使用正则表达式验证 IPv4 地址 [英] Validating IPv4 addresses with regexp

问题描述

我一直试图为 IPv4 验证获得一个有效的正则表达式，但运气不佳.似乎有一次我用 (25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}，但它产生了一些奇怪的结果:

I've been trying to get an efficient regex for IPv4 validation, but without much luck. It seemed at one point I had had it with (25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}, but it produces some strange results:

$ grep --version
grep (GNU grep) 2.7
$ grep -E '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}' <<< 192.168.1.1
192.168.1.1
$ grep -E '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}' <<< 192.168.1.255
192.168.1.255
$ grep -E '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}' <<< 192.168.255.255
$ grep -E '(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?(.|$)){4}' <<< 192.168.1.2555
192.168.1.2555

我进行了搜索，看看是否已经有人问过并回答过这个问题，但其他答案似乎只是简单地展示了如何确定 4 组 1-3 个数字，或者对我不起作用.

I did a search to see if this had already been asked and answered, but other answers appear to simply show how to determine 4 groups of 1-3 numbers, or do not work for me.

推荐答案

^((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]?)$

接受:

127.0.0.1
192.168.1.1
192.168.1.255
255.255.255.255
0.0.0.0
1.1.1.01        # This is an invalid IP address!

拒绝:

30.168.1.255.1
127.1
192.168.1.256
-1.2.3.4
1.1.1.1.
3...3

在线尝试单元测试:https://www.debuggex.com/r/-EDZOqxTxhiTncN6/1

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