Grab hidden extension from Uri

Question

I have an app to watch and download videos online, my code detects mimetype by the url to use as the video extension downloaded.

But some sites have the hidden extension, such as this url: http://www.animeplus.org/inc/video.inc.php?&file=NSwzOWE0YWQ5ODhhNWQ4NWEx&token=rm_uJeMTUxNjA4OTM0OQ== Videoview can play the video, but Downloader takes mimetype as php and saves the video in the php extension, this is the code snippet I use to get the extension:

String mimeType =  '.' + MimeTypeMap.getFileExtensionFromUrl(url.toString());

In this case he takes the video extension as he should: https://www.blogger.com/video-play.mp4?contentId=6bc99b4dd120f980

So how to solve this problem?

Answer 1

Use of the ways to capture the link is by accessing the page with the class Urlconnection and then returning the Header Response: Location . Ex:

import java.net.HttpURLConnection;
import java.net.URL;

public class HelloWorld
{
  public static void main(String[] args)
  {
    try {
        /* Especifica a URL */
        URL url = new URL("http://www.animeplus.org/inc/video.inc.php?&file=NSwzOWE0YWQ5ODhhNWQ4NWEx&token=rm_uJeMTUxNjA4OTM0OQ==");

        /* Abre a conexão */
        HttpURLConnection connection = (HttpURLConnection) url.openConnection();

        /* Define um User-Agent. É necessário, caso não adicione essa linha, você receberá um erro 403 (Não Autorizado) */
        connection.setRequestProperty("user-agent", "Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/63.0.3239.132 Safari/537.36");

        /* Aqui você captura o Header da resposta. */
        System.out.println( connection.getHeaderField("Location") );

    } catch (Exception e) {
      System.out.println("Erro: " + e.getMessage());
    }
  }
}