Regex Pattern to Find and Replace

Question

The Regex should identify texts like '0000046qwerty' and replace it with '46qwerty'.

Other Examples:

0000qw --> 0qw
123 --> 123
0123 --> 123
123qw --> 123qw
12003qw --> 12003qw

So, focus is mainly on the leading zeros and a way to truncate them in the appropriate scenario.

Solution I have is: (calls replaceFirst twice)

text.replaceFirst("^[0]+(\\\\d+)", "$1").replaceFirst("^[0]*(\\\\d{1}\\\\w+)", "$1")

Is there a single line regex to do the operation?

Answer 1

只是“跳过”前导零，在其后留下一位数字和任何符号：

text.replaceAll("^0+(\\d.*)", "$1")

Answer 2

Pattern matching is greedy so 0* will match as many zeroes as it can, leaving at least one digit to be matched with \\d. ^ means that only leading zeroes will be deleted.

text.replaceAll("^0*(\\d.*)", "$1")

Answer 3

Use regex like this : This works for all your cases :P

public static void main(String[] args) {
    String text="0000qw";
    System.out.println(text.replaceAll("^0{2,}(?=0[^0])",""));
}

O/P :

0qw

Answer 4

text.replaceFirst("^[0]+(\\d+)", "$1")

works with

0000qw -> 0qw

12003qw -> 12003qw

why do you call the second replaceFirst?