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".
gethostbynamel
(PHP 4, PHP 5)
gethostbynamel — 指定したインターネットホスト名に対応するIPアドレスのリストを取得 する。
説明
array gethostbynamel ( string $hostname )hostnameで指定したインターネットホストを 検索して得られた IP アドレスのリストを返します。 もし hostname が解決できなかった場合は FALSE を返します。
例 1550. gethostbynamel() の例
<?php
$hosts = gethostbynamel('www.example.com');
print_r($hosts);
?>
上の例の出力は例えば以下のようになります:
Array
(
[0] => 192.0.34.166
)
参考
| gethostbyname() |
| gethostbyaddr() |
| checkdnsrr() |
| getmxrr() |
| named(8) のマニュアルページ |
gethostbynamel
info at methfessel-computers.de
30-Sep-2006 07:27
30-Sep-2006 07:27
webdev at concraption dot com
20-Sep-2005 03:25
20-Sep-2005 03:25
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 ".$hostname." is not tied to any IP.";
}
?>
Skyld at o2 dot co dot uk
25-Sep-2004 05:45
25-Sep-2004 05:45
Obviously, in some cases, not all IPs are likely to be useful while checking a hostname. Sometimes also, not all IPs will work. This code will check for the first WORKING IP from the list. Or at least it should - I haven't had time to test it yet.
Needs domain parameter, and port and max IPs to check are optional.
If port is not set, it will check HTTP port 80, and if max IPs to check is not set, it will only check the first 10 IPs from the list.
Hope it helps someone.
<?php
function checkhostlist($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;
}
}
}
}
?>