我认为这是最好的Ip验证规则。请检查一次!!

^(([01]?[0-9]?[0-9]|2([0-4][0-9]|5[0-5]))\.){3}([01]?[0-9]?[0-9]|2([0-4][0-9]|5[0-5]))$

新的网络框架为结构IPv4Address和结构IPv6Address提供了可失败的初始化器，可以很容易地处理IP地址部分。在IPv6中使用regex实现这一点很困难，因为所有的缩短规则。

不幸的是，对于主机名，我没有一个优雅的答案。

注意，网络框架是最近的，所以它可能会强迫你编译最新的操作系统版本。

import Network
let tests = ["192.168.4.4","fkjhwojfw","192.168.4.4.4","2620:3","2620::33"]

for test in tests {
    if let _ = IPv4Address(test) {
        debugPrint("\(test) is valid ipv4 address")
    } else if let _ = IPv6Address(test) {
        debugPrint("\(test) is valid ipv6 address")
    } else {
        debugPrint("\(test) is not a valid IP address")
    }
}

output:
"192.168.4.4 is valid ipv4 address"
"fkjhwojfw is not a valid IP address"
"192.168.4.4.4 is not a valid IP address"
"2620:3 is not a valid IP address"
"2620::33 is valid ipv6 address"

>>> my_hostname = "testhostn.ame"
>>> print bool(re.match("^(([a-zA-Z]|[a-zA-Z][a-zA-Z0-9\-]*[a-zA-Z0-9])\.)*([A-Za-z]|[A-Za-z][A-Za-z0-9\-]*[A-Za-z0-9])$", my_hostname))
True
>>> my_hostname = "testhostn....ame"
>>> print bool(re.match("^(([a-zA-Z]|[a-zA-Z][a-zA-Z0-9\-]*[a-zA-Z0-9])\.)*([A-Za-z]|[A-Za-z][A-Za-z0-9\-]*[A-Za-z0-9])$", my_hostname))
False
>>> my_hostname = "testhostn.A.ame"
>>> print bool(re.match("^(([a-zA-Z]|[a-zA-Z][a-zA-Z0-9\-]*[a-zA-Z0-9])\.)*([A-Za-z]|[A-Za-z][A-Za-z0-9\-]*[A-Za-z0-9])$", my_hostname))
True

检查主机名，比如…mywebsite.co。In, thangaraj.name, 18thangaraj。thangaraj106。在等,

[a-z\d+].*?\\.\w{2,4}$

这个怎么样?

([0-9]{1,3}\.){3}[0-9]{1,3}

Regarding IP addresses, it appears that there is some debate on whether to include leading zeros. It was once the common practice and is generally accepted, so I would argue that they should be flagged as valid regardless of the current preference. There is also some ambiguity over whether text before and after the string should be validated and, again, I think it should. 1.2.3.4 is a valid IP but 1.2.3.4.5 is not and neither the 1.2.3.4 portion nor the 2.3.4.5 portion should result in a match. Some of the concerns can be handled with this expression:

grep -E '(^|[^[:alnum:]+)(([0-1]?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\.){3}([0-1]?[0-9]{1,2}|2[0-4][0-9]|25[0-5])([^[:alnum:]]|$)' 

The unfortunate part here is the fact that the regex portion that validates an octet is repeated as is true in many offered solutions. Although this is better than for instances of the pattern, the repetition can be eliminated entirely if subroutines are supported in the regex being used. The next example enables those functions with the -P switch of grep and also takes advantage of lookahead and lookbehind functionality. (The function name I selected is 'o' for octet. I could have used 'octet' as the name but wanted to be terse.)

grep -P '(?<![\d\w\.])(?<o>([0-1]?[0-9]{1,2}|2[0-4][0-9]|25[0-5]))(\.\g<o>){3}(?![\d\w\.])'

如果IP地址在一个包含句子形式文本的文件中，那么点号的处理实际上可能会产生错误的否定，因为句号可以跟在后面，而不是点号符号的一部分。上面的一个变体可以修复这个问题:

grep -P '(?<![\d\w\.])(?<x>([0-1]?[0-9]{1,2}|2[0-4][0-9]|25[0-5]))(\.\g<x>){3}(?!([\d\w]|\.\d))'