Validate decimal numbers in JavaScript - IsNumeric

Question

What's the cleanest, most effective way to validate decimal numbers in JavaScript?

Bonus points for:

1. Clarity. Solution should be clean and simple.
2. Cross-platform.

Test cases:

01. IsNumeric('-1')      => true
02. IsNumeric('-1.5')    => true
03. IsNumeric('0')       => true
04. IsNumeric('0.42')    => true
05. IsNumeric('.42')     => true
06. IsNumeric('99,999')  => false
07. IsNumeric('0x89f')   => false
08. IsNumeric('#abcdef') => false
09. IsNumeric('1.2.3')   => false
10. IsNumeric('')        => false
11. IsNumeric('blah')    => false

Answer 1

I realize the original question did not mention jQuery, but if you do use jQuery, you can do:

$.isNumeric(val)

Simple.

https://api.jquery.com/jQuery.isNumeric/ (as of jQuery 1.7)

Answer 2

I'd like to add the following:

1. IsNumeric('0x89f') => true
2. IsNumeric('075') => true

Positive hex numbers start with 0x and negative hex numbers start with -0x. Positive oct numbers start with 0 and negative oct numbers start with -0. This one takes most of what has already been mentioned into consideration, but includes hex and octal numbers, negative scientific, Infinity and has removed decimal scientific (4e3.2 is not valid).

function IsNumeric(input){
  var RE = /^-?(0|INF|(0[1-7][0-7]*)|(0x[0-9a-fA-F]+)|((0|[1-9][0-9]*|(?=[\.,]))([\.,][0-9]+)?([eE]-?\d+)?))$/;
  return (RE.test(input));
}

Answer 3

None of the answers return false for empty strings, a fix for that...

function is_numeric(n)
{
 return (n != '' && !isNaN(parseFloat(n)) && isFinite(n));
}

Answer 4

My solution,

function isNumeric(input) {
    var number = /^\-{0,1}(?:[0-9]+){0,1}(?:\.[0-9]+){0,1}$/i;
    var regex = RegExp(number);
    return regex.test(input) && input.length>0;
}

It appears to work in every situation, but I might be wrong.

Answer 5

Yeah, the built-in isNaN(object) will be much faster than any regex parsing, because it's built-in and compiled, instead of interpreted on the fly.

Although the results are somewhat different to what you're looking for (try it):

                                              // IS NUMERIC
document.write(!isNaN('-1') + "<br />");      // true
document.write(!isNaN('-1.5') + "<br />");    // true
document.write(!isNaN('0') + "<br />");       // true
document.write(!isNaN('0.42') + "<br />");    // true
document.write(!isNaN('.42') + "<br />");     // true
document.write(!isNaN('99,999') + "<br />");  // false
document.write(!isNaN('0x89f') + "<br />");   // true
document.write(!isNaN('#abcdef') + "<br />"); // false
document.write(!isNaN('1.2.3') + "<br />");   // false
document.write(!isNaN('') + "<br />");        // true
document.write(!isNaN('blah') + "<br />");    // false

Answer 6

To check if a variable contains a valid number and not just a String which looks like a number, Number.isFinite(value) can be used.

This is part of the language since ES2015

Examples:

Number.isFinite(Infinity)   // false
Number.isFinite(NaN)        // false
Number.isFinite(-Infinity)  // false

Number.isFinite(0)          // true
Number.isFinite(2e64)       // true

Number.isFinite('0')        // false
Number.isFinite(null)       // false