Unable to fetch numbers from string by java split() method

Question

I have this code which takes an IP address(string) of the format 255.255.255.255 and needs to perform some post processing on those numbers (not posted here) but for which the string must be converted into an array of ints.

I have used here the split() method but it's not giving me the result. I saw other answers on sp doing it with regex but none of them worked for me.

import java.util.Scanner;
public class Main{
    public static void main(String args[]){
        String text;

        Scanner take=new Scanner(System.in);
        text=take.nextLine();
        String data[]=text.split(".",4);

        for(String w:data){
            System.out.println(w);
        }
        take.close();
    }
}

I have tried with the input 12.36.26.25

But it outputs 36.26.25 which was supposed to be like 12 36 26 25

i love the "please dont mark as duplicate" haha. gimme a sec to look over the code — Alan
– Alan, Commented Feb 12, 2019 at 8:35
Try escaping the period (\\.) as it's a special regex symbol (it matches any character). — hensing1
– hensing1, Commented Feb 12, 2019 at 8:35

Veselin Davidov · Accepted Answer · 2019-02-12 08:38:22Z

11

Use it like that:

        String example="12.36.26.25";
        String data[]=example.split("\\.");

        for(String w:data){
            System.out.println(w);
        }

And it will do what you want ;)

Using split(regex,limit) as you did will actually split on any character (since . is the regex for any character) and it will basically remove the first few characters

edited Feb 12, 2019 at 8:38

answered Feb 12, 2019 at 8:36

Veselin Davidov

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5 Comments

Alan Over a year ago

as simple as that lmao

dutta.ari Over a year ago

what is the significance of " \\." can u please explain?

Alan Over a year ago

let me refer to this post aswell stackoverflow.com/questions/3674930/… in conjunction to Veselin Davidov's answer

Aritro Sen Over a year ago

It escapes the '.' character. Basically '.' is the regex form of any character

hensing1 Over a year ago

If you're not familiar with regular expressions, check out stackoverflow.com/questions/2912894/… .. or just do some googling ;)

Sid · Accepted Answer · 2019-02-12 08:53:20Z

2

Solution using java.util.regex package:

import java.util.regex.*;
public class Main{

     public static void main(String args[]){
        String text = "12.36.26.25";
        String separator = ".";
        String[] data = text.split(Pattern.quote(separator));
        System.out.println(data.length);
        for(String w: data){
            System.out.println(w);
        }
    }
}

The Pattern.quote will do the escaping.

edited Feb 12, 2019 at 8:53

answered Feb 12, 2019 at 8:40

Sid

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Comments

Alan · Accepted Answer · 2019-02-12 08:43:15Z

0

Let me give you guys an alternative solution aswell:

public static void main(String args[]){
        String text;

        Scanner take=new Scanner(System.in);
        text=take.nextLine();
        String data[]=text.split("[.]");

        for(String w:data){
            System.out.println(w);
        } 
        take.close();
    }

you can escape the dot using "[.]" as a string, turning the dot from an regex operator into a literal char

answered Feb 12, 2019 at 8:43

Alan

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Comments

javalearner_heaven · Accepted Answer · 2019-02-12 11:48:11Z

0

Just change this line

String data[]=text.split(".",4);

to

String data[]=text.split("\\.");

answered Feb 12, 2019 at 11:48

javalearner_heaven

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1 Comment

javalearner_heaven Over a year ago

For all the special character the escape \\ is required. They are \, ^, $, ., |, ?, *, +, (, ), [, ], {, }

Leo Varghese · Accepted Answer · 2019-02-13 05:39:25Z

0

An alternate solution is to use the sun.net.util.IPAddressUtil to extract the numbers from a given String of an IPAddress as below-

String ipAddressString = "10.20.30.40";
byte[] numbers = IPAddressUtil.textToNumericFormatV4(ipAddressString);

The numbers in ByteArray format can be converted to integer array using this solution.

answered Feb 13, 2019 at 5:39

Leo Varghese

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