Validating Floating Point Precision in Javascript

Question

I'm trying to use Javascript to verify that user input is a valid monetary (dollars and cents) amount, with no more than two digits to the right of the decimal point.

Shifting left, then performing a modulo 1 fails because of inherent floating-point issues:

(parseFloat("19.99") * 100) % 1 // Returns 0.9999999999997726

I'm aware of the BigDecimal library, but I'm not sure that converting to BigDecimal and then doing the validation would be the right approach, since it could cover up invalid inputs as well as floating-point issues.

As it stands, my workaround is to test the raw (string) input against the following regex:

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

Is this sufficient to guarantee that the input is a valid float in currency format? And if not, is there a math-based way to do the validation?

are .7 and 7. allowed?

Casimir et Hippolyte
– Casimir et Hippolyte

2014-05-01 17:05:54 +00:00
Commented May 1, 2014 at 17:05 — Casimir et Hippolyte
– Casimir et Hippolyte, Commented May 1, 2014 at 17:05
Yes, as are "7", "7.7" and "0.7"

Bluesky
– Bluesky

2014-05-01 17:07:30 +00:00
Commented May 1, 2014 at 17:07 — Bluesky
– Bluesky, Commented May 1, 2014 at 17:07
(+(n).toFixed(2) == +n)

dandavis
– dandavis

2014-05-01 17:08:46 +00:00
Commented May 1, 2014 at 17:08 — dandavis
– dandavis, Commented May 1, 2014 at 17:08
Dan, you mind making that a formal answer so I can mark it?

Bluesky
– Bluesky

2014-05-01 17:20:07 +00:00
Commented May 1, 2014 at 17:20 — Bluesky
– Bluesky, Commented May 1, 2014 at 17:20

dandavis · Accepted Answer · 2014-05-01 17:24:50Z

2

if you don't need to use RegExp, let JS's math do the heavy-lifting:

 var n=myInput.value;
 if(!n || +(+n || 0).toFixed(2) != +n) {  
   alert("real numbers only!"); 
   myInput.select(); 
 }

the validation routine part is made up and poor UX, but the math is solid and bullet-proof.

answered May 1, 2014 at 17:24

dandavis

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2 Comments

G. Stoynev Over a year ago

Above will fail for n=0 at !n, no? (And let's not go into "is zero a real number :-))

dandavis Over a year ago

@G.Stoynev I thought that was the desired behavior, but it's easy enough to handle that case if needed...

anubhava · Accepted Answer · 2014-05-01 17:08:04Z

2

Your regex will also match empty input, which is not right.

Try this regex instead:

^(?=[^.])\d*(?:\.\d{0,2})?$

Working Demo

(?=[^.]) is a lookahead that will make sure at least one non-dot character is there in input to avoid matching empty string.

answered May 1, 2014 at 17:08

anubhava

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1 Comment

Bluesky Over a year ago

Matching "12." is desired functionality. I'll use the leading follow quantifier, though, thanks.

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