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gethostbynamel

(PHP 4, PHP 5, PHP 7, PHP 8)

gethostbynamel Obtenha uma lista de endereços IPv4 correspondentes a um determinado nome de host da Internet

Descrição

gethostbynamel(string $hostname): array|false

Retorna uma lista de endereços IPv4 aos quais o host da Internet especificado por hostname resolve.

Parâmetros

hostname

O nome do host.

Valor Retornado

Retorna uma matriz de endereços IPv4 ou false se o hostname não for possivel resolver.

Exemplos

Exemplo #1 gethostbynamel() exemplo

<?php
$hosts
= gethostbynamel('www.example.com');
print_r($hosts);
?>

O exemplo acima produzirá:

Array
(
    [0] => 192.0.34.166
)

Veja Também

  • gethostbyname() - Obtenha o endereço IPv4 correspondente a um determinado nome de host da Internet
  • gethostbyaddr() - Obtém nome do servidor de Internet correspondente ao endereço de IP fornecido
  • checkdnsrr() - Checa os registros DNS correspondentes para o nome do host ou endereço IP informado
  • getmxrr() - Get MX records corresponding to a given Internet host name
  • the named(8) manual page

add a note

User Contributed Notes 5 notes

up
8
ab at null dot ixo dot ca
7 years ago
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
)
up
-1
info at methfessel-computers.de
17 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".
up
-3
webdev at concraption dot com
18 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.";
}
?>
up
-3
Skyld at o2 dot co dot uk
19 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;
}
}
}
}
?>
up
-5
Anonymous
6 years ago
不要使用http协议,gethostbynamel函数中
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