Argthtjtr

In this code, we remove the substring "luna" from email string using the .replaceFirst method. We are removing the characters in between + and @. But this happens only in the first instance because we used .replaceFirst. What if we wanted to target the second instance of + and @ to remove "smith"?
Our output now is alice+@john+smith@steve+oliver@ but we want alice+luna@john+@steve+oliver@

public class Main {



    public static void main(String args) {



        String email = "alice+luna@john+smith@steve+oliver@";



        String newEmail = email.replaceFirst("\+.*?@", "");



        System.out.println(newEmail);



    }

}

You can find the second + like so:

int firstPlus = email.indexOf('+');

int secondPlus = email.indexOf('+', firstPlus + 1);

(You need to handle the case that there aren't two +s to find, if necessary).

Then find the following @:

int at = email.indexOf('@', secondPlus);

Then stitch it back together:

String newEmail = email.substring(0, secondPlus + 1) + email.substring(at);

or

String newEmail2 = new StringBuilder(email).delete(secondPlus + 1, at).toString();

Ideone demo

Unfortunately Java doesn't have methods like replace second, replace third etc. You can either replaceAll (which will replace all occurences) OR invoce replaceFirst again on the already replaced string. That's basically replacing the second. If you want to replace ONLY the second - then you can do it with substrings or do a regex matcher and iterate on results.

  public static void main(String args) {



        String email = "alice+luna@john+smith@steve+oliver@";



        String newEmail = email.replaceFirst("\+.*?@", "");

        newEmail = newEmail .replaceFirst("\+.*?@", ""); //this replaces the second right? :) 

        newEmail = newEmail .replaceFirst("\+.*?@", ""); // replace 3rd etc.



        System.out.println(newEmail);



    }

You can alternate value of parameter n in following replaceNth method to 2, 3 to perform exactly the same operation as that of replaceSecond or replaceThird. ( Note: this method can be applied in any other value of n. If nth pattern do not exist , it simply return given string ).

import java.util.regex.Matcher;

import java.util.regex.Pattern;



public class Main {



    public static String replaceNth(String str, int n, String regex, String replaceWith) {

        Pattern p = Pattern.compile(regex);

        Matcher m = p.matcher(str);

        while (m.find()) {

            n--;

            if (n == 0) {

                return str.substring(0,m.start() + 1)+ replaceWith + str.substring(m.end() - 1);

            }

        }

        return str;

    }



    public static void main(String args) {

        String email = "alice+luna@john+smith@steve+oliver@";

        System.out.println(replaceNth(email, 2, "\+.*?@", ""));

    }

}

I think it is better to encapsulate this logic into separate method, where String and group position are arguments.

private static final Pattern PATTERN = Pattern.compile("([^+@]+)@");



private static String removeSubstringGroup(String str, int pos) {

    Matcher matcher = PATTERN.matcher(str);



    while (matcher.find()) {

        if (pos-- == 0)

            return str.substring(0, matcher.start()) + str.substring(matcher.end() - 1);

    }



    return str;

}

Additionally, you can add more methods, to simplify using this util; like removeFirst() or removeLast()

public static String removeFirst(String str) {

    return removeSubstringGroup(str, 0);

}



public static String removeSecond(String str) {

    return removeSubstringGroup(str, 1);

}

Demo:

String email = "alice+luna@john+smith@steve+oliver@";

System.out.println(email);

System.out.println(removeFirst(email));

System.out.println(removeSecond(email));

System.out.println(removeSubstringGroup(email, 2));

System.out.println(removeSubstringGroup(email, 3));

Output:

alice+luna@john+smith@steve+oliver@

alice+@john+smith@steve+oliver@

alice+luna@john+@steve+oliver@

alice+luna@john+smith@steve+@

alice+luna@john+smith@steve+oliver@

Ideone demo