How to determine a user's IP address in node

How can I determine the IP address of a given request from within a controller? For example (in express):

app.post('/get/ip/address', function (req, res) {
    // need access to IP address here
})

In your request object there is a property called socket, which is a net.Socket object. The net.Socket object has a property remoteAddress, therefore you should be able to get the IP with this call:

request.socket.remoteAddress

(if your node version is below 13, use the deprecated now request.connection.remoteAddress)

EDIT

As @juand points out in the comments, the correct method to get the remote IP, if the server is behind a proxy, is request.headers['x-forwarded-for']

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

Note that sometimes you can get more than one IP address in req.headers['x-forwarded-for']. Also, an x-forwarded-for header will not always be set which may throw an error.

The general format of the field is:

x-forwarded-for: client, proxy1, proxy2, proxy3

where the value is a comma+space separated list of IP addresses, the left-most being the original client, and each successive proxy that passed the request adding the IP address where it received the request from. In this example, the request passed through proxy1, proxy2, and then proxy3. proxy3 appears as remote address of the request.

This is the solution suggested by Arnav Gupta with a fix Martin has suggested below in the comments for cases when x-forwarded-for is not set :

var ip = (req.headers['x-forwarded-for'] || '').split(',').pop().trim() || 
         req.socket.remoteAddress

Suggestion using modern JS:

• processing x-forwarded-for only if set, if so, take the first address
• other parameters use optional chaining (?.)

const parseIp = (req) =>
    req.headers['x-forwarded-for']?.split(',').shift()
    || req.socket?.remoteAddress

console.log(parseIp(req))
// => 127.0.0.1

If using express...

req.ip

I was looking this up then I was like wait, I'm using express. Duh.

You can stay DRY and just use node-ipware that supports both IPv4 and IPv6.

Install:

npm install ipware

In your app.js or middleware:

var getIP = require('ipware')().get_ip;
app.use(function(req, res, next) {
    var ipInfo = getIP(req);
    console.log(ipInfo);
    // { clientIp: '127.0.0.1', clientIpRoutable: false }
    next();
});

It will make the best attempt to get the user's IP address or returns 127.0.0.1 to indicate that it could not determine the user's IP address. Take a look at the README file for advanced options.

You can use request-ip, to retrieve a user's ip address. It handles quite a few of the different edge cases, some of which are mentioned in the other answers.

Disclosure: I created this module

Install:

npm install request-ip

In your app:

var requestIp = require('request-ip');

// inside middleware handler
var ipMiddleware = function(req, res, next) {
    var clientIp = requestIp.getClientIp(req); // on localhost > 127.0.0.1
    next();
};

Hope this helps

request.headers['x-forwarded-for'] || request.connection.remoteAddress

If the x-forwarded-for header is there then use that, otherwise use the .remoteAddress property.

The x-forwarded-for header is added to requests that pass through load balancers (or other types of proxy) set up for HTTP or HTTPS (it's also possible to add this header to requests when balancing at a TCP level using proxy protocol). This is because the request.connection.remoteAddress the property will contain the private IP address of the load balancer rather than the public IP address of the client. By using an OR statement, in the order above, you check for the existence of an x-forwarded-for header and use it if it exists otherwise use the request.connection.remoteAddress.

Following Function has all the cases covered will help

var ip;
if (req.headers['x-forwarded-for']) {
    ip = req.headers['x-forwarded-for'].split(",")[0];
} else if (req.connection && req.connection.remoteAddress) {
    ip = req.connection.remoteAddress;
} else {
    ip = req.ip;
}console.log("client IP is *********************" + ip);

There are two ways to get the ip address :

1. let ip = req.ip

2. let ip = req.connection.remoteAddress;

But there is a problem with above approaches.

If you are running your app behind Nginx or any proxy, every single IP addresses will be 127.0.0.1.

So, the best solution to get the ip address of user is :-

let ip = req.header('x-forwarded-for') || req.connection.remoteAddress;

Warning:

Don't just blindly use this for important rate-limiting:

let ip = request.headers['x-forwarded-for'].split(',')[0];

It's very easy to spoof:

curl --header "X-Forwarded-For: 1.2.3.4" "https://example.com"

In that case ther user's real IP address will be:

let ip = request.headers['x-forwarded-for'].split(',')[1];

I'm surprised that no other answers have mentioned this.

function getCallerIP(request) {
    var ip = request.headers['x-forwarded-for'] ||
        request.connection.remoteAddress ||
        request.socket.remoteAddress ||
        request.connection.socket.remoteAddress;
    ip = ip.split(',')[0];
    ip = ip.split(':').slice(-1); //in case the ip returned in a format: "::ffff:146.xxx.xxx.xxx"
    return ip;
}

I have tried all of them didn't work though,

console.log(clientIp);
console.log(req.ip);

console.log(req.headers['x-forwarded-for']);
console.log(req.connection.remoteAddress);
console.log(req.socket.remoteAddress);
console.log(req.connection.socket.remoteAddress.split(",")[0]);

When running an Express app behind a proxy for me Nginx, you have to set the application variable trust proxy to true. Express offers a few other trust proxy values which you can review in their documentation, but below steps worked for me.

1. app.set('trust proxy', true) in your Express app.

> app.set('trust proxy', true);

2. Add proxy_set_header X-Forwarded-For $remote_addr in the Nginx configuration for your server block.

> location / { > proxy_pass http://localhost:3001; > proxy_http_version 1.1; > proxy_set_header Upgrade $http_upgrade; > proxy_set_header Connection 'upgrade'; > proxy_set_header Host $host; > proxy_set_header X-Forwarded-For $remote_addr; # this line > proxy_cache_bypass $http_upgrade; > }

3. You can now read off the client’s IP address from the req.header('x-forwarded-for') or req.connection.remoteAddress; Full code for ipfilter

> module.exports = function(req, res, next) { > let enable = true; // true/false > let blacklist = ['x.x.x.x']; > let whitelist = ['x.x.x.x']; > let clientIp = req.header('x-forwarded-for') || req.connection.remoteAddress; > if (!clientIp) { > return res.json('Error'); > } > if (enable > && paths.some((path) => (path === req.originalUrl))) { >
> let blacklist = blacklist || []; > if (blacklist.some((ip) => clientIp.match(ip) !== null)) { > return res.json({ status: 401, error: 'Your IP is black-listed !'}); > } > let whitelist = whitelist || []; > if (whitelist.length === 0 || whitelist.some((ip) => clientIp.match(ip) !== null)) { > next(); > return; > } else { > return res.json({ status: 401, error: 'Your IP is not listed !'}); > } > } > next(); > };

In node 10.14 , behind nginx, you can retrieve the ip by requesting it through nginx header like this:

proxy_set_header X-Real-IP $remote_addr;

Then in your app.js:

app.set('trust proxy', true);

After that, wherever you want it to appear:

var userIp = req.header('X-Real-IP') || req.connection.remoteAddress;

If you're using express version 3.x or greater, you can use the trust proxy setting (http://expressjs.com/api.html#trust.proxy.options.table) and it will walk the chain of addresses in the x-forwarded-for header and put the latest ip in the chain that you've not configured as a trusted proxy into the ip property on the req object.

If you get multiple IPs , this works for me:

var ipaddress = (req.headers['x-forwarded-for'] || 
req.connection.remoteAddress || 
req.socket.remoteAddress || 
req.connection.socket.remoteAddress).split(",")[0];

Simple get remote ip in nodejs:

var ip = req.header('x-forwarded-for') || req.connection.remoteAddress;

req.connection has been deprecated since [email protected]. Using req.connection.remoteAddress to get the client IP might still work but is discouraged.

Luckily, req.socket.remoteAddress has been there since [email protected] and is a perfect replacement:

> The string representation of the remote IP address. For example, '74.125.127.100' or '2001:4860:a005::68'. Value may be undefined if the socket is destroyed (for example, if the client disconnected).

If you're using express.js then,

app.post('/get/ip/address', function (req, res) {
      res.send(req.ip);
})

var ipaddress = (req.headers['x-forwarded-for'] || 
req.connection.remoteAddress || 
req.socket.remoteAddress || 
req.connection.socket.remoteAddress).split(",")[0];

I realize this has been answered to death, but here's a modern ES6 version I wrote that follows airbnb-base eslint standards.

const getIpAddressFromRequest = (request) => {
  let ipAddr = request.connection.remoteAddress;

  if (request.headers && request.headers['x-forwarded-for']) {
    [ipAddr] = request.headers['x-forwarded-for'].split(',');
  }

  return ipAddr;
};

The X-Forwarded-For header may contain a comma-separated list of proxy IPs. The order is client,proxy1,proxy2,...,proxyN. In the real world, people implement proxies that may supply whatever they want in this header. If you are behind a load balancer or something, you can at least trust the first IP in the list is at least whatever proxy some request came through.

If you are using Graphql-Yoga you can use the following function:

const getRequestIpAddress = (request) => {
    const requestIpAddress = request.request.headers['X-Forwarded-For'] || request.request.connection.remoteAddress
    if (!requestIpAddress) return null

    const ipv4 = new RegExp("(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)")

    const [ipAddress] = requestIpAddress.match(ipv4)

    return ipAddress
}

I'm using express behind nginx and

req.headers.origin

did the trick for me

There were a lot of great points here but nothing that was comprehensive, so here's what I ended up using:

function getIP(req) {
  // req.connection is deprecated
  const conRemoteAddress = req.connection?.remoteAddress
  // req.socket is said to replace req.connection
  const sockRemoteAddress = req.socket?.remoteAddress
  // some platforms use x-real-ip
  const xRealIP = req.headers['x-real-ip']
  // most proxies use x-forwarded-for
  const xForwardedForIP = (() => {
    const xForwardedFor = req.headers['x-forwarded-for']
    if (xForwardedFor) {
      // The x-forwarded-for header can contain a comma-separated list of
      // IP's. Further, some are comma separated with spaces, so whitespace is trimmed.
      const ips = xForwardedFor.split(',').map(ip => ip.trim())
      return ips[0]
    }
  })()
  // prefer x-forwarded-for and fallback to the others
  return xForwardedForIP || xRealIP || sockRemoteAddress || conRemoteAddress
}

I use this for ipv4 format

req.connection.remoteAddress.split(':').slice(-1)[0]

In a shell, you would just curl https://api.ipify.org

So, let's observe this to port it to node.js!

curl is a application to fetch data from websites, we pass the website "https://api.ipify.org" as the argument. We could use node-fetch to replace curl!

The data we get from the website is our IP address, It's some sort of thing which just fetches your IP.

So to sum it up:

const fetch = require('node-fetch');

fetch('https://api.ipify.org')
  .then(response => {/* whatever */})
  .catch(err => {/* whatever */})

    const express = require('express')
    const app = express()
    const port = 3000

    app.get('/', (req, res) => {
    var ip = req.ip
    console.log(ip);
    res.send('Hello World!')
    })

   // Run as nodejs ip.js
    app.listen(port, () => {
    console.log(`Example app listening at http://localhost:${port}`)
    })

we can check with this code in node js

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

let addresses = [];

for (var k in interfaces) {

    for (var k2 in interfaces[k]) {

        const address = interfaces[k][k2];

        if ( (address.family === 'IPv4' || address.family === 'IPv6')  && 
            !address.internal) {

            addresses.push(address.address);

        }
    }
}
console.log(addresses);

Had the same problem...im also new at javascript but i solved this with req.connection.remoteAddress; that gave me th IP address (but in ipv6 format ::ffff.192.168.0.101 ) and then .slice to remove the 7 first digits.

var ip = req.connection.remoteAddress;

if (ip.length < 15) 
{   
   ip = ip;
}
else
{
   var nyIP = ip.slice(7);
   ip = nyIP;
}