如何发挥从网址视频 [英] how to play video from url
本文介绍了如何发挥从网址视频的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我是初学android开发,并尝试从链接播放视频。但它给错误对不起,我们无法播放此视频
。我试过很多的联系,但所有链接的显示同样的错误。
我的code是以下
公共类VideoDemo延伸活动{
私有静态最后弦乐路径=http://demo.digi-corp.com/S2LWebservice/Resources/SampleVideo.mp4;
私人VideoView视频;
私人的MediaController CTLR;
@覆盖
公共无效的onCreate(包冰柱){
super.onCreate(冰柱);
。getWindow()和setFormat(PixelFormat.TRANSLUCENT);
的setContentView(R.layout.videoview);
视频=(VideoView)findViewById(R.id.video);
video.setVideoPath(路径);
CTLR =新的MediaController(本);
ctlr.setMediaPlayer(视频)
video.setMediaController(CTLR);
video.requestFocus();
}
}
LogCat中显示以下错误信息:
04-12 15:04:54.245:ERROR / PlayerDriver(554):HandleErrorEvent:PVMFErrTimeout
解决方案
它是与你的链接和内容。请尝试以下两个链接:
字符串路径=http://www.ted.com/talks/download/video/8584/talk/761;
字符串PATH1 =http://commonsware.com/misc/test2.3gp;
开放的我们的uri = Uri.parse(路径1);
VideoView视频=(VideoView)findViewById(R.id.VideoView01);
video.setVideoURI(URI);
video.start();
开始路径1,它是一个小重量轻的视频流,然后尝试在路径,这是一个更高的分辨率比路径1,一个完美的高分辨率的手机。
I am beginner in android development and try to play video from link. But it's giving error "sorry,we can't play this video"
. I tried so many links but for all links its show same error.
My code is the following
public class VideoDemo extends Activity {
private static final String path ="http://demo.digi-corp.com/S2LWebservice/Resources/SampleVideo.mp4";
private VideoView video;
private MediaController ctlr;
@Override
public void onCreate(Bundle icicle) {
super.onCreate(icicle);
getWindow().setFormat(PixelFormat.TRANSLUCENT);
setContentView(R.layout.videoview);
video = (VideoView) findViewById(R.id.video);
video.setVideoPath(path);
ctlr = new MediaController(this);
ctlr.setMediaPlayer(video);
video.setMediaController(ctlr);
video.requestFocus();
}
}
Logcat shows following error message:
04-12 15:04:54.245: ERROR/PlayerDriver(554): HandleErrorEvent: PVMFErrTimeout
解决方案
It has something to do with your link and content. Try the following two links:
String path="http://www.ted.com/talks/download/video/8584/talk/761";
String path1="http://commonsware.com/misc/test2.3gp";
Uri uri=Uri.parse(path1);
VideoView video=(VideoView)findViewById(R.id.VideoView01);
video.setVideoURI(uri);
video.start();
Start with "path1", it is a small light weight video stream and then try the "path", it is a higher resolution than "path1", a perfect high resolution for the mobile phone.
这篇关于如何发挥从网址视频的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文