机器人。

第一次很好,但第二次出现这个错误:

   Traceback (most recent call last):
  File "/root/Documents/Projects/Darlene/core/api.py", line 34, in <module>
    main()
  File "/root/Documents/Projects/Darlene/core/api.py", line 28, in main
    speech, action = b.handle(cmd)
  File "/root/Documents/Projects/Darlene/core/api.py", line 15, in handle
    response = self.request.getresponse().read()
  File "/usr/local/lib/python2.7/dist-packages/apiai/requests/request.py", line 133, in getresponse
    return self._connection.getresponse()
  File "/usr/lib/python2.7/httplib.py", line 1123, in getresponse
    raise ResponseNotReady()
httplib.ResponseNotReady

好像是api有问题。但我不确定。 这是我的代码:

import apiai
import json



class Bot(object):
    def __init__(self, client_token='<clientToken>'):

        self.AI = apiai.ApiAI(client_token)
        self.request = self.AI.text_request()
        self.request.lang = 'en'

    def handle(self, text):
        self.request.query = text
        response = self.request.getresponse().read()
        speech = str(json.loads(response)['result']['fulfillment']['speech'])
        action = str(json.loads(response)['result']['action'])
        if action is not '':
            return speech, action
        else:
            return speech, None


def main():
    b = Bot()
    while True:
        cmd = raw_input('me; ')
        speech, action = b.handle(cmd)
        print speech
        if action is not None:
            print 'action'

if __name__ == '__main__':
    main()

有人知道解决这个问题的方法吗?

【问题讨论】:

    标签: python-2.7 httplib dialogflow-es


    【解决方案1】:

    请不要重复使用请求。只是为每个阶段创建新的请求。同样是这样:

    def handle(self, text):
        self.request = self.AI.text_request()
        self.request.lang = 'en'
    
        self.request.query = text
        response = self.request.getresponse().read()
        speech = str(json.loads(response)['result']['fulfillment']['speech'])
        action = str(json.loads(response)['result']['action'])
        if action is not '':
            return speech, action
        else:
            return speech, None
    

    【讨论】:

    • 为什么我不能重复使用请求?
    【解决方案2】:

    在 read() 之后使用 decode('utf-8')。

    由于您是刚开始,现在将代码移至 Python 3.5 而不是 2.7 是个好主意...

    apiai_request.getresponse().read().decode('utf-8'))

    请注意,不幸的是,他们在示例中命名为“请求”,因为它与请求库、烧瓶请求等冲突……但这就是生活。

    【讨论】: