ios - 如何使用 Swift 在 iOS 中为不同的通知响应播放单独的音频？

Question

我有一个场景，我想使用 AVAudioPlayer 播放不同的音频以获得不同的通知响应。

例如，当通知响应为yes时，我想播放audio1，类似地，对于fail，audio2 >。我的代码如下，

func application(application: UIApplication, didReceiveRemoteNotification userInfo: [NSObject : AnyObject], fetchCompletionHandler completionHandler: (UIBackgroundFetchResult) -> Void) {
//Presenting audioController from here
}

audioController.swift

override func viewDidLoad() {
    super.viewDidLoad()
    var player:AVAudioPlayer = AVAudioPlayer()
    if notificationResponse == "yes" {
        audioPath =  NSBundle.mainBundle().pathForResource("audio1", ofType: "mp3")
    } else if notificationResponse == "no" {
        audioPath = NSBundle.mainBundle().pathForResource("audio2", ofType: "mp3")
    }
    player = try AVAudioPlayer(contentsOfURL: NSURL(fileURLWithPath: audioPath!))
    player.numberOfLoops = -1
    player.play()
}

问题是，当两个通知同时到达时， View Controller 会呈现两次并且两个音频都会播放。

当有新的通知到达时，如何停止播放旧的音频？

Answer 1

将player var作为audioController.swift的属性，然后在AppDelegate中将audioController.swift类的对象设置为全局

添加一个检查，检查audioController.swift类的对象是否已经存在，在didReceiveRemoteNotification中呈现之前，检查您正在呈现它的实例上的presentedViewController是否为nil >.

如果它已经呈现，则停止播放器并更改播放器的audioPath，而无需再次呈现audioController。

var audioC : audioController // Its you global var in AppDelegate

//Use this code after adding a check the audioC is already presented or not in didReceiveRemoteNotification
audioC.player.stop() // Stop previous running audio
audioC.player = try AVAudioPlayer(contentsOfURL: NSURL(fileURLWithPath: newAudioPath!)) // newAudioPath will be your new audio files path
audioC.player.play() // Play the new audio.