Сделать запрос к сайту Java [закрыт]

import org.json.JSONArray;
import org.json.JSONObject;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
public class Main {
    public static void main(String[] args) throws IOException {
        URL url = new URL("http://localhost:3000/home");//your web-site
        HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
        urlConnection.setRequestMethod("GET");
        InputStream answer = urlConnection.getInputStream();
        byte[] bytes = new byte[answer.available()];
        answer.read(bytes);
        String s = new String(bytes);
        System.out.println(s);
//If the request return HTML we can write it in some file
        File file = new File(System.getProperty("user.dir") + File.separator + "index.html");
        file.createNewFile();
        FileOutputStream fileOutputStream = new FileOutputStream(file);
        fileOutputStream.write(bytes);
        fileOutputStream.close();
//If the request return JSON we can parse it
        JSONArray jsonArray = new JSONArray(s);
        for (int i = 0; i < jsonArray.length(); i++) {
            JSONObject o = (JSONObject) jsonArray.get(i);
            String man = o.getString("man");
            int age = o.getInt("age");
            System.out.println(new User(man,age));
        }
    }
}
//or
JSONObject jo = (JSONObject) obj;
String man = (String) jo.get("man"); 
int age = (int) jo.get("age");