(Built-in) way in JavaScript to check if a string

Question

I'm hoping there's something in the same conceptual space as the old VB6 IsNumeric() function?

Answer 1

The accepted answer for this question has quite a few flaws (as highlighted by couple of other users). This is one of the easiest & proven way to approach it in javascript:

function isNumeric(n) {
  return !isNaN(parseFloat(n)) && isFinite(n);
}

Below are some good test cases:

console.log(isNumeric(12345678912345678912)); // true
console.log(isNumeric('2 '));                 // true
console.log(isNumeric('-32.2 '));             // true
console.log(isNumeric(-32.2));                // true
console.log(isNumeric(undefined));            // false

// the accepted answer fails at these tests:
console.log(isNumeric(''));                   // false
console.log(isNumeric(null));                 // false
console.log(isNumeric([]));                   // false

Answer 2

Quote:

isNaN(num) // returns true if the variable does NOT contain a valid number

is not entirely true if you need to check for leading/trailing spaces - for example when a certain quantity of digits is required, and you need to get, say, '1111' and not ' 111' or '111 ' for perhaps a PIN input.

Better to use:

var num = /^\d+$/.test(num)

Answer 3

My solution:

// returns true for positive ints; 
// no scientific notation, hexadecimals or floating point dots

var isPositiveInt = function(str) { 
   var result = true, chr;
   for (var i = 0, n = str.length; i < n; i++) {
       chr = str.charAt(i);
       if ((chr < "0" || chr > "9") && chr != ",") { //not digit or thousands separator
         result = false;
         break;
       };
       if (i == 0 && (chr == "0" || chr == ",")) {  //should not start with 0 or ,
         result = false;
         break;
       };
   };
   return result;
 };

You can add additional conditions inside the loop, to fit you particular needs.

Answer 4

Maybe there are one or two people coming across this question who need a much stricter check than usual (like I did). In that case, this might be useful:

if(str === String(Number(str))) {
  // it's a "perfectly formatted" number
}

Beware! This will reject strings like .1, 40.000, 080, 00.1. It's very picky - the string must match the "most minimal perfect form" of the number for this test to pass.

It uses the String and Number constructor to cast the string to a number and back again and thus checks if the JavaScript engine's "perfect minimal form" (the one it got converted to with the initial Number constructor) matches the original string.

Answer 5

I do it like this:

function isString(value)
{
    return value.length !== undefined;
}
function isNumber(value)
{
    return value.NaN !== undefined;
}

Of course isString() will be tripped up here if you pass some other object that has 'length' defined.

Answer 6

Old question, but there are several points missing in the given answers.

Scientific notation.

!isNaN('1e+30') is true, however in most of the cases when people ask for numbers, they do not want to match things like 1e+30.

Large floating numbers may behave weird

Observe (using Node.js):

> var s = Array(16 + 1).join('9')
undefined
> s.length
16
> s
'9999999999999999'
> !isNaN(s)
true
> Number(s)
10000000000000000
> String(Number(s)) === s
false
>

On the other hand:

> var s = Array(16 + 1).join('1')
undefined
> String(Number(s)) === s
true
> var s = Array(15 + 1).join('9')
undefined
> String(Number(s)) === s
true
>

So, if one expects String(Number(s)) === s, then better limit your strings to 15 digits at most (after omitting leading zeros).

Infinity

> typeof Infinity
'number'
> !isNaN('Infinity')
true
> isFinite('Infinity')
false
>

Given all that, checking that the given string is a number satisfying all of the following:

• non scientific notation
• predictable conversion to Number and back to String
• finite

is not such an easy task. Here is a simple version:

  function isNonScientificNumberString(o) {
    if (!o || typeof o !== 'string') {
      // Should not be given anything but strings.
      return false;
    }
    return o.length <= 15 && o.indexOf('e+') < 0 && o.indexOf('E+') < 0 && !isNaN(o) && isFinite(o);
  }

However, even this one is far from complete. Leading zeros are not handled here, but they do screw the length test.