我将整个屏幕拆分为两个视频视图,以便在各自的视频视图上流式传输两个不同的URI,但是视频URI在两个视频视图中加载一个URI。
在vv_left和vv_right流下面都有第一个uri(“xxx.xxx.xx.xxx:1234”)
private void generateUI(){
....
....
loadUri(vv_left,"xxx.xxx.xx.xxx:1234");
loadUri(vv_right,"xxx.xxx.xx.xxx:2345");
....
}
private void loadUri(VideoView vv,String url){
Uri uri= Uri.parse(url);
vv.setVideoURI(uri);
vv.requestFocus();
vv.start();
}发布于 2014-04-28 18:43:40
两个VideoViews都需要分别使用MediaController和请求视频uri。这就是:
VideoView videoView1 = (VideoView) findViewById(R.id.videoView1);
VideoView videoView2 = (VideoView) findViewById(R.id.videoView2);
try
{
MediaController mediaController = new MediaController(this);
mediaController.setAnchorView(videoView);
videoView.setMediaController(mediaController);
videoView.setVideoURI(Uri.parse(xxx.xxx.xx.xxx:1234));
videoView.setOnPreparedListener(new OnPreparedListener()
{
@Override
public void onPrepared(MediaPlayer mp)
{
videoView.start();
videoView.requestFocus();
}
});
}
catch (Exception e)
{
System.out.println("Exception caught: "+e.getMessage());
}
// do the same above try catch code for videoView2 and url2希望这能对你有所帮助。
https://stackoverflow.com/questions/23332832
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