How to remove leading zeros from alphanumeric text?

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)

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

StringUtils.stripStart(yourString,"0");

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";