现在我有一个程序接收一条SMS消息,发布消息正文,然后将SMS转发给另一个号码。但是,我收到Twilio关于“Scheme验证”的错误。代码的功能与它应该完全一样,但我想修复错误。Twilio/Python - 12200架构验证警告
起初,我有以下代码:
import RUAlertsTwilioWEBSERVER
import twilio.twiml
import time
import praw
from flask import Flask, request, redirect
from twilio.rest import TwilioRestClient
from passwords import *
from twilio import twiml
def login():
r = praw.Reddit(app_ua)
r.set_oauth_app_info(app_id, app_secret, app_uri)
r.refresh_access_information(app_refresh)
return r
r=RUAlertsTwilioWEBSERVER.login()
client = TwilioRestClient(account_sid, auth_token)
app = Flask(__name__)
@app.route("/", methods=['GET', 'POST'])
def AlertService():
TheMessage=request.form.get("Body")
if (TheMessage != None):
print(TheMessage)
client.messages.create(to=ePhone,from_=tPhone,body=str(TheMessage))
r.submit(*submit to reddit code*)
return str(TheMessage)
if __name__ == "__main__":
app.run(debug=True, host="0.0.0.0")
的Twilio调试器给我 Content is not allowed in prolog. Warning - 12200 Schema validation warning The provided XML does not conform to the Twilio Markup XML schema.
我试图改变我的代码以下获得所需职位的XML(只有相关部分)
@app.route("/", methods=['GET', 'POST'])
def AlertService():
TheMessage=request.form.get("Body")
if (TheMessage != None):
print(TheMessage)
resp = twiml.Response()
XML = resp.say(TheMessage)
client.messages.create(to=ePhone,from_=tPhone,body=XML)
r.submit(*submit to reddit code*)
return str(resp)
return str(TheMessage)
此代码没有工作,所以我改变body=XML
到body=str(XML)
。但现在它只是发送XML作为正文,并且收到错误: cvc-complex-type.2.4.a: Invalid content was found starting with element 'Say'. One of '{Sms, Message, Redirect}' is expected. Warning - 12200 Schema validation warning The provided XML does not conform to the Twilio Markup XML schema.
如何解决此问题?
对不起,我不清楚这一点。我想将发送给Twilio的SMS正文转发给另一个第三个号码(不是发件人或Twilio号码)。 当我使用“说”只是为了获得XML格式。 这段代码'client.messages.create(to = ePhone,from_ = tPhone,body = TheMessage)'有效,但我仍然从调试器中得到错误。 –
@DevinRader你能检查我的问题吗?谢谢http://stackoverflow.com/questions/43367898/twilio-quick-start-project-is-not-working –