Detect clients with Proxy Servers via PHP [duplicate]

Answer 1

I think that what's happening here is some client-side JavaScript is trying to load something, and that can "see" that the page is being viewed inside a frame. That might be a more fruitful avenue to explore- as other answers have indicated, proxies intentionally make it hard to determine just from the server alone.

Answer 2

There are various paid / free solutions. Most of them looks at the client IP trying to connect to determine if they're on a proxy or not.

Paid:
Maxmind - They focus on fraud detection and have a sub-category for proxy detection. Note that this is now considered as a "legacy" service.

Free:
W I T C H - Able to detect OpenVPN by looking at MSS values that are uniquely identifiable. The code is available on github.

GetIpIntel - Proxy / VPN detection using machine learning, queries via API.

There are a few more free options listed on security stack exchange.

Answer 3

Everything that the client passes to the server can be self-configured. You cannot trust anything, except for an IP address. So you cannot check the header data, if it's a proxy or a normal client.
By the way: It's the intention of a proxy not to show being a proxy :)

For sure, you could take the requester's IP address and send a http request you would send to a proxy. If it reacts, it may be a proxy otherwise, it's a normal client. This method would be very expensive and not reliable. If the proxy your server requested was behind a firewall, you would get no answer and think that it is a normal client.

Answer 4

i feel like answering an 8 year old thread is useless, but i will answer anyways for future reference.
I dont know about proxies, but most VPN's (if you go to them) will bring some sort of error page. if you did something like...

$cont = file_get_contents($_SERVER["REMOTE_ADDR"]); //the users ip
$errormsg = "nginx";
if(strpos($cont, $errormsg)){
    die("No VPN's or Proxies Allowed");
}

this is very basic, you could get an array of the error messages and do it that way.
This will not work if the user has portforwarded their 80 port, but i dont see why the common person would do this.