Question

I need to check if a file is opened “locally” (same machine or network). I’m using:

<?php
if ((substr($_SERVER['REMOTE_ADDR'],0,8) == "192.168.") || ($_SERVER['REMOTE_ADDR'] == "127.0.0.1")) {
    // client is local
} else {
    // client is not local
}

But I’m not sure this is the best way.

What is a more foolproof way of doing this?

Answer 1

“Foolproof,” as always, can be tricky.

If we do restrict ourselves to IPv4, then checking for “127.0.0.1” takes care of the localhost case, but checking against “192.168.” is plain wrong – it will only work if the script is being run on a server which happens to be on the 192.168 network, using a 16-bit subnet mask.

Checking $_SERVER[‘REMOTE_ADDR’] against $_SERVER[‘SERVER_ADDR’] would be a better bet. This still doesn’t take care of the case of a multi-homed host (ie one which has several IP addresses in addition to 127.0.0.1), though.

In order to catch all same-network cases, you’d need to check the combination of SERVER_ADDR and subnet mask against REMOTE_ADDR, but the subnet mask isn’t available in $_SERVER.

BUT I found a function which does pretty much what you want here. It’s a couple of screens down and it’s called clientInSameSubnet. Not my code, but looks right.