根据阵列中每个I时间的持续时间播放快速声音

如何解决根据阵列中每个I时间的持续时间播放快速声音

我希望我的快速代码调用playSound并根据数组playamount中每个项目的持续时间播放声音。因此，我希望用户第一次播放声音10秒钟，然后从头开始播放声音20秒钟，然后播放同一声音30秒钟。因此，声音总是在每次被调用时从头开始。

import UIKit;  import AVFoundation

class ViewController: UIViewController {
    
  

    var player: AVAudioPlayer?

    func playSound() {
        let url = Bundle.mainBundle().URLForResource("rock",withExtension: "mp3")!

        do {
            player = try AVAudioPlayer(contentsOfURL: url)
            guard let player = player else { return }

            player.prepareToPlay()
            player.play()

        } catch let error as NSError {
            print(error.description)
        }
    }
    var playAmount : [Int] = [10,20,30]

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view.
    }


}

解决方法

class ViewController: UIViewController {

    var player: AVAudioPlayer?
    var currentIndex: Int = 0

    func runMusicBox() {
        guard !playAmount.isEmpty else { return }
        currentIndex = 0
        newTimerForIndex(index: 0)
    }
    
    func newTimerForIndex(index: Int) {
        player?.prepareToPlay()
        player?.play()
        Timer.scheduledTimer(withTimeInterval: Double(playAmount[index]),repeats: false) { timer in
            self.player?.stop()
            if self.currentIndex + 1 < self.playAmount.count {
                self.currentIndex += 1
                self.newTimerForIndex(index: self.currentIndex)
            } else {
                self.player?.stop()
            }
        }
    }
    
    func playSound() {
        let url = Bundle.mainBundle().URLForResource("rock",withExtension: "mp3")!

        do {
            player = try AVAudioPlayer(contentsOfURL: url)
            guard let player = player else { return }
            runMusicBox()

        } catch let error as NSError {
            print(error.description)
        }
    }
    var playAmount : [Int] = [10,20,30]

    override func viewDidLoad() {
        super.viewDidLoad()
        // Do any additional setup after loading the view.
    }


}

您好，您可以尝试使用此代码吗？在这里，我有一个计时器，当一个计时器停止时，我们检查数组中是否还有另一个元素，并将运行一个新的计时器。计时器正在工作，但我没有检查播放器。如果它能按预期工作-应该可以为您工作