Remove leading zero's from alphanumeic text

Question

I've seen questions 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 zero's?

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

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

Answer 2

Use Apache Commons` StringUtils class:

StringUtils.html#strip(String, String);

Answer 3

How about the regex way:

String s = "001234-a";
String 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.

Answer 4

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, str.length() - 1);

Answer 5

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

See also

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

Answer 6

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);