android - mediaplayer 播放互联网 mp3 ：java.io.FileNotFoundException: No content provider

Question

这是我的代码：

public void playSound(String uri,int flag)
    {
        Uri uri1=Uri.parse(uri);
        ContestVoteFragment fragment = (ContestVoteFragment) fragments.get(0);
        if (fragment.playSound(flag))
        {
            if (mediaPlayer==null)
            {
                mediaPlayer=new MediaPlayer();
                mediaPlayer.setAudioStreamType(AudioManager.STREAM_MUSIC);
                try {
                    mediaPlayer.setDataSource(uri);
                    mediaPlayer.prepare();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                mediaPlayer.start();

            }else {
                mediaPlayer.stop();
                mediaPlayer.reset();
                mediaPlayer.release();
                mediaPlayer=new MediaPlayer();
            mediaPlayer.setAudioStreamType(AudioManager.STREAM_MUSIC);
                try {
                    mediaPlayer.setDataSource(uri);
                    mediaPlayer.prepare();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                mediaPlayer.start();
            }
        }else {
            if (mediaPlayer!=null)
            {
                if (mediaPlayer.isPlaying())
                {
                    mediaPlayer.stop();
                }
            }
        }
    }
}

但是当我播放互联网 mp3 时，错误消息是无法打开http://p9snltngh.bkt.clouddn.com/Fs3M3Cq_Mkex9AsunJt57N22JQp2 : java.io.FileNotFoundException: No content provider i can't understand because “<a href= “http://p9snltngh.bkt.clouddn.com/FhEbUwGYcmQbXIVMVzd0GyPypDVC” rel="nofollow noreferrer">http://p9snltngh.bkt.clouddn.com/FhEbUwGYcmQbXIVMVzd0GyPypDVC”这个uri可以播放

标签： android

Answer 1

两个链接都有效

public void playSound(String uri){
    MediaPlayer mediaPlayer = new MediaPlayer();
    try {
        mediaPlayer.setDataSource(uri);
        mediaPlayer.prepare();
        mediaPlayer.start();
    } catch (IOException e) {
        e.printStackTrace();
    } 
}