python - Twilio 可编程语音调用立即完成
<p><p>我正在使用带有 Swift 和 Python 的新 Twilio 可编程语音 SDK。我已经开始了各自的快速入门项目，并且大部分情况下都有效。我能够获得有效的访问 token ，能够成功创建调用，甚至能够接听调用。问题出在房子的调用方。</p>
<p>当我尝试通过 Swift SDK 调用电话时，通话在另一端开始响铃之前就已断开。</p>
<p>我在 Twilio 文档中读到，如果您不处理 <code>status_callback</code> 事件，<code>client.calls.create</code> 函数将立即返回完成状态。我试图添加它，但每当我这样做时，我都会收到一条错误消息，指出键 <code>status_callback</code> 不是 <code>client.calls.create</code> 函数的有效参数。此外，我在任何地方都找不到任何实际如何处理通话状态的示例。</p>
<p>我的问题是我在这里做错了什么？任何帮助将不胜感激。</p>
<p>这是我的 Python 代码</p>
<pre><code>@app.route('/outgoing', methods=['GET', 'POST'])
def outgoing():
account_sid = os.environ.get("ACCOUNT_SID", ACCOUNT_SID)
api_key = os.environ.get("API_KEY", API_KEY)
api_key_secret = os.environ.get("API_KEY_SECRET", API_KEY_SECRET)
from_number =
to_number =
client = Client(api_key, api_key_secret, account_sid)
call = client.calls.create(url='http://twimlets.com/holdmusic?Bucket=com.twilio.music.ambient', to=to_number, from_=from_number)
# return str(call.sid)
resp = twilio.twiml.Response()
resp.say("Thank you for calling");
return str(resp)
</code></pre>
<p>这是我的相关 iOS 代码。请记住，这不是我的全部来源。我只提供了在这种情况下应该有的东西。我的完整来源确实包括处理注册表和邀请代表。我也没有包括显示/隐藏我的事件调用 UI 的源，因为这没有问题。这只是为了说明我是如何调用电话和接听电话完成委托(delegate)的。</p>
<pre><code>class VoiceManager: NSObject, PKPushRegistryDelegate, TVONotificationDelegate, TVOCallDelegate, AVAudioPlayerDelegate {
//MARK: - Singleton
static let sharedManager = VoiceManager()
//MARK: - Private Constants
private let baseURLString =
private let accessTokenEndpoint = "/accessToken"
//MARK: - Private Variables
private var deviceTokenString:String?
private var callInvite: TVOCallInvite?
private var call: TVOCall?
private var status: VoiceStatus = .idle
//MARK: - Getters
private func fetchAccessToken() -> String? {
guard let accessTokenURL = URL(string: baseURLString + accessTokenEndpoint) else {
return nil
}
return try? String.init(contentsOf: accessTokenURL, encoding: .utf8)
}
func placeCall(withParameters params: VoiceParameters, completion: @escaping (_ success: Bool, _ error: VAError?) -> Void) {
if (call != nil) {
call?.disconnect()
completion(false, .phoneCallInProgress)
status = .callEnded
hideActiveCallUI()
} else {
guard let accessToken = fetchAccessToken() else {
completion(false, .phoneAccessTokenFetchFailed)
return
}
guard let paramsDict = params.toDictionary() else {
completion(false, .phoneAccessTokenFetchFailed)
return
}
playOutgoingRingtone(completion: { in
if let strongSelf = self {
strongSelf.call = VoiceClient.sharedInstance().call(accessToken, params: [:], delegate: strongSelf) //NOTE: The params here are not necessary as the phone numbers for now are hard coded on the server
if (strongSelf.call == nil) {
strongSelf.status = .callEnded
completion(false, .phoneCallFailed)
return
} else {
strongSelf.status = .callConnecting
self?.showActiveCallUI(withParameters: params)
completion(true, nil)
}
}
})
}
}
// MARK: TVOCallDelegate
func callDidConnect(_ call: TVOCall) {
NSLog("callDidConnect:")
self.call = call
status = .inCall
routeAudioToSpeaker()
}
func callDidDisconnect(_ call: TVOCall) {
NSLog("callDidDisconnect:")
playDisconnectSound()
self.call = nil
status = .callEnded
hideActiveCallUI()
}
func call(_ call: TVOCall, didFailWithError error: Error) {
NSLog("call:didFailWithError: \(error)");
self.call = nil
status = .callEnded
hideActiveCallUI()
}
}
</code></pre></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>这里是 Twilio 开发者宣传员。</p>
<p>您的 Swift 代码表明您的电话号码现在在服务器上是硬编码的。正如 Robert 所说，问题是当您从 Twilio 获得回调到您的 <code>/outbound</code> 端点时，您正在使用 REST API 生成调用。</p>
<p>实际发生的情况是，当您 <a href="https://www.twilio.com/docs/quickstart/client/ios" rel="noreferrer noopener nofollow">generate the call in Swift</a>在启动应用程序调用的设备上。然后，Twilio 向您的 <code>/outbound</code> 端点发出 HTTP 请求，以查看如何处理该调用。所以，而不是 <a href="https://www.twilio.com/docs/api/rest/making-calls" rel="noreferrer noopener nofollow">generating a new call with the REST API</a> ，您需要回复 <a href="https://www.twilio.com/docs/api/twiml" rel="noreferrer noopener nofollow">TwiML</a>告诉 Twilio 下一步如何处理调用。</p>
<p>在这种情况下，听起来您正在尝试 <a href="https://www.twilio.com/docs/api/twiml/dial" rel="noreferrer noopener nofollow">dial</a>直接上另一个 <a href="https://www.twilio.com/docs/api/twiml/number" rel="noreferrer noopener nofollow">number</a> .为此，您应该尝试以下响应:</p>
<pre><code>@app.route('/outgoing', methods=['GET', 'POST'])
def outgoing():
from_number =
to_number =
resp = twilio.twiml.Response()
with resp.dial(callerId=from_number) as r:
r.number(to_number)
return str(resp)
</code></pre>
<p>如果有帮助，请告诉我。</p></p>
<p style="font-size: 20px;">关于python - Twilio 可编程语音调用立即完成，我们在Stack Overflow上找到一个类似的问题：
<a href="https://stackoverflow.com/questions/42673441/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/42673441/
</a>
</p>
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