如何解决警告 - 12200 架构验证警告提供的 XML 不符合 Twilio 标记 XML 架构
我正在按照本教程设置 Twilio-Dialogflow WhatsApp 聊天机器人: https://www.youtube.com/watch?v=r5EMHIQiGWE
代码如下:
from http.client import responses
from flask import Flask,request,jsonify
import os,dialogflow
from flask.globals import session
from google.api_core.exceptions import InvalidArgument
import requests
os.environ["GOOGLE_APPLICATION_CREDENTIALS"] = 'private_key.json'
DIALOGFLOW_PROJECT_ID = 'whatsapp-twilio-flask-dia-svau'
DIALOGFLOW_LANGUAGE_CODE = 'en'
SESSION_ID = 'me'
app = Flask(__name__)
app.config["DEBUG"] = True
@app.route('/')
def root():
return "Hello World"
@app.route('/api/getMessage',methods=['POST'])
def home():
message = request.form.get('Body')
mobnum = request.form.get('From')
session_client = dialogflow.SessionsClient()
session = session_client.session_path(DIALOGFLOW_PROJECT_ID,SESSION_ID)
text_input = dialogflow.types.TextInput(text=message,language_code = DIALOGFLOW_LANGUAGE_CODE)
query_input = dialogflow.types.QueryInput(text=text_input)
try:
response = session_client.detect_intent(session=session,query_input=query_input)
except InvalidArgument:
raise
sendMessage(mobnum,response.query_result.fulfillment_text)
return response.query_result.fulfillment_text
def sendMessage(mobnum,message):
url = "https://api.twilio.com/2010-04-01/Accounts/XXXXXXXXXX/Messages.json"
payload = {'From': 'whatsapp:+14155238886','Body': message,'To': mobnum}
headers = {'Authorization': 'Basic XXXXXXXXXXXXXXXXXXXXXXXXXXX' }
response = requests.request("POST",url,headers = headers,data=payload)
print(response.text.encode('utf8'))
return ""
if __name__ == '__main__':
app.run()
在 Twilio 上运行代码会出现此错误,因此我没有在 WhatsApp 中得到回复
Warning - 12200
Schema validation warning
The provided XML does not conform to the Twilio Markup XML schema.
response.text 返回这个:
print(response.text.encode('utf8')) :
b'{"sid": "XXX","date_created": "XXX","date_updated": "XXX","date_sent": null,"account_sid": "XXX","to": "whatsapp:+XXX","from": "whatsapp:+XXX","messaging_service_sid": null,"body": "Good day! What can I do for you today?","status": "queued","num_segments": "1","num_media": "0","direction": "outbound-api","api_version": "2010-04-01","price": null,"price_unit": null,"error_code": null,"error_message": null,"uri": "XXX","subresource_uris": {"media": "XXX"}}'
有人可以帮忙吗?我是否因为没有使用 twilio.twiml 库而从 Twilio 收到上述错误?但是从教程视频中,我没有看到正在使用这个库。
解决方法
这里是 Twilio 开发者布道者。
您(和视频教程)设置响应的方式不太正确。您已将 WhatsApp 沙盒网络钩子指向您的 /api/getMessage
端点,这是正确的,但您响应该请求的方式不适用于 Twilio。
当 Twilio 向您的应用程序发出 Webhook 请求时,它期望响应是包含 TwiML 的 XML 响应。您甚至可以使用 TwiML 来响应消息,因此您无需向 REST API 发出请求即可发送消息。
为了让事情变得更简单,您可以做的另一件事是install and use the Twilio Python helper library。您可以使用它向 API 发出请求以及构建 TwiML 响应。
因此,我建议您安装 Twilio Python 库:
pip install twilio
然后使用应用程序中的库来构建并返回 TwiML 响应,如下所示:
from http.client import responses
from flask import Flask,request,jsonify,Response
import os,dialogflow
from flask.globals import session
from google.api_core.exceptions import InvalidArgument
from twilio.twiml.messaging_response import MessagingResponse
import requests
os.environ["GOOGLE_APPLICATION_CREDENTIALS"] = 'private_key.json'
DIALOGFLOW_PROJECT_ID = 'whatsapp-twilio-flask-dia-svau'
DIALOGFLOW_LANGUAGE_CODE = 'en'
SESSION_ID = 'me'
app = Flask(__name__)
app.config["DEBUG"] = True
@app.route('/')
def root():
return "Hello World"
@app.route('/api/getMessage',methods=['POST'])
def home():
message = request.form.get('Body')
mobnum = request.form.get('From')
session_client = dialogflow.SessionsClient()
session = session_client.session_path(DIALOGFLOW_PROJECT_ID,SESSION_ID)
text_input = dialogflow.types.TextInput(text=message,language_code = DIALOGFLOW_LANGUAGE_CODE)
query_input = dialogflow.types.QueryInput(text=text_input)
try:
intent = session_client.detect_intent(session=session,query_input=query_input)
except InvalidArgument:
raise
response = MessagingResponse()
response.message(intent.query_result.fulfillment_text)
return Response(str(response),mimetype="application/xml")
if __name__ == '__main__':
app.run()
在这段代码中,我们没有调用原始的 sendMessage
函数,而是从 MessagingResponse
导入 twilio.twiml.messaging_response
并创建一个我们作为 XML 返回的响应,使用 Response
对象来设置内容和 mimetype。
Response(str(response),mimetype="application/xml")
这将简化您的代码并消除架构验证警告。查看 building with Twilio and Flask 上的这篇博文了解更多信息。
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