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.

no, the separators being the same is fine, atleast for the sake of simplicity. — ThatGuy343
– ThatGuy343, Commented Dec 7, 2014 at 22:06

John · Accepted Answer · 2014-12-07 22:20:55Z

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 7, 2014 at 22:20

answered Dec 7, 2014 at 21:58

John

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5 Comments

İlker Korkut Over a year ago

it will check correct ip in this example 11223472938(dot)115(dot)48(dot)582 but this isn't valid.

sol4me Over a year ago

192.1.1 is Invalid IP4 but your regex returns true for it

John Over a year ago

Fixed... Take a look at the original post.

afzalex Over a year ago

It will say 12x23.34-45 as valid.

John Over a year ago

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.

İlker Korkut · Accepted Answer · 2014-12-07 22:20:33Z

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 7, 2014 at 22:20

answered Dec 7, 2014 at 22:03

İlker Korkut

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8 Comments

ThatGuy343 Over a year ago

interesting, im going to test this a bit more

ThatGuy343 Over a year ago

It doesn't work for this: 112<test>115<test>48<test>582 but it works for this 112dot115dot48dot582

ThatGuy343 Over a year ago

this also only works for dot, the ip separator could be any string, not just "dot".

İlker Korkut Over a year ago

All I know seperator should be certain value. It can be cause issues.

İlker Korkut Over a year ago

Added regex101 url, you can check your cases one by one.

|

afzalex · Accepted Answer · 2014-12-07 22:32:31Z

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 7, 2014 at 22:32

answered Dec 7, 2014 at 22:12

afzalex

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6 Comments

İlker Korkut Over a year ago

1231231128.123.23.121212312 , it will say this ip is valid.

afzalex Over a year ago

The above answer is for : How do i check to see if there are repeated patterns in a string

İlker Korkut Over a year ago

you said valid, then what it means 0 to 255 ?

afzalex Over a year ago

please read my above comment. I am saying this is to check repeated pattern.

afzalex Over a year ago

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

|

sol4me · Accepted Answer · 2014-12-07 22:40:17Z

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();
    }
}

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