How to add a trailing zero to a price with jQuery

Question

So I have a script which returns a price for a product. However the price may or may not include trailing zeros so sometimes I might have:

258.22

and other times I might have

258.2

In the later case I need to add the trailing zero with jQuery. How would I go about doing this?

Possible duplicate of [Format number to always show 2 decimal places](https://stackoverflow.com/questions/6134039/format-number-to-always-show-2-decimal-places) — Liam, Aug 06 '18 at 15:28

score 108 · Accepted Answer · edited Apr 03 '17 at 22:08

108

You can use javascript's toFixed method (source), you don't need jQuery. Example:

var number = 258.2;    
var rounded = number.toFixed(2); // rounded = 258.20

Edit: Electric Toolbox link has succumbed to linkrot and blocks the Wayback Machine so there is no working URL for the source.

edited Apr 03 '17 at 22:08

albert

answered Mar 12 '10 at 14:01

rosscj2533

2

it is not working for integers. Think that number is '258' How can we do "258.00" ? – Cihan Küsmez May 02 '15 at 12:18
@CihanKüsmez, 258 should work. Are you trying with '258' (a string)? – rosscj2533 May 02 '15 at 14:33
but how can we converts into number because its giving string . @T.J.Crowder – Mahi May 09 '17 at 11:50
@Mahi: Trailing zeros make no sense for a number, just a string (the *representation* of anumber). – T.J. Crowder May 09 '17 at 12:01

score 12 · Answer 2 · answered Mar 12 '10 at 14:01

12

Javascript has a function - toFixed - that should do what you want ... no JQuery needed.

var n = 258.2;
n.toFixed (2);  // returns 258.20

answered Mar 12 '10 at 14:01

dbrown0708

score 6 · Answer 3 · edited May 23 '17 at 10:30

6

I don't think jQuery itself has any string padding functions (which is what you're looking for). It's trivial to do, though:

function pad(value, width, padchar) {

    while (value.length < width) {
        value += padchar;
    }
    return value;
}

Edit The above is great for strings, but for your specific numeric situation, rosscj2533's answer is the better way to go.

edited May 23 '17 at 10:30

Community

answered Mar 12 '10 at 14:01

T.J. Crowder

3 Answers3