带有星号掩码的ip地址的javascript正则表达式 [英] javascript regex for ip address with asterisk masking

问题描述

我想用星号(*)为IP地址做一个正则表达式 符合以下条件的

I would like to make a regular expression for an IP address with asterisk(*) which matches these below:

数字127.0可以是0到255之间的任何数字.

**[TRUE]**
127.*.*.*
127.0.*.*
127.0.0.*

**[FALSE]**
127.*.*.1
127.*.0.1
127.0.*.1

到目前为止，我所做的是...，当然，未能弄清楚. 我完全迷路了.

What I have made until now is...and of course, failed to make it out. I totally got lost..

_regex = function(value) {
    var _match = /^(?:(\d|1\d\d|2[0-4]\d|25[0-5]))\.(?:(\*|\d|1\d\d|2[0-4]\d|25[0-5]))\.(\*|(?:\d{1,2}(?:.\d{1,3}?)))\.(\*|(?:\d{1,3}(?:\*?)))$
    if(_match.test(value)){
        //do something;
    }
}

如果您有任何学习的机会，我将不胜感激. 谢谢.

If you give me any chance to learn this, would be highly appreciated. Thanks.

推荐答案

我认为您要寻找的是否定性的看法，以确保没有数字出现在星号之后.

I think what you are looking for is a negative look ahead to make sure no number follows an asterisk.

例如:(\*(?!.*\d))

工作示例:

var ips = [
  '127.*.*.*',
  '127.0.*.*',
  '127.0.0.*',
  '127.*.*.1',
  '127.*.0.1',
  '127.0.*.1'
];

var regex = /^((25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|(\*(?!.*\d)))(\.|$)){4}$/;

for(var i = 0; i < ips.length; i++){
  console.log(ips[i] + ': ' + regex.test(ips[i]));
}

这篇关于带有星号掩码的ip地址的javascript正则表达式的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！