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从URL流式传输视频并在VideoView中显示

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从"URL"流式传输视频我没什么问题 . 我正在使用 "Camera Streaming" 应用程序打开相机并给我一个像 "192.168.0.??:8080" 的URL . 我可以在我的电脑上在浏览器中传输该视频 .

现在我的问题是 how can I stream that video in my application into VideoView .

I tried opening video in videoview using "setVideoURI()" 但是它说的是"Can't play this video."并且在logcat中给我错误,如:

1)D / VideoView:错误:1,-2147483648

2)E / MediaPlayer:错误(1,-2147483648)

3)D / MediaPlayer:无法在客户端打开文件,尝试服务器端

你能帮我在VideoView中播放视频吗?

提前致谢! Darshan Popat

1 回答

  • 0

    试试这个代码,它的出色工作:

    public class PlayVideo extends Activity
    {
    
    
     private String videoPath ="url";
    
     private static ProgressDialog progressDialog;
     String videourl;  
        VideoView videoView ;
    
    
     protected void onCreate(Bundle savedInstanceState)
     {
    
      super.onCreate(savedInstanceState);
      setContentView(R.layout.play_video);
    
       videoView = (VideoView) findViewById(R.id.videoView);
    
    
       progressDialog = ProgressDialog.show(PlayVideo.this, "", "Buffering video...", true);
       progressDialog.setCancelable(true);  
    
    
          PlayVideo();
    
     }
     private void PlayVideo()
     {
      try
           {      
                  getWindow().setFormat(PixelFormat.TRANSLUCENT);
                  MediaController mediaController = new MediaController(PlayVideo.this);
                  mediaController.setAnchorView(videoView);           
    
                   Uri video = Uri.parse(videoPath );             
                   videoView.setMediaController(mediaController);
                   videoView.setVideoURI(video);
                   videoView.requestFocus();              
                   videoView.setOnPreparedListener(new OnPreparedListener()
                   {
    
                       public void onPrepared(MediaPlayer mp)
                       {                  
                           progressDialog.dismiss();     
                           videoView.start();
                       }
                   });           
    
    
                }
           catch(Exception e)
           {
                    progressDialog.dismiss();
                    System.out.println("Video Play Error :"+e.toString());
                    finish();
           }   
    
     }
    }
    

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