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gethostbynamel

(PHP 4, PHP 5)

gethostbynamel Get a list of IPv4 addresses corresponding to a given Internet host name

Descrição

array gethostbynamel ( string $hostname )

Returns a list of IPv4 addresses to which the Internet host specified by hostname resolves.

Parâmetros

hostname

The host name.

Valor Retornado

Returns an array of IPv4 addresses or FALSE if hostname could not be resolved.

Exemplos

Exemplo #1 gethostbynamel() example

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

O exemplo acima irá imprimir:

Array
(
    [0] => 192.0.34.166
)

Veja Também

  • gethostbyname() - Get the IPv4 address corresponding to a given Internet host name
  • gethostbyaddr() - Obtém nome do host de Internet correspondendo ao endereçõ de IP fornecido.
  • checkdnsrr() - Check DNS records corresponding to a given Internet host name or IP address
  • getmxrr() - Obtém registros MX correspondendo ao nome do host de Internet fornecido.
  • the named(8) manual page

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User Contributed Notes 3 notes

up
0
webdev at concraption dot com
8 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
0
Skyld at o2 dot co dot uk
9 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
-1
info at methfessel-computers.de
7 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".
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