Remove/ truncate leading zeros by javascript/jquery

Question

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

Answer 1

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

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

Answer 2

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"

Answer 3

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

Answer 4

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

Answer 5

parseInt(value) or parseFloat(value)

This will work nicely.

Answer 6

Simply try to multiply by one as following:

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

Answer 7

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+/, '')

Answer 8

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>

Answer 9

Try this,

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

    var str =ltrim("01545878","0");

More here

Answer 10

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

Answer 11

If number is int use

"" + parseInt(str)

If the number is float use

"" + parseFloat(str)

Answer 12

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

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

Answer 13

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

Answer 14

Use "Math.abs"

eg: Math.abs(003) = 3;

console.log(Math.abs(003))

Answer 15

I wanted to remove all leading zeros for every sequence of digits in a string and to return 0 if the digit value equals to zero.

And I ended up doing so:

str = str.replace(/(0{1,}\d+)/, "removeLeadingZeros('$1')")

function removeLeadingZeros(string) {
    if (string.length == 1) return string
    if (string == 0) return 0
    string = string.replace(/^0{1,}/, '');
    return string
}

Answer 16

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)

Answer 17

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