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(PHP 4, PHP 5)
gethostbynamel — Ermittelt eine Liste von IPv4-Adressen passend zum angegebenen Internet-Hostnamen
Gibt eine Liste von IPv4-Adressen zurück, zu denen der angegebene Internet-Host hostname aufgelöst wurde.
Der Hostname.
Gibt ein Array von IPv4-Adressen zurück. Kann hostname nicht aufgelöst werden, wird FALSE zurückgegeben.
Beispiel #1 gethostbynamel() example
<?php
$hosts = gethostbynamel('www.example.com');
print_r($hosts);
?>
Das oben gezeigte Beispiel erzeugt folgende Ausgabe:
Array
(
[0] => 192.0.34.166
)
The solution is simpel. Just add a . (point) to the end of the URL for correct name resolving.
Without this point PHP thinks it's a subdomain of your local domain and so returns the "local-IP".
In PHP 5.0.4, gethostbynamel returns an empty string instead of false if the lookup fails. A simple workaround for this error is to use is_array() in an IF block:
<?
$hosts = gethostbynamel($hostname);
if (is_array($hosts)) {
echo "Host ".$hostname." resolves to:<br><br>";
foreach ($hosts as $ip) {
echo "IP: ".$ip."<br>";
}
} else {
echo "Host ".$hostname." is not tied to any IP.";
}
?>
Obviously, in some cases, not all IPs are likely to be useful while checking a hostname. Sometimes also, not all IPs will work. This code will check for the first WORKING IP from the list. Or at least it should - I haven't had time to test it yet.
Needs domain parameter, and port and max IPs to check are optional.
If port is not set, it will check HTTP port 80, and if max IPs to check is not set, it will only check the first 10 IPs from the list.
Hope it helps someone.
<?php
function checkhostlist($domain, $port = 80, $maxipstocheck = 10) {
$hosts = gethostbynamel($domain);
for ($chk=0;$chk<$maxipstocheck;$chk++) {
if (isset($hosts[$chk])) {
$th = fsockopen($domain, $port);
if ($th) {
fclose($th);
return $hosts[$chk];
break;
}
}
}
}
?>
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