Question

我在AWS lambda中调用API时包含的代码如下所示。 urlilb3 python库成功上传为zip文件夹。但是当我尝试访问它显示的特定意图时

当我在AWS lambda（python 3.6）中包含API调用时，我得到了

“无法调用远程端点，或者它返回的响应无效”。

为什么会这样？在python 3.6中包含API调用之前要做的先决条件是什么。我使用了urllib3 python库并上传为zip文件夹。是否需要做其他事情？

def get_weather(session):
    should_end_session = False
    speech_output = " "
    reprompt_text = ""
    api = "some url ...."
    http = urllib3.PoolManager()
    response = http.request('GET',api)
    weather_status = json.loads(response.data.decode('utf-8'))
    for weather in weather_status:
        final_weather = weather["WeatherText"]
    return build_response(session_attributes, build_speechlet_response(speech_output, reprompt_text, should_end_session))

Answer 1

尝试打印response.data，以便在日志中看到它。这可能会给你一个线索。我也会尝试切换到Python请求而不是URLLib3。您可能还需要根据要调用的API的实现来设置内容类型。

Answer 2

from __future__ import print_function
import json
from botocore.vendored import requests
def lambda_handler(event, context):
   print('received request: ' + str(event))
   doctor_intent = event['currentIntent']['slots']['doctor']
   email_intent = event['currentIntent']['slots']['email']
   print(doctor_intent, email_intent)
   print(type(doctor_intent), type(email_intent))
   utf8string = doctor_intent.encode("utf-8")
   utf8string1 = email_intent.encode("utf-8")
   print(type(utf8string))
   print(type(utf8string1))
   car1 = {"business_name": utf8string , "customer_email": utf8string1 }  
   r = requests.post('https://postgresheroku.herokuapp.com/update', 
   json=car1)
   #print ("JSON         : ", r.json())
   print(r.json())
   data = str(r.json())
   print(type(data))
   return {
    "dialogAction": {
    "type": "Close",
    "fulfillmentState": "Fulfilled",
    "message": {
        "contentType": "PlainText",
        "content": "Thank you for booking appointment with {doctor} 
{response}".format(doctor=doctor_intent,response=data)
     }
    }
   }

Answer 3

场景：使用第三方API获取天气

 import urllib3

 def get_weather():
    api = "some url ...."
    http = urllib3.PoolManager()
    response = http.request('GET',api)
    weather_status = json.loads(response.data.decode('utf-8'))
    for weather in weather_status:
        final_weather = weather["WeatherText"] ## The attribute "WeatherText" will varies depending upon the weather API you are using.  
    return final_weather 

    get_weather() # simple function call

如何在AWS lambda函数中调用外部API或URL（python代码）？

3 个答案: