Question

我已经围绕这个主题进行了一些搜索，并参考了相关的API参考资料，但我仍然在努力完全实现这一点。

我尝试使用python-twilio库在Python / Flask中构建一个系统，该系统将执行以下操作：

用户来电，听到消息
应用程序同时拨出n组中的第一组，其中组是带有元信息（如名称）的数字列表。在代码中表示为具有数字列表值的字典。（k [v] = [0,1，n]）
如果小组中的某个人（代理人）回答，请他们输入确认数字以继续。确认后，将它们连接到用户。
如果该组中的代理没有接听，请尝试该组中的下一个代理（响铃号码）并返回步骤3.

是的，等等代码。这已被剥离为基本逻辑。不幸的是，我还没能完全掌握Flask应用程序状态，所以我一直在使用Python全局变量访问。我想摆脱这种情况，因为如果向应用发出两个电话，应用可能会无法正常工作。

# excluding flask import/setup
from twilio import twiml
from twilio.rest import TwilioRestClient

agents = OrderedDict()
agents['Foo'] = ["+440000000000", "+440000000001"]
agents['Bar'] = ["+440000000002", "+440000000003"]

caller = None
@app.route('/notify', methods=['POST'])
def notify():
    """Log all call status callbacks"""
    global caller
    called = request.values.get('Called')
    call_status = request.values.get('CallStatus') 
    timestamp = request.values.get('Timestamp') # e.g. Wed, 02 Mar 2016 15:42:30 +0000
    call_statuses = {
        "initiated": "[%s@%s] INITIATED",
        "ringing": "[%s@%s] RINGING",
        "in-progress": "[%s@%s] IN-PROGRESS",
        "answered": "[%s@%s] ANSWERED",
        "failed": "[%s@%s] CALL FAILED",
        "no-answer": "[%s@%s] NO ANSWER",
        "busy": "[%s@%s] BUSY",
        "completed": "[%s@%s] COMPLETED"
        }
    called_name = [i for i in agents.items() if called in i][0][0]
    print("--%s-- %s" % (timestamp, call_statuses[call_status] % (called_name, called)))

    return jsonify({"status": 200})

def new_twilio_client():
    _ = TwilioRestClient(app.config['TWILIO_ACCOUNT_SID'], app.config['TWILIO_AUTH_TOKEN'])
    return _

@app.route('/welcome', methods=['GET'])
def welcome():
    global caller
    response = twiml.Response()
    response.say("Welcome message", voice="alice")
    caller = request.values.get("Caller")
    # log the call/caller here
    response.redirect(url_for(".start"), _external=True, method="GET")
    return str(response)

@app.route('/start', methods=['GET'])
def start():
    response = twiml.Response()
    response.say("Trying first agent")
    response.enqueue("q")
    client = new_twilio_client()
    for number in agents['Foo']:
        client.calls.create(to=number, Timeout=12, ifMachine="Hangup", statusCallbackEvent="initiated ringing answered completed", StatusCallback=url_for(".notify", _external=True), from_=app.config['TWILIO_CALLER_ID'], url=url_for(".whisper", _external=True), method="GET")        
    return str(response)

@app.route("/whisper", methods=['GET'])
def whisper():
    """Ask agent if they want to accept the call"""
    global call_sid
    response = twiml.Response()
    client = new_twilio_client()
    response.say("Press 1 to accept this call")
    response.gather(action=url_for(".did_agent_accept", _external=True), numDigits=1, timeout=3)
    return str(response)

@app.route("/did_agent_accept", methods=['POST'])
def did_agent_accept():
    """Determine if the agent accepted the call"""
    response = twiml.Response()
    digit = request.values.get('Digits', None)
    if digit == "1":
        with response.dial() as dial:
            # drop the agent into the queue we made earlier
            dial.queue("q", url=url_for(".thanks_for_waiting", _external=True))
    else:
        response.reject()
    return str(response)

@app.route("/thanks_for_waiting", methods=['GET'])
def thanks_for_waiting():
    """Thank the user"""
    response = twiml.Response()
    response.say("Thanks for waiting. You will now be connected.")
    return str(response)

@app.route('/sorry', methods=['GET'])
def sorry():
    response = twiml.Response()
    response.say("Sorry, trying again")
    response.redirect(url=url_for(".start", _external=True), method="GET")
    return str(response)

因此，我在这里使用了Twilio的排队功能将调用者放入队列，然后我使用REST API将调用发送到第一个代理的相关号码。当他们拿起并回复聚集动词时，他们就会被放入队列中。

我遇到的障碍：

管理忙碌或无应答（或其他非应答）状态，以确定应用程序是否应该尝试下一组数字很困难，因为异步性质回调功能。我已经尝试了一种解决方案，其中我维护一个全局列表，每当调用notify（）时都会更新，并且将给出非应答状态的最后一个数字添加到所述列表中。然后我将代理[代理]列表的长度与此全局列表进行比较，如果匹配，我会修改调用以尝试下一个组，但最终会出现幻像/错误调用。

我的问题是，我该如何进行这样的实时通话修改，这样我就可以清除以前任何现有的振铃来电？如何改进此代码的设计？有什么明显的东西我错过了吗？任何指针都会受到赞赏。

在环形系统中处理忙/无应答

0 个答案: