gethostbynamel()

(PHP 4, PHP 5, PHP 7)

获取互联网主机名对应的 IPv4 地址列表

说明

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。

范例

Example #1gethostbynamel()例子

以上例程会输出：

Array
(
    [0] => 192.0.34.166
)

参见

• gethostbyname()返回主机名对应的 IPv4地址。
• gethostbyaddr()获取指定的IP地址对应的主机名
• checkdnsrr()给指定的主机（域名）或者IP地址做DNS通信检查
• getmxrr()获取互联网主机名对应的 MX 记录
• Linux 手册页named(8)

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
)

不要使用http协议,gethostbynamel函数中

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:

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.

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".