我想播放mp4视频。所以我试过下面的方法
private void playVideo() {
try {
final String path = s;
Log.v(TAG, "path: " + path);
if (path == null || path.length() == 0) {
Toast.makeText(VideoPlay1.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) {
MediaController mediaController = new MediaController(this);
mediaController.setAnchorView(mVideoView);
Uri video = Uri.parse(getDataSource(path));
Log.e("Uri video",video.toString());
mVideoView.setMediaController(mediaController);
mVideoView.setVideoURI(video);
mVideoView.setVideoPath(getDataSource(path));
mVideoView.setOnPreparedListener(new OnPreparedListener() {
public void onPrepared(MediaPlayer arg0) {
dialog.dismiss();
mVideoView.start();
}
});
mVideoView.requestFocus();
return;
}
current = path;
MediaController mediaController = new MediaController(this);
Uri video = Uri.parse(getDataSource(path));
Log.e("Uri video",video.toString());
mVideoView.setMediaController(mediaController);
mVideoView.setVideoURI(video);
mVideoView.setVideoPath(getDataSource(path));
mVideoView.setOnPreparedListener(new OnPreparedListener() {
public void onPrepared(MediaPlayer arg0) {
dialog.dismiss();
mVideoView.start();
}
});
mVideoView.requestFocus();
}
} catch (Exception e) {
Log.e(TAG, "error123: " + e.getMessage(), e);
if (mVideoView != null) {
mVideoView.stopPlayback();
}
}
}
@Override
public boolean onKeyDown(int keyCode, KeyEvent event) {
if (Integer.parseInt(android.os.Build.VERSION.SDK) < 5
&& keyCode == KeyEvent.KEYCODE_BACK
&& event.getRepeatCount() == 0) {
Log.d("CDA", "onKeyDown Called");
onBackPressed();
}
return super.onKeyDown(keyCode, event);
}
public void onBackPressed() {
Log.d("CDA", "onBackPressed Called");
mVideoView.stopPlayback();
Intent setIntent = new Intent(VideoPlay1.this,VideoPage.class);
startActivity(setIntent);
finish();
return;
}
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", "dat");
temp.deleteOnExit();
String tempPath = temp.getAbsolutePath();
FileOutputStream out = new FileOutputStream(temp);
byte buf[] = new byte[128];
do {
int numread = stream.read(buf);
if (numread <= 0)
break;
out.write(buf, 0, numread);
} while (true);
try {
stream.close();
} catch (IOException ex) {
Log.e(TAG, "error: " + ex.getMessage(), ex);
}
return tempPath;
}
}
上述工作需要更多时间才能播放视频,请告诉我是否还有其他方法。
谢谢。
最诚挚的问候。
答案 0 :(得分:1)
很抱歉,将视频流式传输到移动设备需要时间。您受到连接的限制,而不是代码
答案 1 :(得分:0)
我知道已经有一个已接受的答案,但可能有其他标准可能会有所帮助,尤其是文件的编码。
默认情况下,MP4视频最后都有MOOV ATOM。 “moov原子,也称为电影原子,定义了电影的时间刻度,持续时间,显示特征,以及包含电影中每个轨道信息的子体。” (http://www.adobe.com/devnet/video/articles/mp4_movie_atom.html)。如果您在开头使用moov atom对视频进行编码(可以使用FFMPEG执行此操作),则可以改善您的流媒体体验。