AVPlayerItem.duration即使在player.status == .readyToPlay之后也返回为NaN [英] AVPlayerItem.duration returns as NaN even after player.status == .readyToPlay

问题描述

我有一个在player.status上使用KVO的实现，非常类似于

I have an implementation using KVO on player.status much like the suggestion documented here.

以下是相关位:

var player: AVPlayer?
var url = URL(string: "some video url")!

override func viewDidLoad() {
    super.viewDidLoad()

    initializeVideoPlayer()
}

func initializeVideoPlayer() {
    let playerItem = AVPlayerItem(url: url)
    self.player = AVPlayer(playerItem: playerItem)
    let playerLayer = AVPlayerLayer(player: player)

    playerLayer.frame = self.view.layer.bounds
    videoView.layer.addSublayer(playerLayer)

    player?.addObserver(self, forKeyPath: "status", options: .new, context: nil)
}

override func observeValue(forKeyPath keyPath: String?, of object: Any?, change: [NSKeyValueChangeKey : Any]?, context: UnsafeMutableRawPointer?) {
    if keyPath == "status" {
        if player?.status == .readyToPlay {
            print(player?.currentItem?.duration)
            print(CMTimeGetSeconds((player?.currentItem?.duration)!))

//          let videoLength = CMTimeGetSeconds((player?.currentItem?.duration)!)
//          videoProgressSlider.maximumValue = Float(videoLength)
        }
    }
}

尽管状态正常，但返回的仍然是NaN.

Despite the status, what's returned is still NaN.

解决方案:

因此，引起我注意的是AVPlayer和AVPlayerItem都具有状态属性.通过像我一样，通过KVO'ing AVPlayerItem.status属性而不是AVPlayer.status属性，可以正确返回持续时间.

So it's come to my attention that both AVPlayer AND AVPlayerItem have status properties. By KVO'ing the AVPlayerItem.status property as opposed to the AVPlayer.status like I had done, the duration will be returned properly.

推荐答案

因此，引起他注意的是AVPlayer和AVPlayerItem都具有状态属性.通过像我一样，通过KVO'ing AVPlayerItem.status属性而不是AVPlayer.status属性，可以正确返回持续时间.

So it's come to his attention that both AVPlayer AND AVPlayerItem have status properties. By KVO'ing the AVPlayerItem.status property as opposed to the AVPlayer.status like I had done, the duration will be returned properly.

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