Javascript - PHP Websocket - Getting User IP Address

Question

I'm trying to make a game.

However, I can't get the user ip address on the php side.

I am using plain websocket.

I tried the following codes but it doesn't give what I want. The server gives the ip address.

JAVASCRIPT

websocket = new WebSocket("wss://site.com/game_play/"); 
websocket.binaryType = "arraybuffer";
websocket.onopen = function(event) { 
    
    var send = {
        type: "sign",
        oauth: g_oauth,
        room: g_room,
    };
    websocket.send(binary_encode(send));
}
.....

PHP

$socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
socket_set_option($socket, SOL_SOCKET, SO_REUSEADDR, 1);
socket_set_option($socket, SOL_SOCKET, SO_RCVTIMEO, array('sec' => 1, 'usec' => 0));
socket_set_option($socket, SOL_SOCKET, SO_SNDTIMEO, array('sec' => 1, 'usec' => 0));
socket_bind($socket, 0, $port);

socket_listen($socket);
.....

APACHE

ProxyPass /game_play/ ws://site.com:8090/server.php
ProxyPassReverse /game_play/ ws://site.com:8090/server.php

socket_getsockname, socket_getpeername

I tried some functions but unsuccessful. Because the server gives the ip address. I am trying to get user ip address.

Does $_SERVER['REMOTE_ADDR'] work? Not sure how this will behave with sockets as I haven't used php w/them. — jshrc
– jshrc, Commented Jul 15, 2021 at 17:45
"I tried some functions but unsuccessful." perhaps the wrong functions? or perhaps because that were wild guesses? Maybe a remote address with Websockets works differently than your understanding what a remote address is? — hakre
– hakre, Commented Jul 15, 2021 at 17:48
@jshrc $_SERVER is for http requests. this is a websocket. I'm booting over SSH. Example: php xx.php — CoinAnalyzer
– CoinAnalyzer, Commented Jul 15, 2021 at 17:53
@hakre I don't understand why you're being aggressive. I wrote all the codes myself, but the ip address issue is a complete mystery. — CoinAnalyzer
– CoinAnalyzer, Commented Jul 15, 2021 at 18:09

jshrc · Accepted Answer · 2021-07-15 18:00:11Z

1

According to the docs you may need to use socket_accept() instead of socket_connect():

Note: socket_getsockname() should not be used with AF_UNIX sockets created with socket_connect(). Only sockets created with socket_accept() or a primary server socket following a call to socket_bind() will return meaningful values.

Edit: wrong function called, should be socket_getpeername()

Either way, the second param in socket_getpeername() is passed by reference and will contain the client IP if the socket is good:

<?php 
....
socket_listen($socket);

$address = '';
socket_getpeername($socket, $address); //$address is passed by reference and will contain the remote IP.

edited Jul 15, 2021 at 18:00

answered Jul 15, 2021 at 17:54

jshrc

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15 Comments

CoinAnalyzer Over a year ago

$socket_new = socket_accept($this->server); socket_getsockname($socket_new, $address); echo "#$socket_new - $address CONNECTED!\n"; Not client ip. Server ip. socket_connect trying...

jshrc Over a year ago

My bad, it should be socket_getpeername is for the client. socket_getsockname is for the local side. Edited my answer.

jshrc Over a year ago

You're not running this locally right? The php is running on a totally different machine? (Sorry, I have to ask as the docs make it pretty clear this would give the remote ip)

CoinAnalyzer Over a year ago

I think we succeed. I detected a ip in the headers section. X-Forwarded-For: So how do I just get the ip adress ? ?

CoinAnalyzer Over a year ago

Yes, However, the return value of $ip address is still the server ip address. The ip address in the header information is correct, but.

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