Get local IP address in Node.js

Here's what might be the cleanest, simplest answer without dependencies & that works across all platforms.

const { lookup } = require('dns').promises;
const { hostname } = require('os');

async function getMyIPAddress(options) {
  return (await lookup(hostname(), options))
    .address;
}

I was able to do this using just Node.js.

As Node.js:

var os = require( 'os' );
var networkInterfaces = Object.values(os.networkInterfaces())
    .reduce((r,a) => {
        r = r.concat(a)
        return r;
    }, [])
    .filter(({family, address}) => {
        return family.toLowerCase().indexOf('v4') >= 0 &&
            address !== '127.0.0.1'
    })
    .map(({address}) => address);
var ipAddresses = networkInterfaces.join(', ')
console.log(ipAddresses);

As Bash script (needs Node.js installed)

function ifconfig2 ()
{
    node -e """
        var os = require( 'os' );
        var networkInterfaces = Object.values(os.networkInterfaces())
            .reduce((r,a)=>{
                r = r.concat(a)
                return r;
            }, [])
            .filter(({family, address}) => {
                return family.toLowerCase().indexOf('v4') >= 0 &&
                    address !== '127.0.0.1'
            })
            .map(({address}) => address);
        var ipAddresses = networkInterfaces.join(', ')
        console.log(ipAddresses);
    """
}

I probably came late to this question, but in case someone wants to a get a one liner ES6 solution to get array of IP addresses then this should help you:

Object.values(require("os").networkInterfaces())
    .flat()
    .filter(({ family, internal }) => family === "IPv4" && !internal)
    .map(({ address }) => address)

As

Object.values(require("os").networkInterfaces())

will return an array of arrays, so flat() is used to flatten it into a single array

.filter(({ family, internal }) => family === "IPv4" && !internal)

Will filter the array to include only IPv4 Addresses and if it's not internal

Finally

.map(({ address }) => address)

Will return only the IPv4 address of the filtered array

so result would be [ '192.168.xx.xx' ]

you can then get the first index of that array if you want or change filter condition

OS used is Windows

The following solution works for me

const ip = Object.values(require("os").networkInterfaces())
        .flat()
        .filter((item) => !item.internal && item.family === "IPv4")
        .find(Boolean).address;

var ip = req.headers['x-forwarded-for'] || req.socket.remoteAddress 

Similar to other answers but more succinct:

'use strict';

const interfaces = require('os').networkInterfaces();

const addresses = Object.keys(interfaces)
  .reduce((results, name) => results.concat(interfaces[name]), [])
  .filter((iface) => iface.family === 'IPv4' && !iface.internal)
  .map((iface) => iface.address);

Many times I find there are multiple internal and external facing interfaces available (example: 10.0.75.1, 172.100.0.1, 192.168.2.3) , and it's the external one that I'm really after (172.100.0.1).

In case anyone else has a similar concern, here's one more take on this that hopefully may be of some help...

const address = Object.keys(os.networkInterfaces())
    // flatten interfaces to an array
    .reduce((a, key) => [
        ...a,
        ...os.networkInterfaces()[key]
    ], [])
    // non-internal ipv4 addresses only
    .filter(iface => iface.family === 'IPv4' && !iface.internal)
    // project ipv4 address as a 32-bit number (n)
    .map(iface => ({...iface, n: (d => ((((((+d[0])*256)+(+d[1]))*256)+(+d[2]))*256)+(+d[3]))(iface.address.split('.'))}))
    // set a hi-bit on (n) for reserved addresses so they will sort to the bottom
    .map(iface => iface.address.startsWith('10.') || iface.address.startsWith('192.') ? {...iface, n: Math.pow(2,32) + iface.n} : iface)
    // sort ascending on (n)
    .sort((a, b) => a.n - b.n)
    [0]||{}.address;

Some answers here seemed unnecessarily over-complicated to me. Here's a better approach to it using plain Nodejs.

import os from "os";

const machine = os.networkInterfaces()["Ethernet"].map(item => item.family==="IPv4")

console.log(machine.address) //gives 192.168.x.x or whatever your local address is

See documentation: NodeJS - os module: networkInterfaces

Here's a neat little one-liner for you which does this functionally:

const ni = require('os').networkInterfaces();
Object
  .keys(ni)
  .map(interf =>
    ni[interf].map(o => !o.internal && o.family === 'IPv4' && o.address))
  .reduce((a, b) => a.concat(b))
  .filter(o => o)
  [0];

Using internal-ip:

const internalIp = require("internal-ip")

console.log(internalIp.v4.sync())