What is equivalent to .replaceSecond or .replaceThird?

up vote
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down vote

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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);



    }

}

java string replace

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edited yesterday

Prashant Gupta

739519

asked yesterday

NoobCoder

717

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use a regex to do this?
  – triplem
  yesterday
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

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);



    }

}

java string replace

share|improve this question

edited yesterday

Prashant Gupta

739519

asked yesterday

NoobCoder

717

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use a regex to do this?
  – triplem
  yesterday
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

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);



    }

}

java string replace

share|improve this question

edited yesterday

Prashant Gupta

739519

asked yesterday

NoobCoder

717

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);



    }

}

java string replace

java string replace

share|improve this question

edited yesterday

Prashant Gupta

739519

asked yesterday

NoobCoder

717

share|improve this question

edited yesterday

Prashant Gupta

739519

asked yesterday

NoobCoder

717

share|improve this question

share|improve this question

edited yesterday

Prashant Gupta

739519

edited yesterday

Prashant Gupta

739519

edited yesterday

Prashant Gupta

739519

739519

asked yesterday

NoobCoder

717

asked yesterday

NoobCoder

717

asked yesterday

NoobCoder

717

717

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use a regex to do this?
  – triplem
  yesterday
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Use a regex to do this?
  – triplem
  yesterday
  

  
  
  
  

1

1

Use a regex to do this?
– triplem
yesterday

Use a regex to do this?
– triplem
yesterday

add a comment | 

4 Answers
4

active

oldest

votes

up vote
2
down vote

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

share|improve this answer

edited yesterday

answered yesterday

Andy Turner

78.3k878129

add a comment | 

up vote
1
down vote

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);



    }

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

add a comment | 

up vote
0
down vote

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, "\+.*?@", ""));

    }

}

share|improve this answer

answered yesterday

Bishal Gautam

131211

add a comment | 

up vote
0
down vote

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

share|improve this answer

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

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

share|improve this answer

edited yesterday

answered yesterday

Andy Turner

78.3k878129

add a comment | 

up vote
2
down vote

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

share|improve this answer

edited yesterday

answered yesterday

Andy Turner

78.3k878129

add a comment | 

up vote
2
down vote

up vote
2
down vote

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

share|improve this answer

edited yesterday

answered yesterday

Andy Turner

78.3k878129

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

share|improve this answer

edited yesterday

answered yesterday

Andy Turner

78.3k878129

share|improve this answer

share|improve this answer

edited yesterday

edited yesterday

edited yesterday

answered yesterday

Andy Turner

78.3k878129

answered yesterday

Andy Turner

78.3k878129

answered yesterday

Andy Turner

78.3k878129

78.3k878129

add a comment | 

add a comment | 

up vote
1
down vote

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);



    }

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

add a comment | 

up vote
1
down vote

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);



    }

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

add a comment | 

up vote
1
down vote

up vote
1
down vote

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);



    }

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

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);



    }

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

share|improve this answer

share|improve this answer

answered yesterday

Veselin Davidov

5,4861515

answered yesterday

Veselin Davidov

5,4861515

answered yesterday

Veselin Davidov

5,4861515

5,4861515

add a comment | 

add a comment | 

up vote
0
down vote

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, "\+.*?@", ""));

    }

}

share|improve this answer

answered yesterday

Bishal Gautam

131211

add a comment | 

up vote
0
down vote

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, "\+.*?@", ""));

    }

}

share|improve this answer

answered yesterday

Bishal Gautam

131211

add a comment | 

up vote
0
down vote

up vote
0
down vote

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, "\+.*?@", ""));

    }

}

share|improve this answer

answered yesterday

Bishal Gautam

131211

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, "\+.*?@", ""));

    }

}

share|improve this answer

answered yesterday

Bishal Gautam

131211

share|improve this answer

share|improve this answer

answered yesterday

Bishal Gautam

131211

answered yesterday

Bishal Gautam

131211

answered yesterday

Bishal Gautam

131211

131211

add a comment | 

add a comment | 

up vote
0
down vote

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

share|improve this answer

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

add a comment | 

up vote
0
down vote

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

share|improve this answer

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

add a comment | 

up vote
0
down vote

up vote
0
down vote

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

share|improve this answer

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

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

share|improve this answer

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

share|improve this answer

share|improve this answer

edited yesterday

edited yesterday

edited yesterday

answered yesterday

oleg.cherednik

4,1902916

answered yesterday

oleg.cherednik

4,1902916

answered yesterday

oleg.cherednik

4,1902916

4,1902916

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