Question

Q:-Now a days I am facing problem in getting IP address of android device using programming. Can any one give ma a code for solving that problem. I have already read lot of thread about it but not getting solid answer from that. Please give me any suggesting about it its appreciated.
Thanks in advance.

Answer 1

The problem is that you cannot know if your current in use network device is really has a public IP. You could however check if this is the case but you need to contact a external server.

In this case we can use http://www.whatismyip.com to check (almost copied from another SO question):

public static InetAddress getExternalIp() throws IOException {
    URL url = new URL("http://automation.whatismyip.com/n09230945.asp");
    URLConnection connection = url.openConnection();
    connection.addRequestProperty("Protocol", "Http/1.1");
    connection.addRequestProperty("Connection", "keep-alive");
    connection.addRequestProperty("Keep-Alive", "1000");
    connection.addRequestProperty("User-Agent", "Web-Agent");

    Scanner s = new Scanner(connection.getInputStream());
    try {
        return InetAddress.getByName(s.nextLine());
    } finally {
        s.close();
    }
}

To check that this ip is bound to one of your network interfaces:

public static boolean isIpBoundToNetworkInterface(InetAddress ip) {

    try {
        Enumeration<NetworkInterface> nets = 
            NetworkInterface.getNetworkInterfaces();

        while (nets.hasMoreElements()) {
            NetworkInterface intf = nets.nextElement();
            Enumeration<InetAddress> ips = intf.getInetAddresses();
            while (ips.hasMoreElements())
                if (ip.equals(ips.nextElement()))
                    return true;
        }
    } catch (SocketException e) {
        // ignore
    }
    return false;
}

Test code:

public static void main(String[] args) throws IOException {

    InetAddress ip = getExternalIp();

    if (!isIpBoundToNetworkInterface(ip))
        throw new IOException("Could not find external ip");

    System.out.println(ip);
}