用于检查 IP 地址的 JavaScript 正则表达式
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javascript regular expression to check for IP addresses
提问by KennC.
I have several ip addresses like:
我有几个 IP 地址，例如：
115.42.150.37115.42.150.38115.42.150.50
115.42.150.37115.42.150.38115.42.150.50
What type of regular expression should I write if I want to search for the all the 3 ip addresses? Eg, if I do 115.42.150.*(I will be able to search for all 3 ip addresses)
如果我想搜索所有 3 个 ip 地址，我应该写什么类型的正则表达式？例如，如果我这样做115.42.150.*（我将能够搜索所有 3 个 IP 地址）
What I can do now is something like: /[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}/but it can't seems to work well.
我现在可以做的是：/[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}\.[0-9]{1-3}/但它似乎不能很好地工作。
Thanks.
谢谢。
采纳答案by Spudley
The regex you've got already has several problems:
你的正则表达式已经有几个问题：
Firstly, it contains dots. In regex, a dot means "match any character", where you need to match just an actual dot. For this, you need to escape it, so put a back-slash in front of the dots.
首先，它包含点。在正则表达式中，点表示“匹配任何字符”，您只需要匹配一个实际的点。为此，您需要对其进行转义，因此在点前放置一个反斜杠。
Secondly, but you're matching any three digits in each section. This means you'll match any number between 0 and 999, which obviously contains a lot of invalid IP address numbers.
其次，但您要匹配每个部分中的任意三位数字。这意味着您将匹配 0 到 999 之间的任何数字，这些数字显然包含许多无效的 IP 地址编号。
This can be solved by making the number matching more complex; there are other answers on this site which explain how to do that, but frankly it's not worth the effort -- in my opinion, you'd be much better off splitting the string by the dots, and then just validating the four blocks as numeric integer ranges -- ie:
这可以通过使数字匹配更复杂来解决；这个网站上还有其他答案解释了如何做到这一点，但坦率地说，这不值得付出努力——在我看来，你最好用点分割字符串，然后将四个块验证为数字整数范围——即：
if(block >= 0 && block <= 255) {....}
Hope that helps.
希望有帮助。
回答by ErickBest
May be late but, someone could try:
可能会迟到，但有人可以尝试：
Example of VALID IP address
有效 IP 地址示例
115.42.150.37
192.168.0.1
110.234.52.124
Example of INVALID IP address
无效 IP 地址示例
210.110 – must have 4 octets
255 – must have 4 octets
y.y.y.y – only digits are allowed
255.0.0.y – only digits are allowed
666.10.10.20 – octet number must be between [0-255]
4444.11.11.11 – octet number must be between [0-255]
33.3333.33.3 – octet number must be between [0-255]
JavaScript code to validate an IP address
用于验证 IP 地址的 JavaScript 代码
function ValidateIPaddress(ipaddress) {
if (/^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/.test(ipaddress)) {
return (true)
}
alert("You have entered an invalid IP address!")
return (false)
}
回答by oriadam
Try this one, it's a shorter version:
试试这个，它是一个较短的版本：
^(?!0)(?!.*\.$)((1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$
Explained:
解释：
^ start of string
(?!0) Assume IP cannot start with 0
(?!.*\.$) Make sure string does not end with a dot
(
(
1?\d?\d| A single digit, two digits, or 100-199
25[0-5]| The numbers 250-255
2[0-4]\d The numbers 200-249
)
\.|$ the number must be followed by either a dot or end-of-string - to match the last number
){4} Expect exactly four of these
$ end of string
Unit test for a browser's console:
浏览器控制台的单元测试：
var rx=/^(?!0)(?!.*\.$)((1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$/;
var valid=['1.2.3.4','11.11.11.11','123.123.123.123','255.250.249.0','1.12.123.255','127.0.0.1','1.0.0.0'];
var invalid=['0.1.1.1','01.1.1.1','012.1.1.1','1.2.3.4.','1.2.3\n4','1.2.3.4\n','259.0.0.1','123.','1.2.3.4.5','.1.2.3.4','1,2,3,4','1.2.333.4','1.299.3.4'];
valid.forEach(function(s){if (!rx.test(s))console.log('bad valid: '+s);});
invalid.forEach(function(s){if (rx.test(s)) console.log('bad invalid: '+s);});
回答by Andrew K
If you are using nodejs try:
如果您使用的是 nodejs，请尝试：
require('net').isIP('10.0.0.1')
require('net').isIP('10.0.0.1')
doc net.isIP()
doc net.isIP()
回答by Teja Kantamneni
回答by Ben McCormick
If you want something more readable than regex for ipv4 in modern browsers you can go with
如果你想要一些比现代浏览器中 ipv4 的正则表达式更具可读性的东西，你可以使用
function checkIsIPV4(entry) {
var blocks = entry.split(".");
if(blocks.length === 4) {
return blocks.every(function(block) {
return parseInt(block,10) >=0 && parseInt(block,10) <= 255;
});
}
return false;
}
回答by mikemaccana
Don't write your own regex or copy paste!You probably won't cover all edge ceses (IPv6, but also octal IPs, etc). Use the is-ippackage from npm:
不要编写自己的正则表达式或复制粘贴！您可能不会涵盖所有边缘服务（IPv6，还有八进制 IP 等）。使用is-ip来自 npm的包：
var isIp = require('is-ip');
isIp('192.168.0.1');
isIp('1:2:3:4:5:6:7:8');
Will return a Boolean.
将返回一个布尔值。
Downvoters: care to explain why using an actively maintained library is better than copy pasting from a website?
Downvoters：注意解释为什么使用积极维护的库比从网站复制粘贴更好？
回答by Mahdi Pedram
Below Solution doesn't accept Padding Zeros
下面的解决方案不接受填充零
Here is the cleanest way to validate an IP Address, Let's break it down:
这是验证 IP 地址的最简洁方法，让我们分解一下：
Fact:a valid IP Address is has 4 octets, each octets can be a number between 0 - 255
事实：有效的 IP 地址是 has 4 octets，每个八位字节可以是介于0 - 255
Breakdown of Regex that matches any value between 0 - 255
匹配任何值之间的正则表达式的细分 0 - 255
25[0-5]matches250 - 2552[0-4][0-9]matches200 - 2491[0-9][0-9]matches100 - 199[1-9][0-9]?matches1 - 990matches0
25[0-5]火柴250 - 2552[0-4][0-9]火柴200 - 2491[0-9][0-9]火柴100 - 199[1-9][0-9]?火柴1 - 990火柴0
const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';
Notes:When using new RegExpyou should use \\.instead of \.since string will get escaped twice.
注意：使用时new RegExp你应该使用\\.而不是\.因为字符串会被转义两次。
function isValidIP(str) {
const octet = '(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]?|0)';
const regex = new RegExp(`^${octet}\.${octet}\.${octet}\.${octet}$`);
return regex.test(str);
}
回答by Tom
/^(?!.*\.$)((?!0\d)(1?\d?\d|25[0-5]|2[0-4]\d)(\.|$)){4}$/
Full credit to oriadam. I would have commented below his/her answer to suggest the double zero change I made, but I do not have enough reputation here yet...
完全归功于oriadam。我会在他/她的回答下面发表评论，建议我做出双零更改，但我在这里还没有足够的声誉......
change:
改变：
-(?!0) Because IPv4 addresses starting with zeros ('0.248.42.223') are valid(but not usable)
+(?!0\d) Because IPv4 addresses with leading zeros ('63.14.209.00' and '011.012.013.014') can sometimes be interpreted as octal
-(?!0) 因为 IPv4 地址以零开头 ('0.248.42.223') 是有效的（但不可用）
+(?!0\d) 因为带有前导零的 IPv4 地址（'63.14.209.00' 和 '011.012.013.014'）有时可以解释为八进制
回答by SamWhan
Throwing in a late contribution:
抛出迟到的贡献：
^(?!\.)((^|\.)([1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d))){4}$
Of the answers I checked, they're either longer or incomplete in their verification. Longer, in my experience, means harder to overlook and therefore more prone to be erroneous. And I like to avoid repeating similar patters, for the same reason.
在我检查过的答案中，它们的验证要么更长，要么不完整。根据我的经验，更长的时间意味着更难忽视，因此更容易出错。出于同样的原因，我喜欢避免重复类似的模式。
The main part is, of course, the test for a number - 0 to 255, but also making sure it doesn't allow initial zeroes (except for when it's a single one):
当然，主要部分是测试数字 - 0 到 255，但还要确保它不允许初始零（除非它是单个零）：
[1-9]?\d|1\d\d|2(5[0-5]|[0-4]\d)
Three alternations - one for sub 100: [1-9]?\d, one for 100-199: 1\d\dand finally 200-255: 2(5[0-5]|[0-4]\d).
三种交替 - 一种用于 sub 100: [1-9]?\d，一种用于 100-199:1\d\d最后 200-255: 2(5[0-5]|[0-4]\d)。
This is preceded by a test for start of lineora dot ., and this whole expression is tested for 4 times by the appended {4}.
这之前是对行首或点的.测试，并且整个表达式由附加的{4}.测试 4 次。
This complete test for four byte representations is started by testing for start of line followed by a negative look ahead to avoid addresses starting with a .: ^(?!\.), and ended with a test for end of line ($).
这个对四字节表示的完整测试开始于测试行首，然后是否定前瞻以避免以.:开头的地址^(?!\.)，并以行尾 ( $)测试结束。
