我需要以相同的布局播放youtube视频和录制视频。
要执行此操作，我搜索youtube api，发现android版本需要高于2.2，这没关系，但是，我想使用VideoView。
我在这里看到一些关于此问题的帖子，并决定最终使用此代码在VideoView中观看视频。
videoView = (VideoView) findViewById(R.id.your_video_view);
Log.d(TAG,getUrlVideoRTSP(current_url) + " id yotube1 " );
//here type the url...
videoView.setVideoURI(Uri.parse(getUrlVideoRTSP(current_url)));
videoView.setMediaController(new MediaController(this)); //sets MediaController in the video view
videoView.requestFocus();//give focus to a specific view
videoView.start();//starts the video
此代码有效，但只需使用 rtsp 链接：
String exemple = "rtsp://v4.cache3.c.youtube.com/CjYLENy73wIaLQlW_ji2apr6AxMYDSANFEIJbXYtZ29vZ2xlSARSBXdhdGNoYOr_86Xm06e5UAw=/0/0/0/video.3gp";
我在url中有多个链接，因此我需要将url转换为RTSP的代码，我不能手动执行此操作，我检查一些代码并且所有代码都不起作用...
我试试这个：从这里开始 How to get RTSP URL?
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;
}
我使用上面这样的方法：
getUrlVideoRTSP(current_url)
当要测试的currnt_url是：
current_url = "http://m.youtube.com/#/watch?v=FlsBObg-1BQ"
我尝试使用此代码并且无效
private class Syncyoutube extends AsyncTask <Void , Void , Void>{
@Override
protected void onPostExecute(Void result) {
// TODO Auto-generated method stub
super.onPostExecute(result);
/**
videoView.setMediaController(new MediaController(this)); //sets MediaController in the video view
// MediaController containing controls for a MediaPlayer
videoView.requestFocus();//give focus to a specific view
videoView.start();//starts the video
*/
}
public String getRstpLinks(String code){
String[] urls = new String[3];
String link = "http://gdata.youtube.com/feeds/api/videos/" + code + "?alt=json";
String json = getJsonString(link); // here you request from the server
try {
JSONObject obj = new JSONObject(json);
String entry = obj.getString("entry");
JSONObject enObj = new JSONObject(entry);
String group = enObj.getString("media$group");
JSONObject grObj = new JSONObject(group);
String content = grObj.getString("media$content");
JSONObject cntObj = new JSONObject(group);
JSONArray array = grObj.getJSONArray("media$content");
for(int j=0; j<array.length(); j++){
JSONObject thumbs = array.getJSONObject(j);
String url = thumbs.getString("url");
urls[j] = url;
Log.d(TAG, url);
//data.setThumbUrl(thumbUrl);
}
Log.v(TAG, content);
} catch (Exception e) {
Log.e(TAG, e.toString());
urls[0] = urls[1] = urls[2] = null;
}
return urls[2];
}
public String getJsonString(String url){
Log.e("Request URL", url);
StringBuilder buffer = new StringBuilder();
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet( url );
HttpEntity entity = null;
try {
HttpResponse response = client.execute(request);
if( response.getStatusLine().getStatusCode() == 200 ){
entity = response.getEntity();
InputStream is = entity.getContent();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line = null;
while( (line = br.readLine() )!= null ){
buffer.append(line);
}
br.close();
}
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}finally{
try {
entity.consumeContent();
} catch (Exception e) {
Log.e(TAG, "Exception = " + e.toString() );
}
}
return buffer.toString();
}
@Override
protected Void doInBackground(Void... params) {
// TODO Auto-generated method stub
code = id_current_url(current_url);
//here type the url...
String rstp_url = getRstpLinks(code);
videoView.setVideoURI(Uri.parse(rstp_url));
// the code crech in this line because null exeption
// i chack this and discover that code variable is =tFXS9krT2VY , ok..
// but rstp_url variable in null
Log.d(TAG,getRstpLinks(code) + " idan id yotube1 " );
return null;
}
}
public String id_current_url (String url) {
String c_id = null ;
c_id = url.substring((url.lastIndexOf("=")), url.length());
return c_id ;
}
}
代码崩溃在“videoView.setVideoURI（Uri.parse（rstp_url））;”行因为null exeption 我查这个并发现代码变量= tFXS9krT2VY，好吧.. 但rstp_url变量为null
他的回答：
Element rsp = (Element)entry.getElementsByTagName("media:content").item(1);
String anotherurl=rsp.getAttribute("url");
在gdata api中我们只有这种类型的链接：rtsp：//v3.cache7.c.youtube.com/CiILENy73wIaGQlOCTh0GvUeYRMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp
这些是在VideoView中播放的。
最后我不使用此代码和视频浏览
我使用youtube android api，它的工作来自android 2.2，而不是像我这样的4.2 在我的Q中写道，这是奶油。
使用rtsp的结果是视频质量差，需要处理宽高比。
答案 0 :(得分：7)
以下为我工作：
在你的情况下代码= FlsBObg-1BQ。
你会得到很多网址，我选择返回最好的网站。
private String getRstpLinks(String code){
String[] urls = new String[3];
String link = "http://gdata.youtube.com/feeds/api/videos/" + code + "?alt=json";
String json = getJsonString(link); // here you request from the server
try {
JSONObject obj = new JSONObject(json);
String entry = obj.getString("entry");
JSONObject enObj = new JSONObject(entry);
String group = enObj.getString("media$group");
JSONObject grObj = new JSONObject(group);
String content = grObj.getString("media$content");
JSONObject cntObj = new JSONObject(group);
JSONArray array = grObj.getJSONArray("media$content");
for(int j=0; j<array.length(); j++){
JSONObject thumbs = array.getJSONObject(j);
String url = thumbs.getString("url");
urls[j] = url;
Log.d(TAG, url);
//data.setThumbUrl(thumbUrl);
}
Log.v(TAG, content);
} catch (Exception e) {
Log.e(TAG, e.toString());
urls[0] = urls[1] = urls[2] = null;
}
return urls[2];
}
getJsonString（）方法。
public static String getJsonString(String url){
Log.e("Request URL", url);
StringBuilder buffer = new StringBuilder();
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet( url );
HttpEntity entity = null;
try {
HttpResponse response = client.execute(request);
if( response.getStatusLine().getStatusCode() == 200 ){
entity = response.getEntity();
InputStream is = entity.getContent();
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line = null;
while( (line = br.readLine() )!= null ){
buffer.append(line);
}
br.close();
}
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}finally{
try {
entity.consumeContent();
} catch (Exception e) {
Log.e(TAG, "Exception = " + e.toString() );
}
}
return buffer.toString();
}
由于网络请求，不要忘记将其包装在异步任务中。