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

Answer 1

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 (ie 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+(?!$)", "") + "]");
}

See also

• regular-expressions.info
  • repetitions , lookarounds , and anchors
• String.replaceFirst(String regex)

Answer 2

您可以像这样使用Apache Commons Lang 中的StringUtils类：

StringUtils.stripStart(yourString,"0");

Answer 3

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";

Answer 4

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";
}

Answer 5

如果您使用 Kotlin，这是您唯一需要的代码：

yourString.trimStart('0')

Answer 6

要使用 thelost 的 Apache Commons 答案：使用guava-libraries （我认为 Google 的通用 Java 实用程序库现在应该在任何非平凡的 Java 项目的类路径上），这将使用CharMatcher ：

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

Answer 7

你可以这样做： String s = Integer.valueOf("0001007").toString();

Answer 8

用这个：

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

Answer 9

使用 Apache Commons StringUtils类：

StringUtils.strip(String str, String stripChars);

Answer 10

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); 
}

Answer 11

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;

Answer 12

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());

Answer 13

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

Answer 14

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
    }

Answer 15

       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,""));

Answer 16

您可以用正则表达式将"^0*(.*)"替换为"$1"

Answer 17

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

Answer 18

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.)

Answer 19

使用 kotlin 很容易

value.trimStart('0')

Answer 20

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.