如何解决Python Twilio错误-序言中不允许包含内容模式验证警告
与先前的SO query类似,我有一个侦听POST的webhook接收器。我将WhatsApp消息发送到我的Twilio号码,将Twilio POST发送到webhook接收者,服务器代码处理请求,然后将响应对象返回给Twilio。该代码从Airtable的事件日历中提取了一些项目,并应通过Twilio将它们发送到WhatsApp。
webhook使用Postman测试OK。但是,WhatsApp不会返回POST请求的结果,并且我从Twilio调试器收到以下警告:
sourceComponent "14100"
line "1"
ErrorCode "12200"
LogLevel "WARN"
Msg "Content is not allowed in prolog."
EmailNotification "false"
parserMessage "Content is not allowed in prolog."
cols "1"
我的完整代码在这里:
from flask import Flask,request
import requests
import json
from airtable import Airtable
from datetime import datetime
from twilio.twiml.messaging_response import MessagingResponse
base_key = 'BASE_KEY'
table_name = 'Events Calendar'
api_key = 'API_KEY'
airtable = Airtable(base_key,table_name,api_key)
API_URL = "https://api.airtable.com/v0/BASE_KEY/Events%20Calendar?maxRecords=3&filterByFormula=IS_AFTER(%7BDate%7D,NOW())"
headers = {
"Authorization": "Bearer API_KEY"
}
app = Flask(__name__)
@app.route('/bot',methods=['POST'])
def accessdb():
incoming_msg = request.values.get('Body','').lower()
resp = MessagingResponse()
msg = resp.message()
responded = False
if 'next' in incoming_msg:
def pretty(d):
date = datetime.strptime(d["Date"],"%Y-%m-%dT%H:%M:%S.%fZ")
return f'''Date: {date.strftime("%A,%d. %B %Y %I:%M%p")}
Title: {d['Title']}
Description: {d['Description']}
'''
pages = airtable.get_iter(maxRecords=3,formula="Date >= NOW()",sort=["Date"],fields=('Date','Title','Description'))
mylist = []
for page in pages:
for record in page:
if 'fields' not in record:
continue
fields = record['fields']
mylist.append(pretty(fields))
return "\n".join(mylist)
elif 'what' in incoming_msg:
msg.body("Please enter 'next' to get the next events.")
responded = True
elif not responded:
msg.body("I don't know about that,sorry!")
return str(resp)
if __name__ == '__main__':
app.run()
以上代码已部署到Heroku中,我可以使用curl或Postman调用webhook并成功产生以下结果:
Date: Wednesday,07. October 2020 06:00PM
Title: Social Distancing Dance Party: Slow Motion
Description: Slow Motion is the yang to the Saturday Dance Party's yin
Date: Thursday,08. October 2020 08:00AM
Title: Free Online Meditation Mornings
Description: 20-30 minute meditation. Donations can be made at https://example.com
Date: Saturday,10. October 2020 07:00PM
Title: Social Distancing Dance Party
Description: A party in your very own kitchen
但是从Twilio调用webhook会产生先前给出的错误。
早期的SO答案建议按以下方式返回查询结果:return twilioResponse.ToString()
我对语法没有把握,并尝试了以下方法:
for page in pages:
for record in page:
if 'fields' not in record:
continue
fields = record['fields']
mylist.append(pretty(fields))
return "\n".join.twilioResponse.ToString(mylist)
但是,当我尝试上述操作时,失败如下:
return "\n".join.twilioResponse.ToString(mylist)
AttributeError: 'builtin_function_or_method' object has no attribute 'twilioResponse'
在正确的方向上轻推将不胜感激。预先非常感谢。
解决方法
这里是Twilio开发人员的传播者。
我认为这里的问题是,当您发送机器人“ next”时,它会以一串结果而不是TwiML来响应,这正是Twilio所期望的。如果用户发送“什么”或其他信息,您将返回TwiML,因此没有太多更改。
您需要输入以下代码:
for page in pages:
for record in page:
if 'fields' not in record:
continue
fields = record['fields']
mylist.append(pretty(fields))
return "\n".join(mylist)
而不是在结尾处直接return
,而是将您构建的字符串设置为TwiML响应中的消息,如下所示:
for page in pages:
for record in page:
if 'fields' not in record:
continue
fields = record['fields']
mylist.append(pretty(fields))
msg.body("\n".join(mylist))
现在,您将事件列表设置为要发送的消息,并且方法末尾的行return str(resp)
将事件列表作为TwiML返回给Twilio。
作为补充,我不认为您在整个条件语句中都需要使用responded
布尔值。相反,您可以仅使用else
完成条件,该条件将捕获以前未匹配的所有情况。像这样:
else:
msg.body("I don't know about that,sorry!")
让我知道是否有帮助。
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