Currency validation

Please help with me writing a JavaScript Validation for currency/money field.

So please provide any regular expressions if u have :)

Also, for开发者_如何学JAVA my region, don't need any currency symbols like '$' in the field.

Only decimals are to be included for validation as special chars., along with numbers.

You could use a regexp:

var regex  = /^\d+(?:\.\d{0,2})$/;
var numStr = "123.20";
if (regex.test(numStr))
    alert("Number is valid");

If you're not looking to be as strict with the decimal places you might find it easier to use the unary (+) operator to cast to a number to check it's validity:

var numStr = "123.20";
var numNum = +numStr; // gives 123.20

If the number string is invalid, it will return NaN (Not a Number), something you can test for easily:

var numStr = "ab123c";
var numNum = +numStr;
if (isNaN(numNum))
    alert("numNum is not a number");

It will, of course, allow a user to add more decimal places but you can chop any extra off using number.toFixed(2) to round to 2 decimal places. parseFloat is much less strict with input and will pluck the first number it can find out of a string, as long as that string starts with a number, eg. parseFloat("123abc") would yield 123.

I built my answer from the accepted answer.

var regex = /^[1-9]\d*(((,\d{3}){1})?(\.\d{0,2})?)$/;

^[1-9] The number must start with 1-9
\d* The number can then have any number of any digits
(...)$ look at the next group from the end (...)$
(...)?(...)? Look for two groups optionally. The first is for the comma, the second is for the decimal.
(,\d{3}){1} Look for one occurance of a comma followed by exactly three digits
\.\d{0,2} Look for a decimal followed by zero, one, or two digits.

This regex works off of these rules:

• Valid values are numbers 0-9, comma and decimal point.

• If a customer enters more than one decimal point or more than one comma, the value is invalid and will not be accepted.

• Examples of invalid input values

  • 1.2.3
  • 1,2,4
• Examples of valid input values
  • 1.23
  • 1,000
  • 3967.
  • 23
  • 1.2
  • 999,999.99

An example can be seen here: http://jsfiddle.net/rat141312/Jpxu6/1/

UPDATE

by changing the [1-9] in the regex to [0-9] any number less than 1 can also be validated. Example: 0.42, 007

/[1-9]\d*(?:\.\d{0,2})?/

[1-9] - must start with 1 to 9
\d* - any number of other digits
(?: )? - non capturing optional group
\. - a decimal point
\d{0,2} - 0 to 2 digits

does that work for you? or maybe parseFloat:

var float = parseFloat( input );

 let amount = document.querySelector('#amount'), preAmount = amount.value;
        amount.addEventListener('input', function(){
            if(isNaN(Number(amount.value))){
                amount.value = preAmount;
                return;
            }

            let numberAfterDecimal = amount.value.split(".")[1];
            if(numberAfterDecimal && numberAfterDecimal.length > 3){
                amount.value = Number(amount.value).toFixed(3);;
            }
            preAmount = amount.value;
        })

<input type="text" id="amount">

For me its working fine for Indian currency in INR

var regex = /^[1-9]{0,2}(,{0,1})(\d{2},)*(\d{3})*(?:\.\d{0,2})$/;
var a = '1,111.11';
regex.test(a); 

Now I use this:

let func = function (vStr) {
    let v0 = Number(vStr);
    let v1 = Number(v0.toFixed(2));
    return v0 === v1;
};

Note that, NaN === NaN returns false. Maybe some substitution for '$' and ',' before parsing is needed, for other cases. And there is a problem of precision for very large number, longer than 16 digits. As well as values of '0x3a', '68n' is considered valid.

Nowadays, <input> of type="number", with step='.01' may be more proper.