我创建了一个函数,它以特定的时间间隔触发通知服务.通知可以是文本,图像或视频……
现在,对于视频我首先下载它然后播放它需要更多的时间…所以有任何机制,我可以直接从远程网址播放视频???
请帮我 …
我迫切需要答案……
提前致谢 ……..
看看我的代码片段::
public void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.notificationvideo); mVideoView = (VideoView) findViewById(R.id.video); //pd=ProgressDialog.show(this,"Loading...","Please Wait...",true,false); playVideo(); //pd.dismiss(); img_back = (ImageView) findViewById(R.id.img_back); img_back.setonClickListener(new View.OnClickListener() { public void onClick(View v) { Intent int_back=new Intent(NotificationsVideoActivity.this,MyChannelsActivity.class); startActivity(int_back); finish(); } }); } private void playVideo() { try { path = getIntent().getStringExtra("url"); Log.v(TAG,"path: " + path); if (path == null || path.length() == 0) { Toast.makeText(NotificationsVideoActivity.this,"File URL/path is empty",Toast.LENGTH_LONG).show(); } else { // If the path has not changed,just start the media player if (path.equals(current) && mVideoView != null) { mVideoView.start(); mVideoView.requestFocus(); return; } current = path; mVideoView.setVideoPath(getDataSource(path)); mVideoView.start(); mVideoView.requestFocus(); } } catch (Exception e) { Log.e(TAG,"error: " + e.getMessage(),e); if (mVideoView != null) { mVideoView.stopPlayback(); } } } private String getDataSource(String path) throws IOException { if (!URLUtil.isNetworkUrl(path)) { return path; } else { URL url = new URL(path); URLConnection cn = url.openConnection(); cn.connect(); InputStream stream = cn.getInputStream(); if (stream == null) throw new RuntimeException("stream is null"); File temp = File.createTempFile("mediaplayertmp","mp4"); temp.deleteOnExit(); String tempPath = temp.getAbsolutePath(); FileOutputStream out = new FileOutputStream(temp); byte buf[] = new byte[128]; //byte buf[] = new byte[8192]; do { int numread = stream.read(buf); if (numread <= 0) break; out.write(buf,numread); } while (true); try { stream.close(); } catch (IOException ex) { Log.e(TAG,"error: " + ex.getMessage(),ex); } return tempPath; } }
解决方法
试试这个 –
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();
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