Question

这是我的代码：

class SomeAudioManager: NSObject
{

    class var sharedInstance: SomeAudioManager{
        struct Static {
            static var onceToken: dispatch_once_t = 0
            static var instance: SomeAudioManager? = nil
        }
        dispatch_once(&Static.onceToken) {
            Static.instance = SomeAudioManager()
        }
        return Static.instance!
    }
    func audioView(songname: NSString,format: NSString)
    {
        let audioPlayer:ava
        audioPlayer=try AVAudioPlayer(contentsOfURL: NSURL(fileURLWithPath: NSBundle.mainBundle().pathForResource(songname, ofType:format)!), fileTypeHint: AVFileTypeMPEGLayer3)
        audioPlayer!.delegate=self;
        self.audioPlayer!.play()
    }
}

AVAudioPlayer属于NSObject，但我无法实现它。

键入时让audioPlayer：AVAudio - ＆gt;它没有显示任何东西。

Answer 1

至于你案例中的单身部分（来自评论）你应该使用这样的东西：

MyAudioPlayer.playFile("Breach", type: "mp3")

用法是：

loaders: [{
        test: /\.jsx?$/,
        loader: 'babel',
        query: {
            presets: ['es2015', 'react']
        }
    }]

Answer 2

这并不是很有意义，但这可以为我编译：

import AVFoundation
class SomeAudioManager: NSObject, AVAudioPlayerDelegate
{
    class var sharedInstance: SomeAudioManager {
        struct Static {
            static var onceToken: dispatch_once_t = 0
            static var instance: SomeAudioManager? = nil
        }
        dispatch_once(&Static.onceToken) {
            Static.instance = SomeAudioManager()
        }
        return Static.instance!
    }
    func audioView(songname: String,format: String) {
        let audioPlayer: AVAudioPlayer

        do {
            audioPlayer = try AVAudioPlayer(contentsOfURL: NSURL(fileURLWithPath: NSBundle.mainBundle().pathForResource(songname, ofType:format)!), fileTypeHint: AVFileTypeMPEGLayer3)
            audioPlayer.delegate = self;
            audioPlayer.play()
        } catch {
            // error
        }
    }
}

所以你需要导入框架，swift中的try-catch是do-try-catch。其他一些语法也失败了。

在Swift BTW中没有以这种方式使用单例。

<强>用法：

class someOtherClass {
    func doSomething() {
        SomeAudioManager().audioView("name_here", format: "format_here")
        SomeAudioManager.sharedInstance.audioView("name_here", format: "format_here")
    }
}

如何在Singleton方法中实现AVAudioPlayer？

2 个答案: