How to identify patterns of repeated characters within string?

Question

Lets say you have a string like this:

198<string>12<string>88<string>201

Yes that looks like an IPv4 address, because it is one.

How do i check to see if there are repeated patterns in a string? I've no idea where to start, and im not sure that regex would help since i dont really know what string to look for, just that it will be repeated.

The goal here is to strip <string> from the main string.

Okay lets say the string is:

String test = "112(dot)115(dot)48(dot)582";
if(isIP(test){
   System.out.println("Yep, it's an ip");
}

The output should be:

Yep, it's an ip

The seperator (dot) will always be different.

possible duplicate of [Validating IPv4 string in Java](http://stackoverflow.com/questions/4581877/validating-ipv4-string-in-java) — Matt Coubrough, Dec 07 '14 at 21:54
no, the separators being the same is fine, atleast for the sake of simplicity. — ThatGuy343, Dec 07 '14 at 22:06

John · Accepted Answer · 2014-12-07T22:20:55.540

1

Try this: https://regex101.com/r/oR1gS8/4

/^((?:\d{1,3}[^0-9]+){3}\d{1,3})$/

Matches 198<string>12<string>88<string>201, 112(dot)115(dot)48(dot)582, and 112<test>115<test>48<test>582, among others...

edited Dec 07 '14 at 22:20

answered Dec 07 '14 at 21:58

John

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it will check correct ip in this example 11223472938(dot)115(dot)48(dot)582 but this isn't valid. – İlker Korkut Dec 07 '14 at 22:18
192.1.1 is Invalid IP4 but your regex returns true for it – sol4me Dec 07 '14 at 22:21
Fixed... Take a look at the original post. – John Dec 07 '14 at 22:24
It will say `12x23.34-45` as valid. – afzalex Dec 07 '14 at 22:41
Is it not? @ThatGuy343 implied the delimiter could be any string. The one weakness this regex has is that it'll allow groups with values above 255, which can be fixed if necessary. – John Dec 07 '14 at 22:51

İlker Korkut · Answer 2 · 2014-12-07T22:20:33.757

1

/((((\d{1,3})\D{1,5}){3})(\d{1,3}))$/

112(dot)115(dot)48(dot)582

Matches

1.  [0-26]  `112(dot)115(dot)48(dot)582`
2.  [0-23]  `112(dot)115(dot)48(dot)`
3.  [16-23] `48(dot)`
4.  [16-18] `48`
5.  [23-26] `582`

control one by one your cases in here

edited Dec 07 '14 at 22:20

answered Dec 07 '14 at 22:03

İlker Korkut

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interesting, im going to test this a bit more – ThatGuy343 Dec 07 '14 at 22:07
It doesn't work for this: 11211548582 but it works for this 112dot115dot48dot582 – ThatGuy343 Dec 07 '14 at 22:08
this also only works for dot, the ip separator could be any string, not just "dot". – ThatGuy343 Dec 07 '14 at 22:11
All I know seperator should be certain value. It can be cause issues. – İlker Korkut Dec 07 '14 at 22:16
Added regex101 url, you can check your cases one by one. – İlker Korkut Dec 07 '14 at 22:25
It will say `12x23.34-45` as valid. – afzalex Dec 07 '14 at 22:40
You are right @afzalex, but the seperators must be same. Otherwise all results can be valid. – İlker Korkut Dec 07 '14 at 22:51
But considering the heading of the OP these answers are not solving the problem. @İlkerKorkut – afzalex Dec 07 '14 at 22:55

afzalex · Answer 3 · 2014-12-07T22:32:31.273

0

You will need to capture group to accomplish this and use Non number D.

public boolean isIP(String test) {
    String regex = "\\d+(\\D+)\\d+\\1\\d+\\1\\d+";
    return test.matches(regex);
}

Here I have used regex : \d+(\D+)\d+\1\d+\1\d+ It is equivalent to : -

  \d+          (\D+)       \d+         \1          \d+          \1         \d+
numbers_1 non-numbers_1 numbers_2 non-numbers_1 numbers_3 non-numbers_1 numbers_4

Or you can further reduce the above regex to \d+(\D+)\d+(\1\d+){2}

edited Dec 07 '14 at 22:32

answered Dec 07 '14 at 22:12

afzalex

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1231231128.123.23.121212312 , it will say this ip is valid. – İlker Korkut Dec 07 '14 at 22:20
The above answer is for : **How do i check to see if there are repeated patterns in a string** – afzalex Dec 07 '14 at 22:31
you said valid, then what it means 0 to 255 ? – İlker Korkut Dec 07 '14 at 22:32
please read my above comment. I am saying this is to check repeated pattern. – afzalex Dec 07 '14 at 22:33
And I posted this answer because as you can see nobody marked this, and in their answer they are saying `12x23.34-45` as valid. And I have not edited my answer. it will still say valid to `1231231128.123.23.121212312`. @İlkerKorkut – afzalex Dec 07 '14 at 22:38
Yes, it means this is not valid. Anyway, this is boring subject, Welcome to Network101 :) – İlker Korkut Dec 07 '14 at 22:41

score 0 · Answer 4 · answered Dec 07 '14 at 22:40

This should help

import java.util.regex.Pattern;

public class App
{
    private static String IPV4_REGEX ="^(\\d{1,3})\\.(\\d{1,3})\\.(\\d{1,3})\\.(\\d{1,3})$";
    private static final Pattern IP4_PATTERN = Pattern.compile(IPV4_REGEX);

    public static void main( String[] args ) {
        String test1 = "198<string>12<string>88<string>201";
        String test2 = "198(foo)12(foo)88(foo)201";
        if(isIP(test1)) {
            System.out.println("Yep, it's an ip");
        }
        if(isIP(test2)) {
            System.out.println("Yep, it's an ip");
        }
    }

    public static boolean isIP(String input) {

        String[] chunks = input.replaceAll("[^\\d.]", "x").split("x+");
        if (chunks.length != 4) {
            System.out.println("not valid ");
            return false; 
        }

        String ip = chunks[0] + "." + chunks[1] + "." + chunks[2] + "." + chunks[3];
        return IP4_PATTERN.matcher(ip).matches();
    }
}

How to identify patterns of repeated characters within string?

4 Answers4