我正在尝试在Android视频 View 中播放视频。但是，当我运行代码时，它将打开媒体播放器，但不播放视频，并显示一条错误消息，并且日志目录显示“无法在尝试服务器端的客户端打开文件”。这是我的代码:
public class ExercisesActivity extends Activity {
VideoView videoView;
String videoUrl;
public boolean flage = true;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_exercises);
videoView = (VideoView)findViewById(R.id.videoView);
new performBackgroundTask(ExercisesActivity.this).execute();
}
public void onPause ()
{
super.onPause();
videoView.stopPlayback();
}
private class performBackgroundTask extends AsyncTask < Void, Void, Void >
{
private ProgressDialog Dialog;
public performBackgroundTask(ExercisesActivity context)
{
Dialog = new ProgressDialog(context);
}
protected void onPreExecute()
{
this.Dialog.setMessage("Loading Video...");
this.Dialog.show();
}
protected Void doInBackground(Void... params)
{
try
{
String url = "http://www.youtube.com/watch?v=m5_AKjDdqaU";
videoUrl = getUrlVideoRTSP(url);
Log.e("Video url for playing=========>>>>>", videoUrl);
}
catch (Exception e)
{
Log.e("Login Soap Calling in Exception", e.toString());
}
return null;
}
protected void onPostExecute(Void unused)
{
if(!flage){
if(Dialog.isShowing())
{
Dialog.dismiss();
}
}
else
{
videoView.setVideoURI(Uri.parse(videoUrl));
MediaController mc = new MediaController(ExercisesActivity.this);
videoView.setMediaController(mc);
videoView.requestFocus();
videoView.start();
mc.show();
}
if(Dialog.isShowing())
{
Dialog.dismiss();
}
}
}
public static String getUrlVideoRTSP(String urlYoutube)
{
try
{
String gdy = "http://gdata.youtube.com/feeds/api/videos/";
DocumentBuilder documentBuilder
=DocumentBuilderFactory.newInstance().newDocumentBuilder();
String id = extractYoutubeId(urlYoutube);
URL url = new URL(gdy + id);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
Document doc = documentBuilder.parse(connection.getInputStream());
Element el = doc.getDocumentElement();
NodeList list = el.getElementsByTagName("media:content");///media:content
String cursor = urlYoutube;
for (int i = 0; i < list.getLength(); i++)
{
Node node = list.item(i);
if (node != null)
{
NamedNodeMap nodeMap = node.getAttributes();
HashMap<String, String> maps = new HashMap<String, String>();
for (int j = 0; j < nodeMap.getLength(); j++)
{
Attr att = (Attr) nodeMap.item(j);
maps.put(att.getName(), att.getValue());
}
if (maps.containsKey("yt:format"))
{
String f = maps.get("yt:format");
if (maps.containsKey("url"))
{
cursor = maps.get("url");
}
if (f.equals("1"))
return cursor;
}
}
}
return cursor;
}
catch (Exception ex)
{
Log.e("Get Url Video RTSP Exception======>>", ex.toString());
}
return urlYoutube;
}
protected static String extractYoutubeId(String url) throws MalformedURLException
{ String id = null;
try
{
String query = new URL(url).getQuery();
if (query != null)
{
String[] param = query.split("&");
for (String row : param)
{
String[] param1 = row.split("=");
if (param1[0].equals("v"))
{
id = param1[1];
}
}
}
else
{
if (url.contains("embed"))
{
id = url.substring(url.lastIndexOf("/") + 1);
}
}
}
catch (Exception ex)
{
Log.e("Exception", ex.toString());
}
return id;
}
}
日志猫显示:
01-07 20:36:40.388: E/Video url for playing=========>>>>>(378):
rtsp://r4---sn-4g57kue6.c.youtube.com/CiILENy73wIaGQmlqd0wKsCfmxMYDSANFEgGUgZ2aWRlb3MM
/0/0/0/video.3gp
01-07 20:36:40.398: D/MediaPlayer(378): Couldn't open file on client side, trying
server side
最佳答案
RTSP链接不适用于Youtube API v 2.0提供的任何数据。而且我认为它不会得到解决。
https://code.google.com/p/gdata-issues/issues/detail?id=4858
关于android - 无法在客户端打开文件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20976238/