ipinfo.io filter AS#### out? - php

Question

I am using ipinfo.io for some simple lookups, but I have one little problem with echo $details->org; It outputs "AS15169 Google Inc.", but I want only the ISP part so "Google Inc.".
Example code:
<?
function ip_details($ip) {
$json = file_get_contents("http://ipinfo.io/{$ip}");
$details = json_decode($json);
return $details;
}
$details = ip_details($_SERVER['REMOTE_ADDR']);
echo $details->org;
?>
Example output: http://ipinfo.io/8.8.8.8/org
Need some help, anyone?

Answer 1

If you just want the org field you can query http://ipinfo.io/{$ip}/org which will give you the org as a string, which will save you from having to parse any JSON:
$org = file_get_contents("http://ipinfo.io/{$ip}/org");
We can split the org string into the ASN and name by exploding on the first space:
list($asn, $name) = explode(" ", $org, 2);
Putting it all together we get:
function org_name($ip) {
$org = file_get_contents("http://ipinfo.io/{$ip}/org");
list($asn, $name) = explode(" ", $org, 2);
return $name;
}
echo org_name("8.8.8.8");
// => Google Inc.
echo org_name("189.154.55.170");
// => Uninet S.A. de C.V.
echo org_name("172.250.147.230");
// => Time Warner Cable Internet LLC
See http://ipinfo.io/developers for more details about the different endpoints, and rate limits.

Answer 2

Use regex to find anything between word boundaries that starts with AS and has one or more digits followed by a space and then replace it with a blank string.
I'm not great with regex so someone might come in with a better solution that this. But I tested it at PHP Live Regex and it worked for the few test cases I tried.
<?
function ip_details($ip) {
$json = file_get_contents("http://ipinfo.io/{$ip}");
$details = json_decode($json);
return $details;
}
$details = ip_details($_SERVER['REMOTE_ADDR']);
$org = preg_replace('/\bAS\d+\s\b/i', '', $details->org);
echo $org;
?>