CODEWITHSUNDEEP
javalinuxnetwork-programmingip-address
Bobby Green
Bobby Green

Reputation:

How to get the ip of the computer on linux through Java?

How to get the ip of the computer on linux through Java ?

I searched the net for examples, I found something regarding NetworkInterface class, but I can't wrap my head around how I get the Ip address.

What happens if I have multiple network interfaces running in the same time ? Which Ip address will be returned.

I would really appreciate some code samples.

P.S: I've used until now the InetAddress class which is a bad solution for cross-platform applications. (win/Linux).

Upvotes: 12

Views: 28498

Answers (6)

Adir Dayan
Adir Dayan

Reputation: 1617

The best solution i've found is to run command on linux / ubuntu machine. So I run this command hostname -I | cut -d' ' -f1 using java.

note - second part of method is just to collect the output.

public String getIP() throws IOException
{
    ProcessBuilder pb = new ProcessBuilder("bash", "-c", "hostname -I | cut -d' ' -f1");

    Process process = pb.start();
    InputStream inputStream = process.getInputStream();
    Scanner scanner = new Scanner(inputStream, StandardCharsets.UTF_8);

    StringBuilder outputString = new StringBuilder();
    while (scanner.hasNextLine())
    {
        synchronized (this)
        {
            String message = scanner.nextLine();
            outputString.append(message);
            outputString.append("\n");
            log(message);
        }
    }
    scanner.close();

    return outputString.toString().trim();
}

Upvotes: 0

grepsedawk
grepsedawk

Reputation: 6059

From Java Tutorial

Why is InetAddress not a good solution? I don't see anything in the docs about cross platform compatibility?

This code will enumerate all network interfaces and retrieve their information.

import java.io.*;
import java.net.*;
import java.util.*;
import static java.lang.System.out;

public class ListNets 
{
    public static void main(String args[]) throws SocketException {
        Enumeration<NetworkInterface> nets = NetworkInterface.getNetworkInterfaces();
        for (NetworkInterface netint : Collections.list(nets))
            displayInterfaceInformation(netint);
    }

    static void displayInterfaceInformation(NetworkInterface netint) throws SocketException {
        out.printf("Display name: %s\n", netint.getDisplayName());
        out.printf("Name: %s\n", netint.getName());
        Enumeration<InetAddress> inetAddresses = netint.getInetAddresses();
        for (InetAddress inetAddress : Collections.list(inetAddresses)) {
            out.printf("InetAddress: %s\n", inetAddress);
        }
        out.printf("\n");
     }
}

The following is sample output from the example program:

Display name: bge0
Name: bge0
InetAddress: /fe80:0:0:0:203:baff:fef2:e99d%2
InetAddress: /121.153.225.59

Display name: lo0
Name: lo0
InetAddress: /0:0:0:0:0:0:0:1%1
InetAddress: /127.0.0.1

Upvotes: 13

Maciej Trybiło
Maciej Trybiło

Reputation: 1207

It's not ok to just return the first non-loopback interface as it might have been created by some software like Parallels. It's a better bet to try fishing for the eth0.

static private InetAddress getIPv4InetAddress() throws SocketException, UnknownHostException {

    String os = System.getProperty("os.name").toLowerCase();

    if(os.indexOf("nix") >= 0 || os.indexOf("nux") >= 0) {   
        NetworkInterface ni = NetworkInterface.getByName("eth0");

        Enumeration<InetAddress> ias = ni.getInetAddresses();

        InetAddress iaddress;
        do {
            iaddress = ias.nextElement();
        } while(!(iaddress instanceof Inet4Address));

        return iaddress;
    }

    return InetAddress.getLocalHost();  // for Windows and OS X it should work well
}

Upvotes: 4

FelixM
FelixM

Reputation: 1506

The simplest solution in my case was Socket.getLocalAddress(). I had to open the Socket specifically for that purpose, but with all the NetworkInterfaces on my Ubuntu 10.04 machine it was the only way to get the external IP address.

Upvotes: 0

Bruno
Bruno

Reputation: 51

This code worked 4me: 

import java.net.InetAddress;
import java.net.NetworkInterface;
import java.net.SocketException;
import java.util.Enumeration;


public class ShowIp {

    public static void main(String[] args) throws SocketException {
        NetworkInterface ni = NetworkInterface.getByName("eth0");
        Enumeration<InetAddress> inetAddresses =  ni.getInetAddresses();


        while(inetAddresses.hasMoreElements()) {
            InetAddress ia = inetAddresses.nextElement();
            if(!ia.isLinkLocalAddress()) {
                System.out.println("IP: " + ia.getHostAddress());
            }
        }
    }

}

Upvotes: 5

adrian.tarau
adrian.tarau

Reputation: 3154

Do not forget about loopback addresses, which are not visible outside. Here is a function which extracts the first non-loopback IP(IPv4 or IPv6)

private static InetAddress getFirstNonLoopbackAddress(boolean preferIpv4, boolean preferIPv6) throws SocketException {
    Enumeration en = NetworkInterface.getNetworkInterfaces();
    while (en.hasMoreElements()) {
        NetworkInterface i = (NetworkInterface) en.nextElement();
        for (Enumeration en2 = i.getInetAddresses(); en2.hasMoreElements();) {
            InetAddress addr = (InetAddress) en2.nextElement();
            if (!addr.isLoopbackAddress()) {
                if (addr instanceof Inet4Address) {
                    if (preferIPv6) {
                        continue;
                    }
                    return addr;
                }
                if (addr instanceof Inet6Address) {
                    if (preferIpv4) {
                        continue;
                    }
                    return addr;
                }
            }
        }
    }
    return null;
}

Upvotes: 31

Related Questions