Decimal validation in the textfield using javascript or jquery

Question

i Want to validate a text field in keyup event .

in the field it should accept money type decimal like

(12.23) (.23) (0.26) (5.09) (6.00)

if i enter some wrong value then it should return to the previous value and remove the wrong one

Thats your requirement, what have you tried so far? I would be easy for us to point out the mistake if you show us the code. — Praveen, Aug 28 '13 at 06:20
@abiansh what does it mean when you say "it should return to the previous value" — zzlalani, Aug 28 '13 at 06:33
var theEvent = evt || window.event; var key = theEvent.keyCode || theEvent.which; key = String.fromCharCode(key); var regex = /^(\d{0,2})(\.\d{2})$/; // number with 2 decimal places if (!regex.test(key)) { theEvent.returnValue = false; //--- this prevents the character from being displayed if (theEvent.preventDefault) theEvent.preventDefault(); } — abiansh, Aug 29 '13 at 05:08

Mulan · Answer 1 · 2014-09-29T18:32:31.017

4

I think something like this might be your best bet

var isValidCurrency = function(str) {
  var num = parseFloat(str);
  return !Number.isNaN(num) && num.toFixed(2).toString() === str;
};

Some tests

isValidCurrency("1234.56");   // true
isValidCurrency("1234.565");  // false
isValidCurrency("1234");      // false
isValidCurrency("foo");       // false

edited Sep 29 '14 at 18:32

answered Aug 28 '13 at 06:47

Mulan

129,518
31
228
259

Doesn't work for the edge case "00.00" – ryanVincent Feb 04 '14 at 15:34
I would argue that more than one leading zero is not valid. – Mulan Feb 26 '14 at 06:09
@T.G, "123foo" is not a valid currency. `isValidCurrency("123foo")` returns `false` which is appropriate in this case. – Mulan Sep 29 '14 at 18:30
@T.G if you want, you can delete your comments and I'll delete mine to avoid confusion for other users. – Mulan Sep 30 '14 at 07:51

score 0 · Answer 2 · edited Aug 28 '13 at 07:01

Try this:

function evMoneyFormat(evt) {
    //--- only accepts accepts number and 2 decimal place value
    var theEvent = evt || window.event;
    var key = theEvent.keyCode || theEvent.which;
    key = String.fromCharCode(key);
    var regex = /^[0-9]{1,14}\.[0-9]{0,2}$/; // number with 2 decimal places
    if (!regex.test(key)) {
        theEvent.returnValue = false;
        //--- this prevents the character from being displayed
        if (theEvent.preventDefault) theEvent.preventDefault();
    }
}

The control:

<input type='text' onkeyup='evMoneyFormat( e );'>

zzlalani · Answer 3 · 2013-08-28T06:41:08.163

0

You can use following Regex

val = "2.13"
if (!val.match(/^(\d{0,2})(\.\d{2})$/)) {
    alert("wrong");
} else {
    alert("right");
}

http://jsfiddle.net/rtL3J/1/

EDIT

Please note that if the numbers preceding dot (.) has limit of length to two then the code valid code is

^(\d{0,2})(\.\d{2})$

else if there is no limit then just remove the 2 from the code i.e.

^(\d{0,})(\.\d{2})$

edited Aug 28 '13 at 06:41

answered Aug 28 '13 at 06:29

zzlalani

22,960
16
44
73

Rejayi CS · Answer 4 · 2013-08-28T07:32:40.310

0

Try following code

function validateDecimal(num){

   var dotPosition=num.indexOf(".");

   if(dotPosition=="-1"){
      document.getElementById('cost').value= num+".00"
   }
}

And in html

<input type="text" id='cost' onkeyup="validateDecimal(this.value)" />

edited Aug 28 '13 at 07:32

answered Aug 28 '13 at 06:36

Rejayi CS

1,034
1
10
22

You are multiplying number by hundred? need to add `.` before `00`. – Faisal Sayed Aug 28 '13 at 07:01

Decimal validation in the textfield using javascript or jquery

4 Answers4

Linked