Reputation:

How to remove the last character from a string?

I want to remove the last character from a string. I've tried doing this:

public String method(String str) {
    if (str.charAt(str.length()-1)=='x'){
        str = str.replace(str.substring(str.length()-1), "");
        return str;
    } else{
        return str;
    }
}

Getting the length of the string - 1 and replacing the last letter with nothing (deleting it), but every time I run the program, it deletes middle letters that are the same as the last letter.

For example, the word is "admirer"; after I run the method, I get "admie." I want it to return the word admire.

Upvotes: 610

Answers (30)

Pawan

Reputation: 4

Directly use trim() to remove character from last of string

String a=1,2,3,4,5,;

I want to remove last ,

a.trim(); It'll get the task done

Upvotes: -4

Fahim Parkar

Reputation: 31647

Why not just one liner?

public static String removeLastChar(String str) {
    return removeChars(str, 1);
}

public static String removeChars(String str, int numberOfCharactersToRemove) {
    if(str != null && !str.trim().isEmpty()) {
        return str.substring(0, str.length() - numberOfCharactersToRemove);
    }
    return "";
}

Full Code

public class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        String s1 = "Remove Last CharacterY";
        String s2 = "Remove Last Character2";
        String s3 = "N";
        String s4 = null;
        String s5 = "";
        System.out.println("After removing s1==" + removeLastChar(s1) + "==");
        System.out.println("After removing s2==" + removeLastChar(s2) + "==");
        System.out.println("After removing s3==" + removeLastChar(s3) + "==");
        System.out.println("After removing s4==" + removeLastChar(s4) + "==");
        System.out.println("After removing s5==" + removeLastChar(s5) + "==");
        }
    
    public static String removeLastChar(String str) {
        return removeChars(str, 1);
    }

    public static String removeChars(String str, int numberOfCharactersToRemove) {
        if(str != null && !str.trim().isEmpty()) {
            return str.substring(0, str.length() - numberOfCharactersToRemove);
        }
        return "";
    }
}

Demo

Upvotes: 280

Denilson Anachury

Reputation: 345

Since Java 8 you can use Optional to avoid null pointer exceptions and use functional programming:

public String removeLastCharacter(String string) {
    return Optional.ofNullable(string)
        .filter(str -> !str.isEmpty() && !string.isBlank())
        .map(str -> str.substring(0, str.length() - 1))
        .orElse(""); // Should be another value that need if the {@param: string} is "null"
}

Upvotes: 0

klvnmarshall

Reputation: 57

For kotlin check out

string.dropLast(1)

Upvotes: 2

sten

Reputation: 7486

this is well known problem solved beautifully in Perl via chop() and chomp()

long answer here : Java chop and chomp

here is the code :

  //removing trailing characters
  public static String chomp(String str, Character repl) {
    int ix = str.length();
    for (int i=ix-1; i >= 0; i-- ){
      if (str.charAt(i) != repl) break;
      ix = i;
    }
    return str.substring(0,ix);
  }
  
  //hardcut  
  public static String chop(String str) {
    return str.substring(0,str.length()-1);
  }

Upvotes: 0

Vasanthakumar Jagannathan

Reputation: 863

if we want to remove file extension of the given file,

** Sample code

 public static String removeNCharactersFromLast(String str,int n){
    if (str != null && (str.length() > 0)) {
        return str.substring(0, str.length() - n);
    }

    return "";

}

Upvotes: 1

Baha' Al-Khateib

Reputation: 341

Use StringUtils.Chop(Str) it also takes care of null and empty strings, u need to import common-io:

    <dependency>
        <groupId>commons-io</groupId>
        <artifactId>commons-io</artifactId>
        <version>2.8.0</version>
    </dependency>

Upvotes: -3

tstempko

Reputation: 1193

How can a simple task be made complicated. My solution is:

public String removeLastChar(String s) {
    return s[0..-1]
}

public String removeLastChar(String s) {
    if (s.length() > 0) {
        return s[0..-1]
    }
    return s
}

Upvotes: 3

Donoso

Reputation: 107

 string = string.substring(0, (string.length() - 1));

I'm using this in my code, it's easy and simple. it only works while the String is > 0. I have it connected to a button and inside the following if statement

if (string.length() > 0) {
    string = string.substring(0, (string.length() - 1));
}

Upvotes: 8

RoboC

Reputation: 23

Easy Peasy:

StringBuilder sb= new StringBuilder();
for(Entry<String,String> entry : map.entrySet()) {
        sb.append(entry.getKey() + "_" + entry.getValue() + "|");
}
String requiredString = sb.substring(0, sb.length() - 1);

Upvotes: 0

Jacob Mehnert

Reputation: 63

I had to write code for a similar problem. One way that I was able to solve it used a recursive method of coding.

static String removeChar(String word, char charToRemove)
{
    for(int i = 0; i < word.lenght(); i++)
    {
        if(word.charAt(i) == charToRemove)
        {
            String newWord = word.substring(0, i) + word.substring(i + 1);
            return removeChar(newWord, charToRemove);
        }
    }

    return word;
}

Most of the code I've seen on this topic doesn't use recursion so hopefully I can help you or someone who has the same problem.

Upvotes: 0

C Williams

Reputation: 859

public String removeLastCharacter(String str){
       String result = null;
        if ((str != null) && (str.length() > 0)) {
          return str.substring(0, str.length() - 1);
        }
        else{
            return "";
        }

}

Upvotes: 1

Avinash Khadsan

Reputation: 497

Suppose total length of my string=24 I want to cut last character after position 14 to end, mean I want starting 14 to be there. So I apply following solution.

String date = "2019-07-31T22:00:00.000Z";
String result = date.substring(0, date.length() - 14);

Upvotes: 0

Syed Salman Hassan

Reputation: 32

just replace the condition of "if" like this:

if(a.substring(a.length()-1).equals("x"))'

this will do the trick for you.

Upvotes: 0

MC Emperor

Reputation: 23037

Most answers here forgot about surrogate pairs.

For instance, the character 𝕫 (codepoint U+1D56B) does not fit into a single char, so in order to be represented, it must form a surrogate pair of two chars.

If one simply applies the currently accepted answer (using str.substring(0, str.length() - 1), one splices the surrogate pair, leading to unexpected results.

One should also include a check whether the last character is a surrogate pair:

public static String removeLastChar(String str) {
    Objects.requireNonNull(str, "The string should not be null");
    if (str.isEmpty()) {
        return str;
    }

    char lastChar = str.charAt(str.length() - 1);
    int cut = Character.isSurrogate(lastChar) ? 2 : 1;
    return str.substring(0, str.length() - cut);
}

Upvotes: 2

Mahavir Jain

Reputation: 1571

In Kotlin you can used dropLast() method of the string class. It will drop the given number from string, return a new string

var string1 = "Some Text"
string1 = string1.dropLast(1)

Upvotes: 42

abd ak

Reputation: 1

u can do i hereString = hereString.replace(hereString.chatAt(hereString.length() - 1) ,' whitespeace');

Upvotes: -2

Bachan Joseph

Reputation: 357

 // creating StringBuilder
 StringBuilder builder = new StringBuilder(requestString);
 // removing last character from String
 builder.deleteCharAt(requestString.length() - 1);

Upvotes: 4

Patrick Parker

Reputation: 4967

Here's an answer that works with codepoints outside of the Basic Multilingual Plane (Java 8+).

Using streams:

public String method(String str) {
    return str.codePoints()
            .limit(str.codePoints().count() - 1)
            .mapToObj(i->new String(Character.toChars(i)))
            .collect(Collectors.joining());
}

More efficient maybe:

public String method(String str) {
    return str.isEmpty()? "": str.substring(0, str.length() - Character.charCount(str.codePointBefore(str.length())));
}

Upvotes: 0

simon francis

Reputation: 17

How to make the char in the recursion at the end:

public static String  removeChar(String word, char charToRemove)
    {
        String char_toremove=Character.toString(charToRemove);
        for(int i = 0; i < word.length(); i++)
        {
            if(word.charAt(i) == charToRemove)
            {
                String newWord = word.substring(0, i) + word.substring(i + 1);
                return removeChar(newWord,charToRemove);
            }
        }
        System.out.println(word);
        return word;
    }

for exemple:

removeChar ("hello world, let's go!",'l') → "heo word, et's go!llll"
removeChar("you should not go",'o') → "yu shuld nt goooo"

Upvotes: 0

Nadhu

Reputation: 429

Java 8

import java.util.Optional;

public class Test
{
  public static void main(String[] args) throws InterruptedException
  {
    System.out.println(removeLastChar("test-abc"));
  }

  public static String removeLastChar(String s) {
    return Optional.ofNullable(s)
      .filter(str -> str.length() != 0)
      .map(str -> str.substring(0, str.length() - 1))
      .orElse(s);
    }
}

Output : test-ab

Upvotes: 1

Gyan

Reputation: 12410

replace will replace all instances of a letter. All you need to do is use substring():

public String method(String str) {
    if (str != null && str.length() > 0 && str.charAt(str.length() - 1) == 'x') {
        str = str.substring(0, str.length() - 1);
    }
    return str;
}

Upvotes: 808

Mart135688

Reputation: 1057

The described problem and proposed solutions sometimes relate to removing separators. If this is your case, then have a look at Apache Commons StringUtils, it has a method called removeEnd which is very elegant.

Example:

StringUtils.removeEnd("string 1|string 2|string 3|", "|");

Would result in: "string 1|string 2|string 3"

Upvotes: 99

membersound

Reputation: 86875

Don't try to reinvent the wheel, while others have already written libraries to perform string manipulation: org.apache.commons.lang3.StringUtils.chop()

Upvotes: 43

Kasturi

Reputation: 3333

As far as the readability is concerned, I find this to be the most concise

StringUtils.substring("string", 0, -1);

The negative indexes can be used in Apache's StringUtils utility. All negative numbers are treated from offset from the end of the string.

Upvotes: 11

Tilak Madichetti

Reputation: 4346

Why not use the escape sequence ... !

System.out.println(str + '\b');

Life is much easier now . XD ! ~ A readable one-liner

Upvotes: 0

alagu raja

Reputation: 7

This is the one way to remove the last character in the string:

Scanner in = new Scanner(System.in);
String s = in.nextLine();
char array[] = s.toCharArray();
int l = array.length;
for (int i = 0; i < l-1; i++) {
    System.out.print(array[i]);
}

Upvotes: -2

Nick Louloudakis

Reputation: 6015

A one-liner answer (just a funny alternative - do not try this at home, and great answers already given):

public String removeLastChar(String s){return (s != null && s.length() != 0) ? s.substring(0, s.length() - 1): s;}

Upvotes: 2

Oleksandr Paldypanida

Reputation: 21

// Remove n last characters  
// System.out.println(removeLast("Hello!!!333",3));

public String removeLast(String mes, int n) {
    return mes != null && !mes.isEmpty() && mes.length()>n
         ? mes.substring(0, mes.length()-n): mes;
}

// Leave substring before character/string  
// System.out.println(leaveBeforeChar("Hello!!!123", "1"));

public String leaveBeforeChar(String mes, String last) {
    return mes != null && !mes.isEmpty() && mes.lastIndexOf(last)!=-1
         ? mes.substring(0, mes.lastIndexOf(last)): mes;
}

Upvotes: 2

Moad MEFTAH

Reputation: 251

if you have special character like ; in json just use String.replace(";", "") otherwise you must rewrite all character in string minus the last.

Upvotes: 0

How to remove the last character from a string?

Answers (30)

Demo

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