Question

I want to send this url from an android class to my servlet. I have written the code for servlet where it catches the values of parameter but I am not able to send this url. What is the code to do this?

    double lat = Double.parseDouble(coordinates[0]);
    double lng = Double.parseDouble(coordinates[1]);
    URL url;
    try {
    URL url = new URL("http://localhost:8080/ExtraServ/AssessmentServlet?param1="+lat+lng);

    } catch (MalformedURLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

My servlet code:

            req.setCharacterEncoding("UTF-8");
            resp.setCharacterEncoding("UTF-8");
            final String par1 =  req.getParameter("param1");
            final String par2 = req.getParameter("param2");
            FileWriter fstream = new FileWriter("C:\\Users\\Hitchhiker\\Desktop\\out2.txt");
            BufferedWriter out = new BufferedWriter(fstream);
            out.write(par1);
            out.append(" ");
            out.append(par2);
            out.close();

Answer 1

localhost will be your own device (127.0.0.1).
you’ll have to choose your servlet’s ip and be connected to the same network.

the connection part is missing in your code:

URLConnection urlConnection = url.openConnection();
urlConnection.connect();