IP range comparison using ip2long

Question

I am trying to compage an IP using ip2long as described in this tutorial: How to check an IP address is within a range of two IPs in PHP?

The code is as follows:

$ip2 = ip2long("127.0.0.1");
$high_ip = ip2long("128.0.0.0");
$low_ip = ip2long("126.0.0.1");
if ($ip2 <= $high_ip && $low_ip <= $ip2) {
  echo "in range";
}
else {
    echo "no";
}

I already tried switching the high and low IPs and still get the same result.

Why is my script not able to detect that the IP is in fact within range?

Note: I need to compare the user's IP against wide ranges like: 1.1.2.0 - 1.2.2.0.

Sir, where is $ip2? Looks like there is a typing error there. (didn't downvote, though) — briosheje, Mar 17 '15 at 19:34
When I run your code, correcting for the typo in which $ip becomes $ip2, I get "in range". — kainaw, Mar 17 '15 at 19:36
Same as @kainaw here. Are you sure you fixed **both** $ip2? there are 2 $ip2. http://prntscr.com/6i0o82 — briosheje, Mar 17 '15 at 19:37
if you get "no" then you're not running the code you posted as there is no echo "no"; ... repost your code... — dbinns66, Mar 17 '15 at 19:37

score 3 · Accepted Answer · edited May 23 '17 at 11:50

You have to use sprintf("%u", $ip), like this:

if (sprintf("%u", $ip) <= sprintf("%u", $high_ip)  && sprintf("%u", $low_ip) <= sprintf("%u", $ip))

Why? If you look into the manual: http://php.net/manual/en/function.ip2long.php

You will see it why:

Note: Because PHP's integer type is signed, and many IP addresses will result in negative integers on 32-bit architectures, you need to use the "%u" formatter of sprintf() or printf() to get the string representation of the unsigned IP address.

EDIT:

As from the comments:

Good to know, but for what evil reason is it working properly here? prntscr.com/6i0o82 Because it might not be a 32-bit architecture? – briosheje 12 mins ago

Yes ^ there in the online IDE it works, because it's a 64bit architecture. And as mentioned above in the quote on 32-bit architectures this won't work, because it results in negative interges, so that's why you have to use %u + sprintf() or printf().

Which architecture do I have?

You can simply do this:

echo PHP_INT_MAX;

2147483647          -> 32 bit
9223372036854775807 -> 64 bit

EDIT 2:

Also from the comments:

btw I am using xampp, but it shouldn't matter in this case? – user1721135 22 mins ago

PHP doesn't support 64 bit on windows or at least not official. You can see/read this here: http://windows.php.net/

And a quote from there:

x86_64 Builds The x64 builds of PHP for Windows should be considered experimental, and do not yet provide 64-bit integer or large file support. Please see this post for work ongoing to improve these builds.

So you're probably using xampp on windows and there is no xampp with php 64 bit version:

enter image description here

And if you are not using xampp on windows you can download xampp with 64bit php here for MAC and linux: https://www.apachefriends.org/download.html

Good to know, but for what evil reason is it working properly here? http://prntscr.com/6i0o82 Because it might not be a 32-bit architecture? — briosheje, Mar 17 '15 at 19:39
@Rizier123: perhaps some server-side conversion due to the fact that it is an online editor? lol — briosheje, Mar 17 '15 at 19:45
@briosheje updated my answer how to check what architecture you have and why it works in the online IDE — Rizier123, Mar 17 '15 at 19:56
@user1721135 updated my answer! I assume you are using xampp on windows — Rizier123, Mar 17 '15 at 20:11

score 0 · Answer 2 · answered Mar 17 '15 at 19:34

0

There is an error in your code : $ip2 is undefined. It would be : if ($ip <= $high_ip && $low_ip <= $ip) {

answered Mar 17 '15 at 19:34

Christophe Maggi

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IP range comparison using ip2long

2 Answers2