Question

我希望能够在最多按两次按钮时播放音频。按两次按钮后，即使按下也不再播放音频。我目前有这个代码，但它不起作用，我不确定我哪里出错：

var soundFileURLRef: NSURL!
var audioPlayer = AVAudioPlayer?()
var audioCounter = 0

override func viewDidLoad() {
    super.viewDidLoad()

    // setup for audio
    let playObjects = NSBundle.mainBundle().URLForResource("mathsinfo", withExtension: "mp3")

    self.soundFileURLRef = playObjects

    do {
        audioPlayer = try AVAudioPlayer(contentsOfURL: soundFileURLRef)
    } catch _ {
        audioPlayer = nil
    }
    audioPlayer?.delegate = self
    audioPlayer?.prepareToPlay()
}



//function for counting times audio is played
func countAudio() {
    if ((audioPlayer?.play()) != nil) {
        ++audioCounter
    }

}

//MARK: Actions

//action for button playing the audio
@IBAction func playMathsQuestion(sender: AnyObject) {
    countAudio()
    if audioCounter < 2 {
       audioPlayer?.play()
    }
}

Answer 1

您的代码中的countAudio函数会因为您调用audioPlayer?.play()而异步播放声音。仅当audioCounter为零时，audioPlayer变量才会递增。试试这个版本

var soundFileURLRef: NSURL!
var audioPlayer: AVAudioPlayer?
var audioCounter = 0

override func viewDidLoad() {
    super.viewDidLoad()

    // setup for audio
    let playObjects = NSBundle.mainBundle().URLForResource("mathsinfo", withExtension: "mp3")

    self.soundFileURLRef = playObjects

    audioPlayer = try? AVAudioPlayer(contentsOfURL: soundFileURLRef)
    audioPlayer?.delegate = self
    audioPlayer?.prepareToPlay()
}


//MARK: Actions

//action for button playing the audio
@IBAction func playMathsQuestion(sender: AnyObject) {
    if (audioPlayer != nil) {
        if (audioCounter < 2 && audioPlayer!.play()) {
            ++audioCounter
        }
    }
}

您可以在此处看到我们首先检查以确保audioPlayer不是零。在此之后，我们只在audioPlayer!.play()时进行audioCounter < 2调用。对play()的调用返回一个布尔值，当音频成功排队等待播放时，该布尔值为真（参见documentation），在这种情况下，我们增加audioCounter。

我还使用audioPlayer版本的try简化了try?的初始化，当被调用者抛出时，返回nil。

使用Swift中的AV Audio Player播放条件音频

1 个答案: