Remove/ truncate leading zeros by javascript/jquery

Suggest solution for removing or truncating leading zeros from number(any string) by javascript,jquery.

You can use a regular expression that matches zeroes at the beginning of the string:

s = s.replace(/^0+/, '');

I would use the Number() function:

var str = "00001";
str = Number(str).toString();

>> "1"

Or I would multiply my string by 1

var str = "00000000002346301625363";
str = (str * 1).toString();

>> "2346301625363"

Maybe a little late, but I want to add my 2 cents.

if your string ALWAYS represents a number, with possible leading zeros, you can simply cast the string to a number by using the '+' operator.

e.g.

x= "00005";
alert(typeof x); //"string"
alert(x);// "00005"

x = +x ; //or x= +"00005"; //do NOT confuse with x+=x, which will only concatenate the value
alert(typeof x); //number , voila!
alert(x); // 5 (as number)

if your string doesn't represent a number and you only need to remove the 0's use the other solutions, but if you only need them as number, this is the shortest way.

and FYI you can do the opposite, force numbers to act as strings if you concatenate an empty string to them, like:

x = 5;
alert(typeof x); //number
x = x+"";
alert(typeof x); //string

hope it helps somebody

Since you said "any string", I'm assuming this is a string you want to handle, too.

"00012  34 0000432    0035"

So, regex is the way to go:

var trimmed = s.replace(/\b0+/g, "");

And this will prevent loss of a "000000" value.

var trimmed = s.replace(/\b(0(?!\b))+/g, "")

You can see a working example here

parseInt(value) or parseFloat(value)

This will work nicely.

I got this solution for truncating leading zeros(number or any string) in javascript:

<script language="JavaScript" type="text/javascript">
<!--
function trimNumber(s) {
  while (s.substr(0,1) == '0' && s.length>1) { s = s.substr(1,9999); }
  return s;
}

var s1 = '00123';
var s2 = '000assa';
var s3 = 'assa34300';
var s4 = 'ssa';
var s5 = '121212000';

alert(s1 + '=' + trimNumber(s1));
alert(s2 + '=' + trimNumber(s2));
alert(s3 + '=' + trimNumber(s3));
alert(s4 + '=' + trimNumber(s4));
alert(s5 + '=' + trimNumber(s5));
// end hiding contents -->
</script>

Simply try to multiply by one as following:

> "00123" * 1;              // Get as number
> "00123" * 1 + "";       // Get as string

Try this,

   function ltrim(str, chars) {
        chars = chars || "\\s";
        return str.replace(new RegExp("^[" + chars + "]+", "g"), "");
    }
    
    var str =ltrim("01545878","0");

More here

1. The most explicit is to use parseInt():

parseInt(number, 10)

2. Another way is to use the + unary operator:

+number

3. You can also go the regular expression route, like this:

number.replace(/^0+/, '')

You should use the "radix" parameter of the "parseInt" function : https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/parseInt?redirectlocale=en-US&redirectslug=JavaScript%2FReference%2FGlobal_Objects%2FparseInt

parseInt('015', 10) => 15

if you don't use it, some javascript engine might use it as an octal parseInt('015') => 0

If number is int use

"" + parseInt(str)

If the number is float use

"" + parseFloat(str)

const number = '0000007457841'; console.log(+number) //7457841;

OR number.replace(/^0+/, '')

Regex solution from Guffa, but leaving at least one character

"123".replace(/^0*(.+)/, '$1'); // 123
"012".replace(/^0*(.+)/, '$1'); // 12
"000".replace(/^0*(.+)/, '$1'); // 0

One another way without regex:

function trimLeadingZerosSubstr(str) {
    var xLastChr = str.length - 1, xChrIdx = 0;
    while (str[xChrIdx] === "0" && xChrIdx < xLastChr) {
        xChrIdx++;
    }
    return xChrIdx > 0 ? str.substr(xChrIdx) : str;
}

With short string it will be more faster than regex (jsperf)

const input = '0093';
const match = input.match(/^(0+)(\d+)$/);
const result = match && match[2] || input;

Use "Math.abs"

eg: Math.abs(003) = 3;

console.log(Math.abs(003))