If using gethostbyname against the name of the localhost is always giving you 127.0.0.1 but you want the DNS address instead, just put a dot at the end of the name. E.g.,
$foo = gethostbynamel("myhost.example.com");
print_r($foo);
...is giving you this:
Array
(
[0] => 127.0.0.1
)
Then put a dot at the end of the name:
$foo = gethostbynamel("myhost.example.com.");
print_r($foo);
...and now you get something like:
Array
(
[0] => 172.217.1.99
)
gethostbynamel
(PHP 4, PHP 5, PHP 7, PHP 8)
gethostbynamel — Belirtilen konak adına ait IPv4 adreslerini döndürür
Açıklama
Belirtilen konakadı'na çözümlenen IPv4 adreslerinin
listesini döndürür.
Bağımsız Değişkenler
-
konakadı -
Konak adı.
Dönen Değerler
konakadı çözümlenemezse false, aksi takdirde IPv4
adreslerinin listesini döndürür.
Örnekler
Örnek 1 - gethostbynamel() örneği
<?php
$hosts = gethostbynamel('www.example.com');
print_r($hosts);
?>Yukarıdaki örneğin çıktısı:
Array
(
[0] => 208.77.188.166
)
Ayrıca Bakınız
- gethostbyname() - Belirtilen konak adına ait IPv4 adresini döndürür
- gethostbyaddr() - Belirtilen IP adresine çözümlenen konak ismini döndürür
- checkdnsrr() - Belirtilen bir konak adı veya IP adresi için DNS sorgusu yapar
- getmxrr() - Belirtilen konak adı ile ilgili MX kaydını döndürür
named(8)kılavuz sayfası
add a note
User Contributed Notes 5 notes
ab at null dot ixo dot ca ¶
8 years ago
webdev at concraption dot com ¶
19 years ago
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 ".$hostname." is not tied to any IP.";
}
?>
Skyld at o2 dot co dot uk ¶
20 years ago
Obviously, in some cases, not all IPs are likely to be useful while checking a hostname. Sometimes also, not all IPs will work. This code will check for the first WORKING IP from the list. Or at least it should - I haven't had time to test it yet.
Needs domain parameter, and port and max IPs to check are optional.
If port is not set, it will check HTTP port 80, and if max IPs to check is not set, it will only check the first 10 IPs from the list.
Hope it helps someone.
<?php
function checkhostlist($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;
}
}
}
}
?>
info at methfessel-computers.de ¶
18 years ago
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".
