How do I get the remote address of a client in servlet? - servlets

Is there any way that I could get the original IP address of the client coming to the server?
I can use request.getRemoteAddr(), but I always seem to get the IP of the proxy or the web server.
I would want to know the IP address that the client is using to connect to me. Is there anyway that I could get it?

try this:
public static String getClientIpAddr(HttpServletRequest request) {
String ip = request.getHeader("X-Forwarded-For");
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("Proxy-Client-IP");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("WL-Proxy-Client-IP");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("HTTP_CLIENT_IP");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("HTTP_X_FORWARDED_FOR");
}
if (ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getRemoteAddr();
}
return ip;
}

Why don't use a more elegant solution like this?
private static final List<String> IP_HEADERS = Arrays.asList("X-Forwarded-For", "Proxy-Client-IP", "WL-Proxy-Client-IP", "HTTP_CLIENT_IP", "HTTP_X_FORWARDED_FOR");
public static String getClientIpAddr(HttpServletRequest request) {
return IP_HEADERS.stream()
.map(request::getHeader)
.filter(Objects::nonNull)
.filter(ip -> !ip.isEmpty() && !ip.equalsIgnoreCase("unknown"))
.findFirst()
.orElseGet(request::getRemoteAddr);
}
Deduplicate your code!

request.getRemoteAddr() is the way. It appears your proxy changes the source IP. When some proxies do that they add the original IP in some custom http header. Use request.getHeaderNames() and request.getHeaders(name) and print all of them to see if there isn't anything of interest. Like X-CLIENT-IP (made that one up, but they look like this)

As this is usually a deployment concern, rather than an application concern, another approach would be to configure the application container appropriately. Once configured, the container takes care of inspecting the appropriate header and your application continues to use request.getRemoteAddr().
For example, in Tomcat you can use the Remote IP Valve. I would assume most application servers have similar functionality.
The container could also take care of understanding if your front-end load balancer is terminating SSL connections, forwarding the request to the app server over HTTP. This is important when your application needs to generate URLs to itself.

The best solution I've ever used
public String getIpAddr(HttpServletRequest request) {
String ip = request.getHeader("x-forwarded-for");
if(ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("Proxy-Client-IP");
}
if(ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getHeader("WL-Proxy-Client-IP");
}
if(ip == null || ip.length() == 0 || "unknown".equalsIgnoreCase(ip)) {
ip = request.getRemoteAddr();
}
return ip;
}

You cannot do this in a meaningful way.
The proxy may or may not add a proxied-for header, but in many cases this will be an internal only address anyway, so it will be meaningless to you. Most proxies at the edge of an organization are configured to reveal as little as possible about the internals of the network anyway.
What are you intending to use this information for?

"x-forwarded-for" request header contains the original client IP if using a proxy or a load balancer. But I think not all proxies/lb adds this header.
Here some java code to parse the header:
http://www.codereye.com/2010/01/get-real-ip-from-request-in-java.html
If this header is not present then I would proceed as #Bozho suggests

String ipAddress = request.getHeader("x-forwarded-for");
if (ipAddress == null) {
ipAddress = request.getHeader("X_FORWARDED_FOR");
if (ipAddress == null){
ipAddress = request.getRemoteAddr();
}
}

Why I think we should try to get IP from header 'X-Forwarded-For' first? If you get from request.getRemoteAddr(), it could be client's real ip or last proxy's ip which forwards the request. Thus we can't tell which condition it belongs to. However, if 'X-Forwarded-For' is set into the header, client ip is bound to be the left-most part of what you get from it.
/**
* Try to get real ip from request:
* <ul>
* <li>try X-Forwarded-For</li>
* <li>try remote address</li>
* </ul>
*
* #param request request
* #return real ip or ""
*/
private String tryGetRealIp(HttpServletRequest request) {
// X-Forwarded-For: <client>, <proxy1>, <proxy2>
// If a request goes through multiple proxies, the IP addresses of each successive proxy is listed.
// This means, the right-most IP address is the IP address of the most recent proxy and
// the left-most IP address is the IP address of the originating client.
String forwards = request.getHeader("X-Forwarded-For");
if (StringUtils.isNotBlank(forwards)) {
// The left-most IP must be client ip
String ip = StringUtils.substringBefore(forwards, ",");
return ip;
} else if (StringUtils.isNotBlank(request.getRemoteAddr())) {
// this could be real client ip or last proxy ip which forwards the request
return request.getRemoteAddr();
}
return "";
}

request.getHeader("True-Client-IP")
it will return the client IP address

InetAddress inetAddress = InetAddress.getLocalHost();
String ip = inetAddress.getHostAddress();

Related

How can I get the IP of a client device connected through terminal services?

Problem
I need to get the client IP of a user connected through RDP/TS from an ASP site.
Sample data
The server hosting the site is at 10.1.0.1.
The user's machine is connected to a terminal server whose IP is 10.2.0.1.
The user is at a physical client whose IP is 10.3.0.1.
Attempted solutions
I have tried pulling the data from headers.
Request.ServerVariables["REMOTE_HOST"] 10.2.0.1
Request.ServerVariables["HTTP_X_FORWARDED_FOR"] [blank]
Request.ServerVariables["REMOTE_ADDR"] 10.2.0.1
Request.UserHostAddress 10.2.0.1
I also tried these methods.
public static string getIP()
{
string ips = string.Empty;
string hostName = HttpContext.Current.Request.UserHostAddress.ToString();
IPAddress[] ipAddresses = Dns.GetHostAddresses(hostName);
foreach (IPAddress ipAddress in ipAddresses)
{
ips += ipAddress + "<br />";
}
return ips;
}
Returns 10.2.0.1.
public static string GetIPAddress()
{
string ipAddress = "";
string Hostname = Environment.MachineName;
IPHostEntry Host = Dns.GetHostEntry(Hostname);
foreach (IPAddress IP in Host.AddressList)
{
if (IP.AddressFamily == AddressFamily.InterNetwork)
{
ipAddress= Convert.ToString(IP);
}
}
return ipAddress;
}
Returns 10.1.0.1.
Question
Is it possible to get 10.3.0.1 returned since there is an extra hop from the user's physical client through the TS to the web server? Is this possible with another language--e.g., PHP?

IP Address always is the same for all users

Before anything I know this question is duplicated but let me tell about my prob
As This Link demonstrated the Request.UserHostAddress only captures the IP address of the proxy server or router
for this reason I have used this code to return users IP Address :
string ip = String.IsNullOrEmpty(Request.ServerVariables["HTTP_X_FORWARDED_FOR"]) ? Request.ServerVariables["REMOTE_ADDR"] : Request.ServerVariables["HTTP_X_FORWARDED_FOR"].Split(',')[0];
but it always return 172.30.1.1 when I connect with my PC,
in the other side my Phone is connected to Telecommunication 3G network
when I connected by phone that returns 172.30.1.1 again.
I also use this code:
public static string GetIP4Address()
{
string IP4Address = String.Empty;
foreach (IPAddress IPA in Dns.GetHostAddresses(HttpContext.Current.Request.UserHostAddress))
{
if (IPA.AddressFamily.ToString() == "InterNetwork")
{
IP4Address = IPA.ToString();
break;
}
}
if (IP4Address != String.Empty)
{
return IP4Address;
}
foreach (IPAddress IPA in Dns.GetHostAddresses(Dns.GetHostName()))
{
if (IPA.AddressFamily.ToString() == "InterNetwork")
{
IP4Address = IPA.ToString();
break;
}
}
return IP4Address;
}
but results are same.
so How can I return the real IP address of users?
Thanks in advance.
Edit: Misunderstood the question previously.
This is indeed an incorrect lookup and why it is returning the proxy/router ip addresses Request.ServerVariables("REMOTE_ADDR") is what sites like rapidshare use to limit downloads as a group, and not really useful.
What you want to be doing is comparing the HTTP_X_FORWARDED_FOR variable against it, to check that there isn’t a non-transparent proxy in the way.
Like so:
' Look for a proxy address first
Dim _ip As String = Request.ServerVariables("HTTP_X_FORWARDED_FOR")
' If there is no proxy, get the standard remote address
If (_ip = "" Or _ip.ToLower = "unknown") Then _
_ip = Request.ServerVariables("REMOTE_ADDR")
Some more samples:
C#
// Look for a proxy address first
String _ip = Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
// If there is no proxy, get the standard remote address
If (_ip == "" || _ip.ToLower == "unknown")
_ip = Request.ServerVariables["REMOTE_ADDR"];
ASP/VBScript
ipaddress = Request.ServerVariables("HTTP_X_FORWARDED_FOR")
if ipaddress = "" then
ipaddress = Request.ServerVariables("REMOTE_ADDR")
end if

ASP.net / VB.net get the ip of a client [duplicate]

We have Request.UserHostAddress to get the IP address in ASP.NET, but this is usually the user's ISP's IP address, not exactly the user's machine IP address who for example clicked a link. How can I get the real IP Address?
For example, in a Stack Overflow user profile it is: "Last account activity: 4 hours ago from 86.123.127.8", but my machine IP address is a bit different. How does Stack Overflow get this address?
In some web systems there is an IP address check for some purposes. For example, with a certain IP address, for every 24 hours can the user just have only 5 clicks on download links? This IP address should be unique, not for an ISP that has a huge range of clients or Internet users.
Did I understand well?
Often you will want to know the IP address of someone visiting your website. While ASP.NET has several ways to do this one of the best ways we've seen is by using the "HTTP_X_FORWARDED_FOR" of the ServerVariables collection.
Here's why...
Sometimes your visitors are behind either a proxy server or a router and the standard Request.UserHostAddress only captures the IP address of the proxy server or router. When this is the case the user's IP address is then stored in the server variable ("HTTP_X_FORWARDED_FOR").
So what we want to do is first check "HTTP_X_FORWARDED_FOR" and if that is empty we then simply return ServerVariables("REMOTE_ADDR").
While this method is not foolproof, it can lead to better results. Below is the ASP.NET code in VB.NET, taken from James Crowley's blog post "Gotcha: HTTP_X_FORWARDED_FOR returns multiple IP addresses"
C#
protected string GetIPAddress()
{
System.Web.HttpContext context = System.Web.HttpContext.Current;
string ipAddress = context.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrEmpty(ipAddress))
{
string[] addresses = ipAddress.Split(',');
if (addresses.Length != 0)
{
return addresses[0];
}
}
return context.Request.ServerVariables["REMOTE_ADDR"];
}
VB.NET
Public Shared Function GetIPAddress() As String
Dim context As System.Web.HttpContext = System.Web.HttpContext.Current
Dim sIPAddress As String = context.Request.ServerVariables("HTTP_X_FORWARDED_FOR")
If String.IsNullOrEmpty(sIPAddress) Then
Return context.Request.ServerVariables("REMOTE_ADDR")
Else
Dim ipArray As String() = sIPAddress.Split(New [Char]() {","c})
Return ipArray(0)
End If
End Function
As others have said you can't do what you are asking. If you describe the problem you are trying to solve maybe someone can help?
E.g.
are you trying to uniquely identify your users?
Could you use a cookie, or the session ID perhaps instead of the IP address?
Edit The address you see on the server shouldn't be the ISP's address, as you say that would be a huge range. The address for a home user on broadband will be the address at their router, so every device inside the house will appear on the outside to be the same, but the router uses NAT to ensure that traffic is routed to each device correctly. For users accessing from an office environment the address may well be the same for all users. Sites that use IP address for ID run the risk of getting it very wrong - the examples you give are good ones and they often fail. For example my office is in the UK, the breakout point (where I "appear" to be on the internet) is in another country where our main IT facility is, so from my office my IP address appears to be not in the UK. For this reason I can't access UK only web content, such as the BBC iPlayer). At any given time there would be hundreds, or even thousands, of people at my company who appear to be accessing the web from the same IP address.
When you are writing server code you can never be sure what the IP address you see is referring to. Some users like it this way. Some people deliberately use a proxy or VPN to further confound you.
When you say your machine address is different to the IP address shown on StackOverflow, how are you finding out your machine address? If you are just looking locally using ipconfig or something like that I would expect it to be different for the reasons I outlined above. If you want to double check what the outside world thinks have a look at whatismyipaddress.com/.
This Wikipedia link on NAT will provide you some background on this.
UPDATE:
Thanks to Bruno Lopes. If several ip addresses could come then need to use this method:
private string GetUserIP()
{
string ipList = Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrEmpty(ipList))
{
return ipList.Split(',')[0];
}
return Request.ServerVariables["REMOTE_ADDR"];
}
If is c# see this way, is very simple
string clientIp = (Request.ServerVariables["HTTP_X_FORWARDED_FOR"] ??
Request.ServerVariables["REMOTE_ADDR"]).Split(',')[0].Trim();
What else do you consider the user IP address? If you want the IP address of the network adapter, I'm afraid there's no possible way to do it in a Web app. If your user is behind NAT or other stuff, you can't get the IP either.
Update: While there are Web sites that use IP to limit the user (like rapidshare), they don't work correctly in NAT environments.
I think I should share my experience with you all. Well I see in some situations REMOTE_ADDR will NOT get you what you are looking for. For instance, if you have a Load Balancer behind the scene and if you are trying to get the Client's IP then you will be in trouble. I checked it with my IP masking software plus I also checked with my colleagues being in different continents. So here is my solution.
When I want to know the IP of a client, I try to pick every possible evidence so I could determine if they are unique:
Here I found another sever-var that could help you all if you want to get exact IP of the client side. so I am using : HTTP_X_CLUSTER_CLIENT_IP
HTTP_X_CLUSTER_CLIENT_IP always gets you the exact IP of the client. In any case if its not giving you the value, you should then look for HTTP_X_FORWARDED_FOR as it is the second best candidate to get you the client IP and then the REMOTE_ADDR var which may or may not return you the IP but to me having all these three is what I find the best thing to monitor them.
I hope this helps some guys.
You can use:
System.Net.Dns.GetHostEntry(System.Net.Dns.GetHostName()).AddressList.GetValue(0).ToString();
All of the responses so far take into account the non-standardized, but very common, X-Forwarded-For header. There is a standardized Forwarded header which is a little more difficult to parse out. Some examples are as follows:
Forwarded: for="_gazonk"
Forwarded: For="[2001:db8:cafe::17]:4711"
Forwarded: for=192.0.2.60;proto=http;by=203.0.113.43
Forwarded: for=192.0.2.43, for=198.51.100.17
I have written a class that takes both of these headers into account when determining a client's IP address.
using System;
using System.Web;
namespace Util
{
public static class IP
{
public static string GetIPAddress()
{
return GetIPAddress(new HttpRequestWrapper(HttpContext.Current.Request));
}
internal static string GetIPAddress(HttpRequestBase request)
{
// handle standardized 'Forwarded' header
string forwarded = request.Headers["Forwarded"];
if (!String.IsNullOrEmpty(forwarded))
{
foreach (string segment in forwarded.Split(',')[0].Split(';'))
{
string[] pair = segment.Trim().Split('=');
if (pair.Length == 2 && pair[0].Equals("for", StringComparison.OrdinalIgnoreCase))
{
string ip = pair[1].Trim('"');
// IPv6 addresses are always enclosed in square brackets
int left = ip.IndexOf('['), right = ip.IndexOf(']');
if (left == 0 && right > 0)
{
return ip.Substring(1, right - 1);
}
// strip port of IPv4 addresses
int colon = ip.IndexOf(':');
if (colon != -1)
{
return ip.Substring(0, colon);
}
// this will return IPv4, "unknown", and obfuscated addresses
return ip;
}
}
}
// handle non-standardized 'X-Forwarded-For' header
string xForwardedFor = request.Headers["X-Forwarded-For"];
if (!String.IsNullOrEmpty(xForwardedFor))
{
return xForwardedFor.Split(',')[0];
}
return request.UserHostAddress;
}
}
}
Below are some unit tests that I used to validate my solution:
using System.Collections.Specialized;
using System.Web;
using Microsoft.VisualStudio.TestTools.UnitTesting;
namespace UtilTests
{
[TestClass]
public class IPTests
{
[TestMethod]
public void TestForwardedObfuscated()
{
var request = new HttpRequestMock("for=\"_gazonk\"");
Assert.AreEqual("_gazonk", Util.IP.GetIPAddress(request));
}
[TestMethod]
public void TestForwardedIPv6()
{
var request = new HttpRequestMock("For=\"[2001:db8:cafe::17]:4711\"");
Assert.AreEqual("2001:db8:cafe::17", Util.IP.GetIPAddress(request));
}
[TestMethod]
public void TestForwardedIPv4()
{
var request = new HttpRequestMock("for=192.0.2.60;proto=http;by=203.0.113.43");
Assert.AreEqual("192.0.2.60", Util.IP.GetIPAddress(request));
}
[TestMethod]
public void TestForwardedIPv4WithPort()
{
var request = new HttpRequestMock("for=192.0.2.60:443;proto=http;by=203.0.113.43");
Assert.AreEqual("192.0.2.60", Util.IP.GetIPAddress(request));
}
[TestMethod]
public void TestForwardedMultiple()
{
var request = new HttpRequestMock("for=192.0.2.43, for=198.51.100.17");
Assert.AreEqual("192.0.2.43", Util.IP.GetIPAddress(request));
}
}
public class HttpRequestMock : HttpRequestBase
{
private NameValueCollection headers = new NameValueCollection();
public HttpRequestMock(string forwarded)
{
headers["Forwarded"] = forwarded;
}
public override NameValueCollection Headers
{
get { return this.headers; }
}
}
}
IP addresses are part of the Network layer in the "seven-layer stack". The Network layer can do whatever it wants to do with the IP address. That's what happens with a proxy server, NAT, relay, or whatever.
The Application layer should not depend on the IP address in any way. In particular, an IP Address is not meant to be an identifier of anything other than the idenfitier of one end of a network connection. As soon as a connection is closed, you should expect the IP address (of the same user) to change.
If you are using CloudFlare,
you can try this Extension Method:
public static class IPhelper
{
public static string GetIPAddress(this HttpRequest Request)
{
if (Request.Headers["CF-CONNECTING-IP"] != null) return Request.Headers["CF-CONNECTING-IP"].ToString();
if (Request.ServerVariables["HTTP_X_FORWARDED_FOR"] != null) return Request.ServerVariables["HTTP_X_FORWARDED_FOR"].ToString();
return Request.UserHostAddress;
}
}
then
string IPAddress = Request.GetIPAddress();
string IP = HttpContext.Current.Request.Params["HTTP_CLIENT_IP"] ?? HttpContext.Current.Request.UserHostAddress;
What you can do is store the router IP of your user and also the forwarded IP and try to make it reliable using both the IPs [External Public and Internal Private]. But again after some days client may be assigned new internal IP from router but it will be more reliable.
Combining the answers from #Tony and #mangokun, I have created the following extension method:
public static class RequestExtensions
{
public static string GetIPAddress(this HttpRequest Request)
{
if (Request.Headers["CF-CONNECTING-IP"] != null) return Request.Headers["CF-CONNECTING-IP"].ToString();
if (Request.ServerVariables["HTTP_X_FORWARDED_FOR"] != null)
{
string ipAddress = Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrEmpty(ipAddress))
{
string[] addresses = ipAddress.Split(',');
if (addresses.Length != 0)
{
return addresses[0];
}
}
}
return Request.UserHostAddress;
}
}
public static class Utility
{
public static string GetClientIP(this System.Web.UI.Page page)
{
string _ipList = page.Request.Headers["CF-CONNECTING-IP"].ToString();
if (!string.IsNullOrWhiteSpace(_ipList))
{
return _ipList.Split(',')[0].Trim();
}
else
{
_ipList = page.Request.ServerVariables["HTTP_X_CLUSTER_CLIENT_IP"];
if (!string.IsNullOrWhiteSpace(_ipList))
{
return _ipList.Split(',')[0].Trim();
}
else
{
_ipList = page.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrWhiteSpace(_ipList))
{
return _ipList.Split(',')[0].Trim();
}
else
{
return page.Request.ServerVariables["REMOTE_ADDR"].ToString().Trim();
}
}
}
}
}
Use;
string _ip = this.GetClientIP();
use in ashx file
public string getIP(HttpContext c)
{
string ips = c.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrEmpty(ips))
{
return ips.Split(',')[0];
}
return c.Request.ServerVariables["REMOTE_ADDR"];
}
In NuGet package install Microsoft.AspNetCore.HttpOverrides
Then try:
public class ClientDeviceInfo
{
private readonly IHttpContextAccessor httpAccessor;
public ClientDeviceInfo(IHttpContextAccessor httpAccessor)
{
this.httpAccessor = httpAccessor;
}
public string GetClientLocalIpAddress()
{
return httpAccessor.HttpContext.Connection.LocalIpAddress.ToString();
}
public string GetClientRemoteIpAddress()
{
return httpAccessor.HttpContext.Connection.RemoteIpAddress.ToString();
}
public string GetClientLocalPort()
{
return httpAccessor.HttpContext.Connection.LocalPort.ToString();
}
public string GetClientRemotePort()
{
return httpAccessor.HttpContext.Connection.RemotePort.ToString();
}
}
Its easy.Try it:
var remoteIpAddress = Request.HttpContext.Connection.RemoteIpAddress;
just it :))
use this
Dns.GetHostEntry(Dns.GetHostName())
Hello guys Most of the codes you will find will return you server ip address not client ip address .however this code returns correct client ip address.Give it a try.
For More info just check this
https://www.youtube.com/watch?v=Nkf37DsxYjI
for getting your local ip address using javascript you can use
put this code inside your script tag
<script>
var RTCPeerConnection = /*window.RTCPeerConnection ||*/
window.webkitRTCPeerConnection || window.mozRTCPeerConnection;
if (RTCPeerConnection) (function () {
var rtc = new RTCPeerConnection({ iceServers: [] });
if (1 || window.mozRTCPeerConnection) {
rtc.createDataChannel('', { reliable: false });
};
rtc.onicecandidate = function (evt) {
if (evt.candidate)
grepSDP("a=" + evt.candidate.candidate);
};
rtc.createOffer(function (offerDesc) {
grepSDP(offerDesc.sdp);
rtc.setLocalDescription(offerDesc);
}, function (e) { console.warn("offer failed", e); });
var addrs = Object.create(null);
addrs["0.0.0.0"] = false;
function updateDisplay(newAddr) {
if (newAddr in addrs) return;
else addrs[newAddr] = true;
var displayAddrs = Object.keys(addrs).filter(function
(k) { return addrs[k]; });
document.getElementById('list').textContent =
displayAddrs.join(" or perhaps ") || "n/a";
}
function grepSDP(sdp) {
var hosts = [];
sdp.split('\r\n').forEach(function (line) {
if (~line.indexOf("a=candidate")) {
var parts = line.split(' '),
addr = parts[4],
type = parts[7];
if (type === 'host') updateDisplay(addr);
} else if (~line.indexOf("c=")) {
var parts = line.split(' '),
addr = parts[2];
updateDisplay(addr);
}
});
}
})(); else
{
document.getElementById('list').innerHTML = "<code>ifconfig| grep inet | grep -v inet6 | cut -d\" \" -f2 | tail -n1</code>";
document.getElementById('list').nextSibling.textContent = "In Chrome and Firefox your IP should display automatically, by the power of WebRTCskull.";
}
</script>
<body>
<div id="list"></div>
</body>
and For getting your public ip address you can use
put this code inside your script tag
function getIP(json) {
document.write("My public IP address is: ", json.ip);
}
<script type="application/javascript" src="https://api.ipify.org?format=jsonp&callback=getIP"></script>
Simply
var ip = Request.UserHostAddress;
That's all...
Try:
using System.Net;
public static string GetIpAddress() // Get IP Address
{
string ip = "";
IPHostEntry ipEntry = Dns.GetHostEntry(GetCompCode());
IPAddress[] addr = ipEntry.AddressList;
ip = addr[2].ToString();
return ip;
}
public static string GetCompCode() // Get Computer Name
{
string strHostName = "";
strHostName = Dns.GetHostName();
return strHostName;
}

Getting the user's local IP - Always getting 192.168.2.1

I need to get the user's local IP for my ASP.NET application and I'm using this method:
protected string GetIPAddress()
{
System.Web.HttpContext context = System.Web.HttpContext.Current;
string ipAddress = context.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (!string.IsNullOrEmpty(ipAddress))
{
string[] addresses = ipAddress.Split(',');
if (addresses.Length != 0)
{
return addresses[0];
}
}
return context.Request.ServerVariables["REMOTE_ADDR"];
}
However, when I publish my website I always get 192.168.2.1 no matter where the user is opening the website from.
Does anyone know how to solve this issue?
Some network devices make use of the X-forwarded-for header. You should check if the requests hitting your application have this header.
You can get the client's IP address from HTTP_X_FORWARDED_FOR or REMOTE_ADDR.
var ipAddress = HttpContext.Current.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (string.IsNullOrEmpty(ipAddress ))
{
ipAddress = HttpContext.Current.Request.ServerVariables["REMOTE_ADDR"];
}
return ipAddress;

Ip address of the client machine

Please let me know how to get the client IP address,
I have tried all of the below things , but I am getting the same output: 127.0.0.1
string strClientIP;
strClientIP = Request.UserHostAddress.ToString();
string strHostName = System.Net.Dns.GetHostName();
string clientIPAddress = System.Net.Dns.GetHostAddresses(strHostName).GetValue(0).ToString();
string ipaddress = string.Empty ;
ipaddress = Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if (ipaddress == "" || ipaddress == null)
ipaddress = Request.ServerVariables["REMOTE_ADDR"];
How can I get the correct IP?
You are on the right track with REMOTE_ADDR, but it might not work if you are accessing the site locally, it will show local host.
REMOTE_ADDR is the header that contains the clients ip address you should check that first.
You should also check to for the HTTP_X_FORWARDED header in case you're visitor is going through a proxy. Be aware that HTTP_X_FORWARDED is an array that can contain multiple comma separated values depending on the number of proxies.
Here is a small c# snippet that shows determining the client's ip:
string clientIp = context.Request.ServerVariables["HTTP_X_FORWARDED_FOR"];
if( !string.IsNullOrEmpty(clientIp) ) {
string[] forwardedIps = clientIp.Split( new char[] { ',' }, StringSplitOptions.RemoveEmptyEntries );
clientIp = forwardedIps[forwardedIps.Length - 1];
} else {
clientIp = context.Request.ServerVariables["REMOTE_ADDR"];
}
If you connect via the localhost address, then your client address will also report as localhost (as it will route through the loopback adapter)

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