4

通过使用 Twilio ( VideoCallKitQuickStart ) 提供的示例视频通话应用程序之一，我试图通过向应用程序发送 VoIP 通知来触发来电。但该应用程序不会触发来电。我还尝试在发送 VoIP 通知时保持应用程序打开并且应用程序崩溃，方法是抛出以下异常

NSInvalidArgumentException：尝试为密钥有效负载插入非属性列表对象“PKPushPayload：0x16e44af0”

当收到 VoIP 通知时，有人可以帮助我或指出正确的方向，了解如何在应用程序中触发来电。

下面是我在 ViewController.swift 文件中的代码

 func pushRegistry(registry: PKPushRegistry!, didReceiveIncomingPushWithPayload payload: PKPushPayload!, forType type: String!) {
        // Process the received push

        self.reportIncomingCall(uuid: UUID(), roomName: "testRoom", completion: nil)
    } 

func reportIncomingCall(uuid: UUID, roomName: String?, completion: ((NSError?) -> Void)? = nil) {

    let callHandle = CXHandle(type: .generic, value: roomName ?? "")
    let callUpdate = CXCallUpdate()
    callUpdate.remoteHandle = callHandle
    callUpdate.supportsDTMF = false
    callUpdate.supportsHolding = true
    callUpdate.supportsGrouping = false
    callUpdate.supportsUngrouping = false
    callUpdate.hasVideo = true

    callKitProvider.reportNewIncomingCall(with: uuid, update: callUpdate) { error in
        if error == nil {
            NSLog("Incoming call successfully reported.")
        } else {
            NSLog("Failed to report incoming call successfully: \(error?.localizedDescription).")
        }
        completion?(error as? NSError)
    }
}
4

3 回答 3

1

发布迟到的答案，但它可能对某人有帮助。

以下代码我处理语音来电。

func pushRegistry(_ registry: PKPushRegistry, didReceiveIncomingPushWith payload: PKPushPayload, for type: PKPushType) {
    NSLog("pushRegistry:didReceiveIncomingPushWithPayload:forType:")
    print(payload)
    if (type == PKPushType.voIP) {
        TwilioVoice.handleNotification(payload.dictionaryPayload, delegate: self)

        pushKitPushReceivedWithPayload(payload: payload)
    }
}
func pushRegistry(_ registry: PKPushRegistry, didReceiveIncomingPushWith payload: PKPushPayload, for type: PKPushType, completion: @escaping () -> Void) {
    NSLog("pushRegistry:didReceiveIncomingPushWithPayload:forType:completion:")

    if (type == PKPushType.voIP) {
        TwilioVoice.handleNotification(payload.dictionaryPayload, delegate: self)

        pushKitPushReceivedWithPayload(payload: payload)
    }

    completion()
}

func pushKitPushReceivedWithPayload(payload: PKPushPayload){
    if UIApplication.shared.applicationState != .active{
        let msgType = payload.dictionaryPayload["twi_message_type"] as? String
        if let messageType = msgType{
            if messageType == "twilio.voice.call"{
                fireLocalNotificationForVoiceCall(didStart: true)
            }else if messageType == "twilio.voice.cancel"{
                fireLocalNotificationForVoiceCall(didStart: false)
            }
        }
    }
}

以下是我添加的调用工具包的委托方法

extension AppDelegate : TVONotificationDelegate, TVOCallDelegate
{
  func callInviteReceived(_ callInvite: TVOCallInvite) 
  {
   if (callInvite.state == .pending) 
   {
        //code
   } 
   else if (callInvite.state == .canceled) 
   {
        //code
   }
  }
  func handleCallInviteReceived(_ callInvite: TVOCallInvite) 
  {
        //code
  }

  func handleCallInviteCanceled(_ callInvite: TVOCallInvite) 
  {
        //code
  }
}

我遵循了 twilio 提供的本教程 - https://github.com/twilio/voice-quickstart-swift

阅读本教程，它将起作用。

于 2018-10-23T06:04:01.227 回答
0

Twilio 开发人员布道者在这里。

我对 iOS 不是特别擅长，但快速查看文档， PKPushRegistryDelegate您的函数定义似乎pushRegistry不正确。

它应该是

func pushRegistry(_ registry: PKPushRegistry, 
    didReceiveIncomingPushWith payload: PKPushPayload, 
    forType type: PKPushType)

也就是说，didReceiveIncomingPushWith而不是didReceiveIncomingPushWithPayload.

forType或者，这与您正在投射的事实有什么关系String吗？

于 2017-01-06T10:57:44.610 回答
0

斯威夫特 3.0

func pushRegistry(_ registry: PKPushRegistry, didReceiveIncomingPushWith payload: PKPushPayload, forType type: PKPushType) {
    NSLog("pushRegistry:didReceiveIncomingPushWithPayload:forType:")

    if (type == PKPushType.voIP) {
        print(payload.dictionaryPayload)
        VoiceClient.sharedInstance().handleNotification(payload.dictionaryPayload, delegate: self)
    }
}

并且请不要在没有修改的情况下对payload进行任何更改，以便SDK从payload中提取来电信息，以便SDK可以通知应用程序有来电

于 2017-04-24T10:11:34.850 回答