使用正则表达式验证输入字符串是否为 0-255 之间的数字

Question

I am facing problem while matching input string with Regex. I want to validate input number is between 0-255 and length should be up to 3 characters long. code is working fine but when I input 000000 up to any length is shows true instead false.

Here is my code :-

String IP = "000000000000000";
        System.out.println(IP.matches("(0*(?:[0-9][0-9]?|[0-2][0-5][0-5]))"));

Answer 1

Tested this:

static String pattern = "^(([0-1]?[0-9]?[0-9]?|2[0-4][0-9]|25[0-5])\\.){3}([0-1]?[0-9]?[0-9]?|2[0-4][0-9]|25[0-5]){1}$";

It works for the following:

• IP Addresses xxx.xxx.xxx.xxx / xx.xx.xx.xx / xxxx / mix of these.
• Leading zeros are allowed.
• Range 0-255 / maximum 3 digts.

Answer 2

You can use this regex:

boolean valid = IP.matches("\\b(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\\b");

RegEx Demo

Answer 3

You can use this pattern which matches "0" , "1" , ... "255" :

"([0-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])"

Demo on Ideone

Answer 4

Using boundary tags to ensure only (0 to 255) numbers is matched, the optimized pattern that I have to offer is:

\b(?:1\d{2}|2[0-4]\d|[1-9]?\d|25[0-5])\b

Pattern Demo (in PHP/PCRE to show step count)

4010 steps when checking a list from 0 to 256 .

This pattern will not match 01 or 001 . (no match on one or more leading zeros)

Considerations:

1. Use quantifiers on consecutive duplicate characters.
2. Organize the alternatives not in sequential order ( single-digit, double-digit, 100's, under-249, 250-255 ) but with quickest mis-matches first.
3. Avoid non-essential capture (or non-capture) groups. (despite seeming logical to condense the "two hundreds" portion of the pattern)

Answer 5

Please try this

"^(((\d|0\d|00\d|\d{2}|0\d{2}|1\d{2}|2[0-4]\d|2[0-5]{2})\.){3})(\d|0\d|00\d|\d{2}|0\d{2}|1\d{2}|2[0-4]\d|2[0-5]{2})$"

It works also with leading zeroes

Answer 6

boolean valid = IP.matches("(0?[0-9]{1,2}|1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])");

Answer 7

Complete ip inet4 match :

JS

/(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\.(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\.(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\.(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])/g.exec(myIp);

https://regex101.com/r/tU3gC3/12

Minified :

/(1?(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])\.){3}(1?[0-9]{1,2}|2[0-4][0-9]|25[0-5])/g.exec(myIp);

https://regex101.com/r/tU3gC3/13

Answer 8

这将适用于以下包含初始零的模式和 ip，例如：023.45.12.56

pattern=(\\d{1,2}|(0|1)\\d{2}|2[0-4]\\d|25[0-5]);

Answer 9

If you need leading zeros, try this:

"((\\d{1,2}|[01]\\d{1,2}|[0-2][0-4]\\d|25[0-5])\\.){3}(\\d{1,2}|[01]\\d{1,2}|[0-2][0-4]\\d|25[0-5])"

It satisfies following conditions: IP address is a string in the form "ABCD", where the value of A, B, C, and D may range from 0 to 255. Leading zeros are allowed. The length of A, B, C, or D can't be greater than 3.

Maybe somebody can help with additional simplifying?

Answer 10

If you want to validate ip4 with 'ip/mask', so regex looks like this:

^((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]|[1-2][0-9]|3[0-2]))$

Just ip

^((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]?)$

JS code to test it

function isMatchByRegexp(stringToValidate, regexp) {
var re = new RegExp(regexp);
return re.test(stringToValidate);
}

Answer 11

(2[0-4][0-9])|([0-1]?[0-9]?[0-9])

To match 0 to 249 specifically

Answer 12

You can simplify it by thinking in four conditions that might happen

 String zeroTo255 = "((0|1)\\d{2}|2[0-4]\\d|25[0-5]|\\d{1,2})";
 String validIpPattern = zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255 + "\\." + zeroTo255;

1. (0|1)\\d{2} catches any three digit number starting with 0 or 1.
2. 2[0-4]\\d catches numbers between 200 and 249.
3. 25[0-5] catches numbers between 250 and 255.
4. \\d{1,2} catches any one or two digit number

you can test it in https://regexr.com/

To test it here in regexr you need to remove one backslash

(\\\\d --> \\d)

((0|1)\d{2}|2[0-4]\d|25[0-5]|\d{1,2})

Note that \\d represents digits in regular expressions, same as [0-9]