gethostbyname
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 — 주어진 인터넷 호스트 이름에 대응이 되는 IP 어드레스의 목록을 가져옵니다.
Description
array gethostbynamel
( string $hostname
)
hostname 분석에 정의된 인터넷 호스트의 IP 어드레스의 목록을 반환합니다.
참조: gethostbyname(), gethostbyaddr(), checkdnsrr(), getmxrr(), 그리고 named(8) 기능설명(manual) 페이지.
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User Contributed Notesgethostbynamel
info at methfessel-computers.de
29-Sep-2006 03:27
29-Sep-2006 03:27
webdev at concraption dot com
19-Sep-2005 11:25
19-Sep-2005 11:25
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 ".$hostname." is not tied to any IP.";
}
?>
Skyld at o2 dot co dot uk
25-Sep-2004 01:45
25-Sep-2004 01:45
Obviously, in some cases, not all IPs are likely to be useful while checking a hostname. Sometimes also, not all IPs will work. This code will check for the first WORKING IP from the list. Or at least it should - I haven't had time to test it yet.
Needs domain parameter, and port and max IPs to check are optional.
If port is not set, it will check HTTP port 80, and if max IPs to check is not set, it will only check the first 10 IPs from the list.
Hope it helps someone.
<?php
function checkhostlist($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;
}
}
}
}
?>