在 VideoView 中播放流,将 url 转换为 rtsp [英] play streaming in VideoView, convert url to rtsp

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问题描述

我需要以相同的布局播放 YouTube 视频和录制视频.

I need to play youtube video and record video in the same layout.

为了执行此操作,我搜索了 youtube api,发现 android 版本需要高于 2.2,这可以,但是,我想使用 VideoView.

To perform this I search for youtube api and found that android version need to be higher than 2.2, this ok but, i want to use VideoView.

我在这里看到了一些关于此问题的帖子,并最终决定使用此代码在 VideoView 中观看视频.

I saw some post here about this issue and decide eventually to use this code to watch video in VideoView.

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,但仅使用 rtsp 链接,如下所示:

this code works, but just with rtsp link like this :

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,因此我需要代码将 url 转换为 RTSP,我无法手动执行此操作,我检查了一些代码,但它们都不起作用...

I have multiple link in url therefore I need code to convert url to RTSP, I can't do this manually, I check some code and all of them do not work...

我试试这个:从这里开始如何获取 RTSP 网址?

I try this: from here 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;
}

我像这样使用上面的方法:

I use above method like this :

getUrlVideoRTSP(current_url)

当要测试的 currnt_url 是:

when currnt_url to test is :

current_url = "http://m.youtube.com/#/watch?v=FlsBObg-1BQ"

<小时>

我尝试了这段代码,但它不起作用


I try this code and it does not work

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 ;
}   













}

"videoView.setVideoURI(Uri.parse(rstp_url)); "行中的代码崩溃,因为空异常我查了一下,发现代码变量是 =tFXS9krT2VY ,好吧..但是 rstp_url 变量为 null

the code crash in " videoView.setVideoURI(Uri.parse(rstp_url)); " line because null exeption i chack this and discover that code variable is =tFXS9krT2VY , ok.. but rstp_url variable in null

UdayKiran 写了这个回答类似的 Q ,有人可以解释他的意思吗?我不明白他的回答

UdayKiran wrote this answer to similar Q , someone could explain what he means? i dont understand his answer

他的回答:

Element rsp = (Element)entry.getElementsByTagName("media:content").item(1);

                              String anotherurl=rsp.getAttribute("url");

只有在 gdata api 中,我们才能获得此类链接:rtsp://v3.cache7.c.youtube.com/CiILENy73wIaGQlOCTh0GvUeYRMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp

In gdata api only we are getting this type of links : rtsp://v3.cache7.c.youtube.com/CiILENy73wIaGQlOCTh0GvUeYRMYDSANFEgGUgZ2aWRlb3MM/0/0/0/video.3gp

这些正在 VideoView 中播放.

These are playing in VideoView.

最后我不使用此代码和视频视图进行蒸

finally i dont use this code and videoview for steaming

我使用 youtube android api,它是从 android 2.2 开始工作,而不是像我一样从 4.2 开始在我的 Q 中写道,它是 mastake.

i use youtube android api , it's work from android 2.2 not from 4.2 like i wrote in my Q , it was mastake.

使用rtsp的结果是视频质量很差,需要处理纵横比.

the result of using rtsp it's poor video quality and need to deal with aspect ratio.

推荐答案

以下对我有用:
在您的情况下,代码 = FlsBObg-1BQ.
你会得到很多网址,我选择返回最好的质量.

The following works for me:
code = FlsBObg-1BQ in your case.
You will get lots of urls, and I choose to return the best quality.

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.

getJsonString() method.

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.

Don't forget to wrap it inside async task because of a network request.

这篇关于在 VideoView 中播放流,将 url 转换为 rtsp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!

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