java正则表达式匹配IP地址和端口号作为捕获组 [英] java regex matching ip address and port number as captured groups

问题描述

可以请任何人告诉我这个正则表达式有什么问题吗？

could please anybody tell me what is wrong with this regexp ?

((?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?))\\:([0-9]{2,5})

匹配此：assfasfas> 192.168.1.1:8080192.168.222.43:8286

我需要192.168.1.1和8080被捕获组

for matching this: assfasfas>192.168.1.1:8080192.168.222.43:8286

I need 192.168.1.1 and 8080 to be captured groups

谢谢

推荐答案

除非你真的，真的还必须进行IP地址验证，我建议你简化正则表达式，因为这个野兽太复杂了，只能匹配IP部分和端口部分。我的建议是

Unless you really, really have to do IP adress validation, as well, I suggest you simplify the regular expression, because this beast is far too complex for only matching "IP part" and "port part". My suggestion would be

(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}):(\d{1,5})

第1组和第2组将分别保留IP和端口。以上情况已经变得非常复杂了，恕我直言，即使这样简单就足够了：

Groups 1 and 2 will hold IP and port, respectively. And the above is already more complex that it needs to be, IMHO even something as simple as this would be enough:

(\d+\.\d+\.\d+\.\d+):(\d+)

请注意，双重反斜杠是Java字符串的要求，而不是正则表达式的要求，所以我把它们排除了。

Note that double backslashes are are requirement of Java strings, not of regex, so I left them out.

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