Need to match the digit using regex by removing the initial zeroes

I want to match initial character in the string by using regex to ignore the symbol "+" or "-" and consuming only 0 before any digit between 1 to 9.

Example

+004200
004200

Here, in the above example, I want to match only +4200 and 4200 respectively, by removing the initial zeroes.

I tried to solve it, by using the expression ^[^-+]\0+, but it is not matching anything. By analysing further, I figured that the expression [^-+] is still consuming the value. Can anybody suggest the correct approach?

If you want to remove optional zeroes and keep at least a single digit 1-9 after it:

([+-]?)0*([1-9]\d*)

Explanation

• ([+-]?) Capture group 1, match an optional + or -
• 0* Match optional zeroes
• ([1-9]\d*) Capture group 2, match a digit 1-9 and optional digits 0-9

See a regex101 demo.

In the replacement use group 1 and group 2

If there should be at least 1 zero present:

([+-]?)0+([1-9]\d*)

See another regex101 demo.

I would look at this:

/([+\-])*[0]*(\d+)/g

Your first capture group ($1 or \\1) will contain the +/- if present, and the second group ($2 or \\2) will contain the number without leading 0s

As a note - you have to escape - with \ when it's in a character group [] since it is used to denote sequences - [a-z] means all letters between a and z.

See this example here:

https://regexr.com/78tih