[Resolved] if ($array[] IS LIKE '$user_ip%')...??

vaaaska

Yeah, I know that that's no correct but it's what I'm trying to do. I can't quite figure out how, or if, this can be written.

It's for an IP ban function. And what I would like to do is compare array values to the user IP and if any of the array values are LIKE the user IP then they will be redirected.

if ($array[] IS LIKE '$user_ip%')

The reason I'm trying to do it this way is that some banned IP values will be longer than others, like:

123.123.123.12
123.123.12

So if there was a way to use a LIKE condition in here I think it would be pretty efficient.

Any ideas regarding how this might be achieved?

Thanks...v

bubblenut

Only way I can think of is using [man]usort()[/man] with a custom built function to do the comparison or use a database of course.

Prikid

You could take the visitors ip address, split it using [man]explode()[/man] compare it against the array with full address and without the last digits...

remnant

preg_match() might be the best way to go about this.

vaaaska

Yeah, preg_match might do also...I might try that out as this is just a draft...but it does work as long as I have at least three characters (ie. 62.) in the ban ip list...

I've reduced things down so that we could compare an exact length $user_ip with whatever lenght the $banned_ip is.

function bannedIP($ip) {
$clip_ip = substr($ip, 0, 2);

$qry = mysql_query ("SELECT ip FROM banip WHERE ip LIKE '$clip_ip%'");
$rs = mysql_fetch_array($qry);

do {
	$chars = strlen($rs['ip']);
	$user_ip = substr($ip, 0, $chars);
		if ($user_ip == $rs['ip']) {
		header("location:http://www.yahoo.com/");
		}

} while($rs = mysql_fetch_array($qry));
}