gethostbynamel
(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
参数
hostname-
主机名
返回值
返回 IPv4 地址数组，或在 hostname
无法解析时返回 false。
示例
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
)
参见
- gethostbyname() - 返回主机名对应的 IPv4地址。
- gethostbyaddr() - 获取指定 IP 地址对应的 Internet 主机名
- checkdnsrr() - 给指定的主机（域名）或者IP地址做DNS通信检查
- getmxrr() - 获取 Internet 主机名对应的 MX 记录
named(8)手册页
＋添加备注
用户贡献的备注 5 notes
ab at null dot ixo dot ca ¶
8 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
)
info at methfessel-computers.de ¶
18 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".
Skyld at o2 dot co dot uk ¶
20 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;
}
}
}
}
?>
webdev at concraption dot com ¶
19 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.";
}
?>备份地址：http://www.lvesu.com/blog/php/function.gethostbynamel.php