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)
gethostbynamel — Obtener una lista de direcciones IPv4 que corresponde a un nombre de host de Internet dado
Descripción
array gethostbynamel
( string
$hostname
)
Devuelve una lista de direcciones IPv4 a la que el host de
Internet especificado por hostname resuelve.
Parámetros
-
hostname -
El nombre de host.
Valores devueltos
Devuelve un array de direcciones IPv4 o FALSE si
hostname no se pudo resolver.
Ejemplos
Ejemplo #1 Ejemplo de gethostbynamel()
<?php
$hosts = gethostbynamel('www.example.com');
print_r($hosts);
?>
El resultado del ejemplo serÃa:
Array
(
[0] => 192.0.34.166
)
Ver también
- gethostbyname() - Obtener la dirección IPv4 que corresponde a un nombre de host de Internet dado
- gethostbyaddr() - Obtener el nombre del host de Internet correspondiente a una dirección IP dada
- checkdnsrr() - Comprueba registros DNS correspondientes a un nombre de host de Internet dado o dirección IP
- getmxrr() - Obtener los registros MX correspondientes a un nombre de host de Internet
- the named(8) manual page
add a note
User Contributed Notes 4 notes
ab at null dot ixo dot ca ¶
1 year ago
webdev at concraption dot com ¶
11 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 ".$hostname." is not tied to any IP.";
}
?>
Skyld at o2 dot co dot uk ¶
12 years ago
Obviously, in some cases, not all IPs are likely to be useful while checking a hostname. Sometimes also, not all IPs will work. This code will check for the first WORKING IP from the list. Or at least it should - I haven't had time to test it yet.
Needs domain parameter, and port and max IPs to check are optional.
If port is not set, it will check HTTP port 80, and if max IPs to check is not set, it will only check the first 10 IPs from the list.
Hope it helps someone.
<?php
function checkhostlist($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 ¶
10 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".
