Web студия "GrandView"

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array gethostbynamel ( string hostname )

Returns a list of IP addresses to which the Internet host specified by hostname resolves. Returns FALSE if hostname could not be resolved.

Пример 1. gethostbynamel() example


<?php

  $hosts = gethostbynamel('www.example.com');

  print_r($hosts);

?>

The printout of the above program will be:

Array
(
    [0] => 192.0.34.166
)

add a note User Contributed Notes
gethostbynamel

info at methfessel-computers.de
29-Sep-2006 03:27


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".

webdev at concraption dot com
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


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) {

gethostbynamel

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