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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

说明

gethostbynamel(string $hostname): array|false

返回互联网主机名 hostname 解析出来的 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

up
9
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
)
up
-2
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".
up
-2
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;
}
}
}
}
?>
up
-3
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.";
}
?>
up
-5
Anonymous
6 years ago
不要使用http协议,gethostbynamel函数中
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