How to remove leading zeros from alphanumeric text?

Question

I've seen questions on how to prefix zeros here in SO. But not the other way!

Can you guys suggest me how to remove the leading zeros in alphanumeric text? Are there any built-in APIs or do I need to write a method to trim the leading zeros?

Example:

01234 converts to 1234
0001234a converts to 1234a
001234-a converts to 1234-a
101234 remains as 101234
2509398 remains as 2509398
123z remains as 123z
000002829839 converts to 2829839

score 756 · Accepted Answer · answered May 10 '10 at 06:53

756

Regex is the best tool for the job; what it should be depends on the problem specification. The following removes leading zeroes, but leaves one if necessary (i.e. it wouldn't just turn "0" to a blank string).

s.replaceFirst("^0+(?!$)", "")

The ^ anchor will make sure that the 0+ being matched is at the beginning of the input. The (?!$) negative lookahead ensures that not the entire string will be matched.

Test harness:

String[] in = {
    "01234",         // "[1234]"
    "0001234a",      // "[1234a]"
    "101234",        // "[101234]"
    "000002829839",  // "[2829839]"
    "0",             // "[0]"
    "0000000",       // "[0]"
    "0000009",       // "[9]"
    "000000z",       // "[z]"
    "000000.z",      // "[.z]"
};
for (String s : in) {
    System.out.println("[" + s.replaceFirst("^0+(?!$)", "") + "]");
}

38

Thank you. And you have tested ruthlessly ;) Great !! +1 for the tests. – jai May 10 '10 at 09:17
replaceFirst is not a function. Is it not available in FireFox? – Greg Dec 09 '11 at 21:45
5

@Greg: This question is about Java, not JavaScript. Java SE has had the method [String.replaceFirst()](http://bit.ly/MCZfrt) since version 1.4. – Jonik May 22 '12 at 10:37
How about below 1.4 ? – meddlesome Oct 17 '13 at 10:19
7

adding trim() to s.replaceFirst("^0+(?!$)", "") (ie. s.trim().replaceFirst("^0+(?!$)", "") will help in removing padded spaces! – AVA Mar 12 '14 at 11:05
2

isn't regex a bit expensive for such a simple task? – demongolem Jul 24 '15 at 15:18
finest answer,excelent – diego matos - keke Aug 08 '16 at 18:19
20

This does not work in Kotlin, you need to be explicit about the Regex `.replaceFirst("^0+(?!$)".toRegex(), "")` – mkabatek Sep 28 '18 at 21:27
2

for kotlin it is more clear to use: `yourString.trimStart('0')` – F.Mysir Jan 04 '21 at 13:46
yourString.trimStart('0') removes all zeros at the first position from a string even the string "0" will get replaced by an empty string. If you want to keep a string like "0" use @mkabatek's Answer : .replaceFirst("^0+(?!$)".toRegex(), "") – Abhijith Brumal Jul 04 '22 at 17:12
1

How can this answer be modified so that it leaves a single zero before the decimal point, like `0.5` instead of `.5`? – Raj Narayanan Dec 15 '22 at 11:57

score 167 · Answer 2 · answered Dec 17 '13 at 19:16

167

You can use the StringUtils class from Apache Commons Lang like this:

StringUtils.stripStart(yourString,"0");

answered Dec 17 '13 at 19:16

Hamilton Rodrigues

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1

Does this have a problem with "0" alone? @Hamilton Rodrigues – PhoonOne Feb 17 '15 at 17:26
3

If using this on "0" alone, it gives back "". So care if this is not the desired effect required. – dARKpRINCE Oct 21 '15 at 15:10
Upvoted cause it works for the use cases in the question asked. Helped me here too for a quick solution. Thx! – Gabriel Amazonas Mesquita Oct 27 '16 at 19:36

paxdiablo · Answer 3 · 2012-12-07T09:03:03.670

How about the regex way:

String s = "001234-a";
s = s.replaceFirst ("^0*", "");

The ^ anchors to the start of the string (I'm assuming from context your strings are not multi-line here, otherwise you may need to look into \A for start of input rather than start of line). The 0* means zero or more 0 characters (you could use 0+ as well). The replaceFirst just replaces all those 0 characters at the start with nothing.

And if, like Vadzim, your definition of leading zeros doesn't include turning "0" (or "000" or similar strings) into an empty string (a rational enough expectation), simply put it back if necessary:

String s = "00000000";
s = s.replaceFirst ("^0*", "");
if (s.isEmpty()) s = "0";

It has problem with "0" alone. – Vadzim Nov 27 '12 at 13:17 — Vadzim, Nov 27 '12 at 13:17

score 40 · Answer 4 · answered Aug 19 '19 at 14:00

40

If you are using Kotlin This is the only code that you need:

yourString.trimStart('0')

answered Aug 19 '19 at 14:00

Saman Sattari

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magiccrafter · Answer 5 · 2018-02-23T11:30:12.313

30

A clear way without any need of regExp and any external libraries.

public static String trimLeadingZeros(String source) {
    for (int i = 0; i < source.length(); ++i) {
        char c = source.charAt(i);
        if (c != '0') {
            return source.substring(i);
        }
    }
    return ""; // or return "0";
}

edited Feb 23 '18 at 11:30

answered Aug 17 '11 at 08:29

magiccrafter

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Although your check for space isn't according to the question, nevertheless I think your answer would execute the quickest. – John Fowler Oct 29 '15 at 23:00
@JohnFowler 10x for the catch, fixed after 2+ years – magiccrafter Jan 31 '18 at 12:04
1

And the method needs a return at the end if the loop finds only zeros. return ""; or return "0"; if you want at least one zero – slipperyseal Feb 23 '18 at 00:19
@slipperyseal I left it open so that you can change it based on your needs but since people tend to copy/paste, it is not a bad idea to always have a default behaviour. thanks for the comment – magiccrafter Feb 23 '18 at 11:32

score 15 · Answer 6 · answered May 10 '10 at 10:08

15

To go with thelost's Apache Commons answer: using guava-libraries (Google's general-purpose Java utility library which I would argue should now be on the classpath of any non-trivial Java project), this would use CharMatcher:

CharMatcher.is('0').trimLeadingFrom(inputString);

answered May 10 '10 at 10:08

Cowan

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+1, the correct answer for any project using Guava. (And now in 2012 that *should* mean pretty much any Java project.) – Jonik May 22 '12 at 09:18
1

@Cowan Does this have a problem with "0" alone? Will CharMatcher.is('0').trimLeadingFrom("0"); Return the "0" or empty String? – PhoonOne Feb 17 '15 at 17:28
@PhoonOne: I just tested this; it returns the empty string. – Stephan202 Dec 14 '16 at 09:48

score 13 · Answer 7 · edited Oct 11 '18 at 12:54

13

You could just do: String s = Integer.valueOf("0001007").toString();

edited Oct 11 '18 at 12:54

Kao

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answered Oct 11 '18 at 10:40

Damjan

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7

Won't handle alphanumeric. – slaman Dec 30 '19 at 21:18
1

It will also fail if the value is too big for Integer like "00087878787878787878" – Shashank Garg May 18 '21 at 19:16

score 6 · Answer 8 · answered Jan 17 '13 at 19:42

6

Use this:

String x = "00123".replaceAll("^0*", ""); // -> 123

answered Jan 17 '13 at 19:42

Eduardo Cuomo

17,828
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score 5 · Answer 9 · edited May 28 '13 at 09:15

5

Use Apache Commons StringUtils class:

StringUtils.strip(String str, String stripChars);

edited May 28 '13 at 09:15

Pang

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answered May 10 '10 at 06:32

thelost

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5

WARNING! This will strip leading and ending zeroes, which may not be what you want. – Jens Bannmann Nov 12 '10 at 10:20
24

You can strip only leading zeroes using StringUtils.stripStart(). – Josh Rosen Dec 14 '11 at 22:53

score 2 · Answer 10 · edited Feb 25 '17 at 01:05

2

Using regex as some of the answers suggest is a good way to do that. If you don't want to use regex then you can use this code:

String s = "00a0a121";

while(s.length()>0 && s.charAt(0)=='0')
{
   s = s.substring(1); 
}

edited Feb 25 '17 at 01:05

Gene Bo

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answered Mar 11 '16 at 22:45

Hüseyin Burhan ÖZKAN

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4

This could create a lot of `String`... use [magiccrafter approch](https://stackoverflow.com/a/7089841/4391450) instead. – AxelH Jan 12 '18 at 12:06

score 2 · Answer 11 · answered May 19 '17 at 16:08

2

If you (like me) need to remove all the leading zeros from each "word" in a string, you can modify @polygenelubricants' answer to the following:

String s = "003 d0g 00ss 00 0 00";
s.replaceAll("\\b0+(?!\\b)", "");

which results in:

3 d0g ss 0 0 0

answered May 19 '17 at 16:08

xdhmoore

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score 2 · Answer 12 · answered Apr 19 '21 at 04:38

2

Using kotlin it is easy

value.trimStart('0')

answered Apr 19 '21 at 04:38

Shaon

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score 2 · Answer 13 · answered Mar 15 '12 at 06:48

Using Regexp with groups:

Pattern pattern = Pattern.compile("(0*)(.*)");
String result = "";
Matcher matcher = pattern.matcher(content);
if (matcher.matches())
{
      // first group contains 0, second group the remaining characters
      // 000abcd - > 000, abcd
      result = matcher.group(2);
}

return result;

score 1 · Answer 14 · edited Sep 22 '13 at 19:46

1

I think that it is so easy to do that. You can just loop over the string from the start and removing zeros until you found a not zero char.

int lastLeadZeroIndex = 0;
for (int i = 0; i < str.length(); i++) {
  char c = str.charAt(i);
  if (c == '0') {
    lastLeadZeroIndex = i;
  } else {
    break;
  }
}

str = str.subString(lastLeadZeroIndex+1, str.length());

edited Sep 22 '13 at 19:46

Community

1
1

answered May 10 '10 at 06:34

vodkhang

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score 1 · Answer 15 · answered Oct 01 '17 at 06:10

Without using Regex or substring() function on String which will be inefficient -

public static String removeZero(String str){
        StringBuffer sb = new StringBuffer(str);
        while (sb.length()>1 && sb.charAt(0) == '0')
            sb.deleteCharAt(0);
        return sb.toString();  // return in String
    }

score 0 · Answer 16 · answered Jan 23 '13 at 10:45

       String s="0000000000046457657772752256266542=56256010000085100000";      
    String removeString="";

    for(int i =0;i<s.length();i++){
      if(s.charAt(i)=='0')
        removeString=removeString+"0";
      else 
        break;
    }

    System.out.println("original string - "+s);

    System.out.println("after removing 0's -"+s.replaceFirst(removeString,""));

score 0 · Answer 17 · answered May 10 '10 at 06:32

0

You could replace "^0*(.*)" to "$1" with regex

answered May 10 '10 at 06:32

YOU

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1

The only issue I see here is this might replace a lone zero '0' to a blank. – Dilipkumar J Feb 28 '12 at 15:25

score 0 · Answer 18 · answered Nov 05 '15 at 08:49

0

If you don't want to use regex or external library. You can do with "for":

String input="0000008008451"
String output = input.trim();
for( ;output.length() > 1 && output.charAt(0) == '0'; output = output.substring(1));

System.out.println(output);//8008451

answered Nov 05 '15 at 08:49

kfatih

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Too many `String` generated during this loop... if there is 1000 `0` ... – AxelH Jan 12 '18 at 12:08

score 0 · Answer 19 · answered Feb 18 '16 at 15:48

0

I made some benchmark tests and found, that the fastest way (by far) is this solution:

    private static String removeLeadingZeros(String s) {
      try {
          Integer intVal = Integer.parseInt(s);
          s = intVal.toString();
      } catch (Exception ex) {
          // whatever
      }
      return s;
    }

Especially regular expressions are very slow in a long iteration. (I needed to find out the fastest way for a batchjob.)

answered Feb 18 '16 at 15:48

Robert Fornesdale

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1
5

1

This doesn't work for alphanumerics, but it does work for my uses. – Brett Slocum Feb 15 '21 at 22:16

score -2 · Answer 20 · edited Nov 01 '16 at 14:02

-2

And what about just searching for the first non-zero character?

[1-9]\d+

This regex finds the first digit between 1 and 9 followed by any number of digits, so for "00012345" it returns "12345". It can be easily adapted for alphanumeric strings.

edited Nov 01 '16 at 14:02

shas123

27
7

answered Mar 04 '16 at 12:01

Sol Garlic

1

1

This will not allow zero afterwards as well. – Nishant Dongare Apr 28 '20 at 11:49

score -2 · Answer 21 · answered Mar 27 '23 at 17:31

-2

  const removeFirstZero = (ele) => parseInt(ele).toString()
  
  console.log('raw ' + '0776211121')
  console.log('removedZero ' + removeFirstZero('0776211121'))

answered Mar 27 '23 at 17:31

ravishankar balasubramaniam

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1

That does not answer the question, because it does not work if the input contains other characters. – Hulk Mar 30 '23 at 11:24
The question was asked about Java (refer to question tags), also the answer expected shall be covering all the cases and not just one particular case. Also, check if the question has been already answered. – shashi009 Apr 01 '23 at 16:01
As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Apr 01 '23 at 16:01

How to remove leading zeros from alphanumeric text?

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