播放声音的代码非常冗长,如果可能,希望创建AVAudioSession的扩展。我这样做的方法是将对象分配给变量,但是需要有关如何设置此函数的帮助/建议,以便其可重用和优化。
这是我所拥有的:
func playSound(name: String, extension: String = "mp3") {
let sound = NSBundle.mainBundle().URLForResource(name, withExtension: extension)
do {
try AVAudioSession.sharedInstance().setCategory(AVAudioSessionCategoryPlayback)
try AVAudioSession.sharedInstance().setActive(true)
UIApplication.sharedApplication().beginReceivingRemoteControlEvents()
audioPlayer = try AVAudioPlayer(contentsOfURL: sound!)
audioPlayer?.prepareToPlay()
audioPlayer?.play()
} catch { }
}
我想我必须在此函数之外创建
audioPlayer
变量,否则我很难演奏任何东西。也许可以自我约束?我希望使用这样的东西:AVAudioSession.sharedInstance().play("bebop")
最佳答案
直接从您的示例中获取代码,我看到两个选择:
1)是否有任何特定原因要将您作为AVAudioSession的扩展?如果没有,请自己做服务!
class AudioPlayerService {
static let sharedInstance = AudioPlayerService()
var audioPlayer: AVAudioPlayer?
func playSound(name: String, extension: String = "mp3") {
let sound = NSBundle.mainBundle().URLForResource(name, withExtension: extension)
do {
try AVAudioSession.sharedInstance().setCategory(AVAudioSessionCategoryPlayback)
try AVAudioSession.sharedInstance().setActive(true)
UIApplication.sharedApplication().beginReceivingRemoteControlEvents()
audioPlayer = try AVAudioPlayer(contentsOfURL: sound!)
audioPlayer?.prepareToPlay()
audioPlayer?.play()
} catch { }
}
}
2)如果您使用做,则需要将其作为AVAudioSession的扩展,然后看看associated objects
extension AVAudioSession {
private struct AssociatedKeys {
static var AudioPlayerTag = "AudioPlayerTag"
}
var audioPlayer: AVAudioPlayer? {
get {
return objc_getAssociatedObject(self, &AssociatedKeys.AudioPlayerTag) as? AVAudioPlayer
}
set {
if let newValue = newValue {
objc_setAssociatedObject(
self,
&AssociatedKeys.AudioPlayerTag,
newValue as AVAudioPlayer?,
.OBJC_ASSOCIATION_RETAIN_NONATOMIC
)
}
}
}
func playSound(name: String, extension: String = "mp3") {
let sound = NSBundle.mainBundle().URLForResource(name, withExtension: extension)
do {
try AVAudioSession.sharedInstance().setCategory(AVAudioSessionCategoryPlayback)
try AVAudioSession.sharedInstance().setActive(true)
UIApplication.sharedApplication().beginReceivingRemoteControlEvents()
audioPlayer = try AVAudioPlayer(contentsOfURL: sound!)
audioPlayer?.prepareToPlay()
audioPlayer?.play()
} catch { }
}
}