(PHP 4, PHP 5, PHP 7, PHP 8)
gethostbynamel — 获取互联网主机名对应的 IPv4 地址列表
hostname主机名
返回 IPv4 地址数组，或在 hostname
无法解析时返回 false。
示例 #1 gethostbynamel() 例子
<?php
$hosts = gethostbynamel('www.example.com');
print_r($hosts);
?>以上示例会输出：
Array
(
[0] => 192.0.34.166
)
named(8) 手册页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
)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".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;
}
}
}
}
?>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.";
}
?>