regex to strip leading zeros treated as string

Question

I have numbers like this that need leading zero\'s removed.

Here is what I need:

00000004334300343 -> 4334300343

000

Answer 1

You're almost there. You just need quantifier:

str = str.replaceAll("^0+", "");

It replaces 1 or more occurrences of 0 (that is what + quantifier is for. Similarly, we have * quantifier, which means 0 or more), at the beginning of the string (that's given by caret - ^), with empty string.

Answer 2

The correct regex to strip leading zeros is

str = str.replaceAll("^0+", "");

This regex will match 0 character in quantity of one and more at the string beginning. There is not reason to worry about replaceAll method, as regex has ^ (begin input) special character that assure the replacement will be invoked only once.

Ultimately you can use Java build-in feature to do the same:

String str = "00000004334300343";
long number = Long.parseLong(str);
// outputs 4334300343

The leading zeros will be stripped for you automatically.

Answer 3

\b0+\B will do the work. See demo \b anchors your match to a word boundary, it matches a sequence of one or more zeros 0+, and finishes not in a word boundary (to not eliminate the last 0 in case you have only 00...000)

Answer 4

Another solution (might be more intuitive to read)

str = str.replaceFirst("^0+", "");

^ - match the beginning of a line
0+ - match the zero digit character one or more times

A exhausting list of pattern you can find here Pattern.

Answer 5

Accepted solution will fail if you need to get "0" from "00". This is right one:

str = str.replaceAll("0+(?!$)", "");

"^0+(?!$)" means match one or more zeros if it is not followed by end of string.

Answer 6

If you know input strings are all containing digits then you can do:

String s = "00000004334300343";
System.out.println(Long.valueOf(s));
// 4334300343

Code Demo

By converting to Long it will automatically strip off all leading zeroes.