How can I check if a string is a valid number?

Question

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

Answer 1

2nd October 2020: note that many bare-bones approaches are fraught with subtle bugs (eg. whitespace, implicit partial parsing, radix, coercion of arrays etc.) that many of the answers here fail to take into account. The following implementation might work for you, but note that it does not cater for number separators other than the decimal point ".":

function isNumeric(str) {
  if (typeof str != "string") return false // we only process strings!  
  return !isNaN(str) && // use type coercion to parse the _entirety_ of the string (`parseFloat` alone does not do this)...
         !isNaN(parseFloat(str)) // ...and ensure strings of whitespace fail
}

---

To check if a variable (including a string) is a number, check if it is not a number:

This works regardless of whether the variable content is a string or number.

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

Examples

isNaN(123)         // false
isNaN('123')       // false
isNaN('1e10000')   // false (This translates to Infinity, which is a number)
isNaN('foo')       // true
isNaN('10px')      // true
isNaN('')          // false
isNaN(' ')         // false
isNaN(false)       // false

Of course, you can negate this if you need to. For example, to implement the IsNumeric example you gave:

function isNumeric(num){
  return !isNaN(num)
}

To convert a string containing a number into a number:

Only works if the string only contains numeric characters, else it returns NaN.

+num               // returns the numeric value of the string, or NaN 
                   // if the string isn't purely numeric characters

Examples

+'12'              // 12
+'12.'             // 12
+'12..'            // NaN
+'.12'             // 0.12
+'..12'            // NaN
+'foo'             // NaN
+'12px'            // NaN

To convert a string loosely to a number

Useful for converting '12px' to 12, for example:

parseInt(num)      // extracts a numeric value from the 
                   // start of the string, or NaN.

Examples

parseInt('12')     // 12
parseInt('aaa')    // NaN
parseInt('12px')   // 12
parseInt('foo2')   // NaN      These last three may
parseInt('12a5')   // 12       be different from what
parseInt('0x10')   // 16       you expected to see.

Floats

Bear in mind that, unlike +num, parseInt (as the name suggests) will convert a float into an integer by chopping off everything following the decimal point (if you want to use parseInt() because of this behaviour, you're probably better off using another method instead):

+'12.345'          // 12.345
parseInt(12.345)   // 12
parseInt('12.345') // 12

Empty strings

Empty strings may be a little counter-intuitive. +num converts empty strings or strings with spaces to zero, and isNaN() assumes the same:

+''                // 0
+'   '             // 0
isNaN('')          // false
isNaN('   ')       // false

But parseInt() does not agree:

parseInt('')       // NaN
parseInt('   ')    // NaN

Answer 2

If you're just trying to check if a string is a whole number (no decimal places), regex is a good way to go. Other methods such as isNaN are too complicated for something so simple.

function isNumeric(value) {
    return /^-?\d+$/.test(value);
}

console.log(isNumeric('abcd'));         // false
console.log(isNumeric('123a'));         // false
console.log(isNumeric('1'));            // true
console.log(isNumeric('1234567890'));   // true
console.log(isNumeric('-23'));          // true
console.log(isNumeric(1234));           // true
console.log(isNumeric(1234n));          // true
console.log(isNumeric('123.4'));        // false
console.log(isNumeric(''));             // false
console.log(isNumeric(undefined));      // false
console.log(isNumeric(null));           // false

To only allow positive whole numbers use this:

function isNumeric(value) {
    return /^\d+$/.test(value);
}

console.log(isNumeric('123'));          // true
console.log(isNumeric('-23'));          // false

Answer 3

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 4

And you could go the RegExp-way:

var num = "987238";

if(num.match(/^-?\d+$/)){
  //valid integer (positive or negative)
}else if(num.match(/^\d+\.\d+$/)){
  //valid float
}else{
  //not valid number
}

Answer 5

If you really want to make sure that a string contains only a number, any number (integer or floating point), and exactly a number, you cannot use parseInt()/ parseFloat(), Number(), or !isNaN() by themselves. Note that !isNaN() is actually returning true when Number() would return a number, and false when it would return NaN, so I will exclude it from the rest of the discussion.

The problem with parseFloat() is that it will return a number if the string contains any number, even if the string doesn't contain only and exactly a number:

parseFloat("2016-12-31")  // returns 2016
parseFloat("1-1") // return 1
parseFloat("1.2.3") // returns 1.2

The problem with Number() is that it will return a number in cases where the passed value is not a number at all!

Number("") // returns 0
Number(" ") // returns 0
Number(" \u00A0   \t\n\r") // returns 0

The problem with rolling your own regex is that unless you create the exact regex for matching a floating point number as Javascript recognizes it you are going to miss cases or recognize cases where you shouldn't. And even if you can roll your own regex, why? There are simpler built-in ways to do it.

However, it turns out that Number() (and isNaN()) does the right thing for every case where parseFloat() returns a number when it shouldn't, and vice versa. So to find out if a string is really exactly and only a number, call both functions and see if they both return true:

function isNumber(str) {
  if (typeof str != "string") return false // we only process strings!
  // could also coerce to string: str = ""+str
  return !isNaN(str) && !isNaN(parseFloat(str))
}

Answer 6

2019: Including ES3, ES6 and TypeScript Examples

Maybe this has been rehashed too many times, however I fought with this one today too and wanted to post my answer, as I didn't see any other answer that does it as simply or thoroughly:

ES3

var isNumeric = function(num){
    return (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);  
}

ES6

const isNumeric = (num) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);

Typescript

const isNumeric = (num: any) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num as number);

This seems quite simple and covers all the bases I saw on the many other posts and thought up myself:

// Positive Cases
console.log(0, isNumeric(0) === true);
console.log(1, isNumeric(1) === true);
console.log(1234567890, isNumeric(1234567890) === true);
console.log('1234567890', isNumeric('1234567890') === true);
console.log('0', isNumeric('0') === true);
console.log('1', isNumeric('1') === true);
console.log('1.1', isNumeric('1.1') === true);
console.log('-1', isNumeric('-1') === true);
console.log('-1.2354', isNumeric('-1.2354') === true);
console.log('-1234567890', isNumeric('-1234567890') === true);
console.log(-1, isNumeric(-1) === true);
console.log(-32.1, isNumeric(-32.1) === true);
console.log('0x1', isNumeric('0x1') === true);  // Valid number in hex
// Negative Cases
console.log(true, isNumeric(true) === false);
console.log(false, isNumeric(false) === false);
console.log('1..1', isNumeric('1..1') === false);
console.log('1,1', isNumeric('1,1') === false);
console.log('-32.1.12', isNumeric('-32.1.12') === false);
console.log('[blank]', isNumeric('') === false);
console.log('[spaces]', isNumeric('   ') === false);
console.log('null', isNumeric(null) === false);
console.log('undefined', isNumeric(undefined) === false);
console.log([], isNumeric([]) === false);
console.log('NaN', isNumeric(NaN) === false);

You can also try your own isNumeric function and just past in these use cases and scan for "true" for all of them.

Or, to see the values that each return:

Answer 7

Try the isNan function:

The isNaN() function determines whether a value is an illegal number (Not-a-Number).

This function returns true if the value equates to NaN. Otherwise it returns false.

This function is different from the Number specific Number.isNaN() method.

  The global isNaN() function, converts the tested value to a Number, then tests it.

Number.isNan() does not convert the values to a Number, and will not return true for any value that is not of the type Number...

Answer 8

TL;DR

It depends largely on what you want to parse as a number.

Comparison Between Built-in Functions

As none of the existing sources satisfied my soul, I tried to figure out what actually was happening with these functions.

Three immediate answers to this question felt like:

1. !isNaN(input) (which gives the same output as +input === +input)
2. !isNaN(parseFloat(input))
3. isFinite(input)

But are any of them correct in every scenario?

I tested these functions in several cases, and generated output as markdown. This is what it looks like:

input	!isNaN(input) or +input===+input	!isNaN( parseFloat( input))	isFinite( input)	Comment
123	✔️	✔️	✔️	-
'123'	✔️	✔️	✔️	-
12.3	✔️	✔️	✔️	-
'12.3'	✔️	✔️	✔️	-
'   12.3   '	✔️	✔️	✔️	Empty whitespace trimmed, as expected.
1_000_000	✔️	✔️	✔️	Numeric separator understood, also expected.
'1_000_000'	❌	✔️	❌	Surprise! JS just won't parse numeric separator inside a string. For details, check this issue. (Why then parsing as float worked though? Well, it didn't. )
'0b11111111'	✔️	✔️	✔️	Binary form understood, as it should've.
'0o377'	✔️	✔️	✔️	Octal form understood too.
'0xFF'	✔️	✔️	✔️	Of course hex is understood. Did anybody think otherwise?
''	✔️	❌	✔️	Should empty string be a number?
'    '	✔️	❌	✔️	Should a whitespace-only string be a number?
'abc'	❌	❌	❌	Everybody agrees, not a number.
'12.34Ab!@#$'	❌	✔️	❌	Ah! Now it's quite understandable what parseFloat() does. Not impressive to me, but may come handy in certain cases.
'10e100'	✔️	✔️	✔️	10100 is a number indeed. But caution! It's way more larger than the maximum safe integer value 253 (about 9×1015). Read this for details.
'10e1000'	✔️	✔️	❌	Say with me, help! Though not as crazy as it may seem. In JavaScript, a value larger than ~10308 is rounded to infinity, that's why. Look here for details. And yes, isNaN() considers infinity as a number, and parseFloat() parses infinity as infinity.
null	✔️	❌	✔️	Now this is awkward. In JS, when a conversion is needed, null becomes zero, and we get a finite number. Then why parseFloat(null) should return a NaN here? Someone please explain this design concept to me.
undefined	❌	❌	❌	As expected.
Infinity	✔️	✔️	❌	As explained before, isNaN() considers infinity as a number, and parseFloat() parses infinity as infinity.

So...which of them is "correct"?

Should be clear by now, it depends largely on what we need. For example, we may want to consider a null input as 0. In that case isFinite() will work fine.

Again, perhaps we will take a little help from isNaN() when 10^10000000000 is needed to be considered a valid number (although the question remains—why would it be, and how would we handle that)!

Of course, we can manually exclude any of the scenarios.

Like in my case, I needed exactly the outputs of isFinite(), except for the null case, the empty string case, and the whitespace-only string case. Also I had no headache about really huge numbers. So my code looked like this:

/**
 * My necessity was met by the following code.
 */

if (input === null) {
    // Null input
} else if (input.trim() === '') {
    // Empty or whitespace-only string
} else if (isFinite(input)) {
    // Input is a number
} else {
    // Not a number
}

And also, this was my JavaScript to generate the table:

/**
 * Note: JavaScript does not print numeric separator inside a number.
 * In that single case, the markdown output was manually corrected.
 * Also, the comments were manually added later, of course.
 */

let inputs = [
    123, '123', 12.3, '12.3', '   12.3   ',
    1_000_000, '1_000_000',
    '0b11111111', '0o377', '0xFF',
    '', '    ',
    'abc', '12.34Ab!@#$',
    '10e100', '10e1000',
    null, undefined, Infinity];

let markdownOutput = `| \`input\` | \`!isNaN(input)\` or <br>\`+input === +input\` | \`!isNaN(parseFloat(input))\` | \`isFinite(input)\` | Comment |
| :---: | :---: | :---: | :---: | :--- |\n`;

for (let input of inputs) {
    let outputs = [];
    outputs.push(!isNaN(input));
    outputs.push(!isNaN(parseFloat(input)));
    outputs.push(isFinite(input));

    if (typeof input === 'string') {
        // Output with quotations
        console.log(`'${input}'`);
        markdownOutput += `| '${input}'`;
    } else {
        // Output without quotes
        console.log(input);
        markdownOutput += `| ${input}`;
    }

    for (let output of outputs) {
        console.log('\t' + output);
        if (output === true) {
            markdownOutput += ` | <div style="color:limegreen">true</div>`;
            // markdownOutput += ` | ✔️`; // for stackoverflow
        } else {
            markdownOutput += ` | <div style="color:orangered">false</div>`;
            // markdownOutput += ` | ❌`; // for stackoverflow
        }
    }

    markdownOutput += ` ||\n`;
}

// Replace two or more whitespaces with $nbsp;
markdownOutput = markdownOutput.replaceAll(`  `, `&nbsp;&nbsp;`);

// Print markdown to console
console.log(markdownOutput);

Answer 9

The JavaScript global isFinite() checks if a value is a valid (finite) number.

See MDN for the difference between Number.isFinite() and global isFinite().

let a = isFinite('abc') // false
let b = isFinite('123') // true
let c = isFinite('12a') // false
let d = isFinite(null)  // true
console.log(a, b, c, d)

Answer 10

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.

Answer 11

Someone may also benefit from a regex based answer. Here it is:

One liner isInteger:

const isInteger = num => /^-?[0-9]+$/.test(num+'');

One liner isNumeric: Accepts integers and decimals

const isNumeric = num => /^-?[0-9]+(?:\.[0-9]+)?$/.test(num+'');

Answer 12

I have tested and Michael's solution is best. Vote for his answer above (search this page for "If you really want to make sure that a string" to find it). In essence, his answer is this:

function isNumeric(num){
  num = "" + num; //coerce num to be a string
  return !isNaN(num) && !isNaN(parseFloat(num));
}

It works for every test case, which I documented here: https://jsfiddle.net/wggehvp9/5/

Many of the other solutions fail for these edge cases: ' ', null, "", true, and []. In theory, you could use them, with proper error handling, for example:

return !isNaN(num);

or

return (+num === +num);

with special handling for /\s/, null, "", true, false, [] (and others?)

Answer 13

You can use the result of Number when passing an argument to its constructor.

If the argument (a string) cannot be converted into a number, it returns NaN, so you can determinate if the string provided was a valid number or not.

Notes: Note when passing empty string or '\t\t' and '\n\t' as Number will return 0; Passing true will return 1 and false returns 0.

    Number('34.00') // 34
    Number('-34') // -34
    Number('123e5') // 12300000
    Number('123e-5') // 0.00123
    Number('999999999999') // 999999999999
    Number('9999999999999999') // 10000000000000000 (integer accuracy up to 15 digit)
    Number('0xFF') // 255
    Number('Infinity') // Infinity  

    Number('34px') // NaN
    Number('xyz') // NaN
    Number('true') // NaN
    Number('false') // NaN

    // cavets
    Number('    ') // 0
    Number('\t\t') // 0
    Number('\n\t') // 0

Answer 14

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 15

Why is jQuery's implementation not good enough?

function isNumeric(a) {
    var b = a && a.toString();
    return !$.isArray(a) && b - parseFloat(b) + 1 >= 0;
};

Michael suggested something like this (although I've stolen "user1691651 - John"'s altered version here):

function isNumeric(num){
    num = "" + num; //coerce num to be a string
    return !isNaN(num) && !isNaN(parseFloat(num));
}

The following is a solution with most likely bad performance, but solid results. It is a contraption made from the jQuery 1.12.4 implementation and Michael's answer, with an extra check for leading/trailing spaces (because Michael's version returns true for numerics with leading/trailing spaces):

function isNumeric(a) {
    var str = a + "";
    var b = a && a.toString();
    return !$.isArray(a) && b - parseFloat(b) + 1 >= 0 &&
           !/^\s+|\s+$/g.test(str) &&
           !isNaN(str) && !isNaN(parseFloat(str));
};

The latter version has two new variables, though. One could get around one of those, by doing:

function isNumeric(a) {
    if ($.isArray(a)) return false;
    var b = a && a.toString();
    a = a + "";
    return b - parseFloat(b) + 1 >= 0 &&
            !/^\s+|\s+$/g.test(a) &&
            !isNaN(a) && !isNaN(parseFloat(a));
};

I haven't tested any of these very much, by other means than manually testing the few use-cases I'll be hitting with my current predicament, which is all very standard stuff. This is a "standing-on-the-shoulders-of-giants" situation.

Answer 16

I like the simplicity of this.

Number.isNaN(Number(value))

The above is regular Javascript, but I'm using this in conjunction with a typescript typeguard for smart type checking. This is very useful for the typescript compiler to give you correct intellisense, and no type errors.

Typescript typeguards

Warning: See Jeremy's comment below. This has some issues with certain values and I don't have time to fix it now, but the idea of using a typescript typeguard is useful so I won't delete this section.

isNotNumber(value: string | number): value is string {
    return Number.isNaN(Number(this.smartImageWidth));
}
isNumber(value: string | number): value is number {
    return Number.isNaN(Number(this.smartImageWidth)) === false;
}

Let's say you have a property width which is number | string. You may want to do logic based on whether or not it's a string.

var width: number|string;
width = "100vw";

if (isNotNumber(width)) 
{
    // the compiler knows that width here must be a string
    if (width.endsWith('vw')) 
    {
        // we have a 'width' such as 100vw
    } 
}
else 
{
    // the compiler is smart and knows width here must be number
    var doubleWidth = width * 2;    
}

The typeguard is smart enough to constrain the type of width within the if statement to be ONLY string. This permits the compiler to allow width.endsWith(...) which it wouldn't allow if the type was string | number.

You can call the typeguard whatever you want isNotNumber, isNumber, isString, isNotString but I think isString is kind of ambiguous and harder to read.

Answer 17

2019: Practical and tight numerical validity check

Often, a 'valid number' means a Javascript number excluding NaN and Infinity, ie a 'finite number'.

To check the numerical validity of a value (from an external source for example), you can define in ESlint Airbnb style :

/**
 * Returns true if 'candidate' is a finite number or a string referring (not just 'including') a finite number
 * To keep in mind:
 *   Number(true) = 1
 *   Number('') = 0
 *   Number("   10  ") = 10
 *   !isNaN(true) = true
 *   parseFloat('10 a') = 10
 *
 * @param {?} candidate
 * @return {boolean}
 */
function isReferringFiniteNumber(candidate) {
  if (typeof (candidate) === 'number') return Number.isFinite(candidate);
  if (typeof (candidate) === 'string') {
    return (candidate.trim() !== '') && Number.isFinite(Number(candidate));
  }
  return false;
}

and use it this way:

if (isReferringFiniteNumber(theirValue)) {
  myCheckedValue = Number(theirValue);
} else {
  console.warn('The provided value doesn\'t refer to a finite number');
}

Answer 18

parseInt(), but be aware that this function is a bit different in the sense that it for example returns 100 for parseInt("100px").

Answer 19

It is not valid for TypeScript as:

declare function isNaN(number: number): boolean;

For TypeScript you can use:

/^\d+$/.test(key)

Answer 20

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 21

When guarding against empty strings and null

// Base cases that are handled properly
Number.isNaN(Number('1')); // => false
Number.isNaN(Number('-1')); // => false
Number.isNaN(Number('1.1')); // => false
Number.isNaN(Number('-1.1')); // => false
Number.isNaN(Number('asdf')); // => true
Number.isNaN(Number(undefined)); // => true

// Special notation cases that are handled properly
Number.isNaN(Number('1e1')); // => false
Number.isNaN(Number('1e-1')); // => false
Number.isNaN(Number('-1e1')); // => false
Number.isNaN(Number('-1e-1')); // => false
Number.isNaN(Number('0b1')); // => false
Number.isNaN(Number('0o1')); // => false
Number.isNaN(Number('0xa')); // => false

// Edge cases that will FAIL if not guarded against
Number.isNaN(Number('')); // => false
Number.isNaN(Number(' ')); // => false
Number.isNaN(Number(null)); // => false

// Edge cases that are debatable
Number.isNaN(Number('-0b1')); // => true
Number.isNaN(Number('-0o1')); // => true
Number.isNaN(Number('-0xa')); // => true
Number.isNaN(Number('Infinity')); // => false 
Number.isNaN(Number('INFINITY')); // => true  
Number.isNaN(Number('-Infinity')); // => false 
Number.isNaN(Number('-INFINITY')); // => true  

When NOT guarding against empty strings and null

Using parseInt:

// Base cases that are handled properly
Number.isNaN(parseInt('1')); // => false
Number.isNaN(parseInt('-1')); // => false
Number.isNaN(parseInt('1.1')); // => false
Number.isNaN(parseInt('-1.1')); // => false
Number.isNaN(parseInt('asdf')); // => true
Number.isNaN(parseInt(undefined)); // => true
Number.isNaN(parseInt('')); // => true
Number.isNaN(parseInt(' ')); // => true
Number.isNaN(parseInt(null)); // => true

// Special notation cases that are handled properly
Number.isNaN(parseInt('1e1')); // => false
Number.isNaN(parseInt('1e-1')); // => false
Number.isNaN(parseInt('-1e1')); // => false
Number.isNaN(parseInt('-1e-1')); // => false
Number.isNaN(parseInt('0b1')); // => false
Number.isNaN(parseInt('0o1')); // => false
Number.isNaN(parseInt('0xa')); // => false

// Edge cases that are debatable
Number.isNaN(parseInt('-0b1')); // => false
Number.isNaN(parseInt('-0o1')); // => false
Number.isNaN(parseInt('-0xa')); // => false
Number.isNaN(parseInt('Infinity')); // => true 
Number.isNaN(parseInt('INFINITY')); // => true  
Number.isNaN(parseInt('-Infinity')); // => true 
Number.isNaN(parseInt('-INFINITY')); // => true 

Using parseFloat:

// Base cases that are handled properly
Number.isNaN(parseFloat('1')); // => false
Number.isNaN(parseFloat('-1')); // => false
Number.isNaN(parseFloat('1.1')); // => false
Number.isNaN(parseFloat('-1.1')); // => false
Number.isNaN(parseFloat('asdf')); // => true
Number.isNaN(parseFloat(undefined)); // => true
Number.isNaN(parseFloat('')); // => true
Number.isNaN(parseFloat(' ')); // => true
Number.isNaN(parseFloat(null)); // => true

// Special notation cases that are handled properly
Number.isNaN(parseFloat('1e1')); // => false
Number.isNaN(parseFloat('1e-1')); // => false
Number.isNaN(parseFloat('-1e1')); // => false
Number.isNaN(parseFloat('-1e-1')); // => false
Number.isNaN(parseFloat('0b1')); // => false
Number.isNaN(parseFloat('0o1')); // => false
Number.isNaN(parseFloat('0xa')); // => false

// Edge cases that are debatable
Number.isNaN(parseFloat('-0b1')); // => false
Number.isNaN(parseFloat('-0o1')); // => false
Number.isNaN(parseFloat('-0xa')); // => false
Number.isNaN(parseFloat('Infinity')); // => false 
Number.isNaN(parseFloat('INFINITY')); // => true  
Number.isNaN(parseFloat('-Infinity')); // => false 
Number.isNaN(parseFloat('-INFINITY')); // => true

Notes:

• Only string, empty, and uninitialized values are considered in keeping with addressing the original question. Additional edge cases exist if arrays and objects are the values being considered.
• Characters in binary, octal, hexadecimal, and exponential notation are not case-sensitive (ie: '0xFF', '0XFF', '0xfF' etc. will all yield the same result in the test cases shown above).
• Unlike with Infinity (case-sensitive) in some cases, constants from the Number and Math objects passed as test cases in string format to any of the methods above will be determined to not be numbers.
• See here for an explanation of how arguments are converted to a Number and why the edge cases for null and empty strings exist.

Answer 22

This is built on some of the previous answers and comments. It covers all the edge cases and also handles scientific notation optionally:

const NUMBER_REG_EXP = /^-?\d+(?:\.\d+)?$/;
const SCIENTIFIC_NOTATION_REG_EXP = /^-?\d+(?:\.\d+)?(?:[eE]\d+)?$/;

const isNumeric = (n, allowScientificNotation = false) => (
    (typeof n === 'number' && !Number.isNaN(n)) || 
    (typeof n === 'string' && (allowScientificNotation ?
        SCIENTIFIC_NOTATION_REG_EXP : NUMBER_REG_EXP).test(n))
);

Answer 23

This way it works for me.

function isNumeric(num){
    let value1 = num.toString();
    let value2 = parseFloat(num).toString();
    return (value1 === value2);
}

---

console.log(
    isNumeric(123),     //true
    isNumeric(-123),    //true
    isNumeric('123'),   //true
    isNumeric('-123'),  //true
    isNumeric(12.2),    //true
    isNumeric(-12.2),   //true
    isNumeric('12.2'),  //true
    isNumeric('-12.2'), //true
    isNumeric('a123'),  //false
    isNumeric('123a'),  //false
    isNumeric(' 123'),  //false
    isNumeric('123 '),  //false
    isNumeric('a12.2'), //false
    isNumeric('12.2a'), //false
    isNumeric(' 12.2'), //false
    isNumeric('12.2 '), //false
)

Answer 24

If anyone ever gets this far down, I spent some time hacking on this trying to patch moment.js (https://github.com/moment/moment). Here's something that I took away from it:

function isNumeric(val) {
    var _val = +val;
    return (val !== val + 1) //infinity check
        && (_val === +val) //Cute coercion check
        && (typeof val !== 'object') //Array/object check
}

Handles the following cases:

True! :

isNumeric("1"))
isNumeric(1e10))
isNumeric(1E10))
isNumeric(+"6e4"))
isNumeric("1.2222"))
isNumeric("-1.2222"))
isNumeric("-1.222200000000000000"))
isNumeric("1.222200000000000000"))
isNumeric(1))
isNumeric(0))
isNumeric(-0))
isNumeric(1010010293029))
isNumeric(1.100393830000))
isNumeric(Math.LN2))
isNumeric(Math.PI))
isNumeric(5e10))

False! :

isNumeric(NaN))
isNumeric(Infinity))
isNumeric(-Infinity))
isNumeric())
isNumeric(undefined))
isNumeric('[1,2,3]'))
isNumeric({a:1,b:2}))
isNumeric(null))
isNumeric([1]))
isNumeric(new Date()))

Ironically, the one I am struggling with the most:

isNumeric(new Number(1)) => false

Any suggestions welcome. :]

Answer 25

I recently wrote an article about ways to ensure a variable is a valid number: https://github.com/jehugaleahsa/artifacts/blob/master/2018/typescript_num_hack.md The article explains how to ensure floating point or integer, if that's important (+x vs ~~x).

The article assumes the variable is a string or a number to begin with and trim is available/polyfilled. It wouldn't be hard to extend it to handle other types, as well. Here's the meat of it:

// Check for a valid float
if (x == null
    || ("" + x).trim() === ""
    || isNaN(+x)) {
    return false;  // not a float
}

// Check for a valid integer
if (x == null
    || ("" + x).trim() === ""
    || ~~x !== +x) {
    return false;  // not an integer
}

Answer 26

function isNumberCandidate(s) {
  const str = (''+ s).trim();
  if (str.length === 0) return false;
  return !isNaN(+str);
}

console.log(isNumberCandidate('1'));       // true
console.log(isNumberCandidate('a'));       // false
console.log(isNumberCandidate('000'));     // true
console.log(isNumberCandidate('1a'));      // false 
console.log(isNumberCandidate('1e'));      // false
console.log(isNumberCandidate('1e-1'));    // true
console.log(isNumberCandidate('123.3'));   // true
console.log(isNumberCandidate(''));        // false
console.log(isNumberCandidate(' '));       // false
console.log(isNumberCandidate(1));         // true
console.log(isNumberCandidate(0));         // true
console.log(isNumberCandidate(NaN));       // false
console.log(isNumberCandidate(undefined)); // false
console.log(isNumberCandidate(null));      // false
console.log(isNumberCandidate(-1));        // true
console.log(isNumberCandidate('-1'));      // true
console.log(isNumberCandidate('-1.2'));    // true
console.log(isNumberCandidate(0.0000001)); // true
console.log(isNumberCandidate('0.0000001')); // true
console.log(isNumberCandidate(Infinity));    // true
console.log(isNumberCandidate(-Infinity));    // true

console.log(isNumberCandidate('Infinity'));  // true

if (isNumberCandidate(s)) {
  // use +s as a number
  +s ...
}

Answer 27

PFB the working solution:

 function(check){ 
    check = check + "";
    var isNumber =   check.trim().length>0? !isNaN(check):false;
    return isNumber;
    }

Answer 28

Save yourself the headache of trying to find a "built-in" solution.

There isn't a good answer, and the hugely upvoted answer in this thread is wrong.

npm install is-number

In JavaScript, it's not always as straightforward as it should be to reliably check if a value is a number. It's common for devs to use +, -, or Number() to cast a string value to a number (for example, when values are returned from user input, regex matches, parsers, etc). But there are many non-intuitive edge cases that yield unexpected results:

console.log(+[]); //=> 0
console.log(+''); //=> 0
console.log(+'   '); //=> 0
console.log(typeof NaN); //=> 'number'

Answer 29

This appears to catch the seemingly infinite number of edge cases:

function isNumber(x, noStr) {
    /*

        - Returns true if x is either a finite number type or a string containing only a number
        - If empty string supplied, fall back to explicit false
        - Pass true for noStr to return false when typeof x is "string", off by default

        isNumber(); // false
        isNumber([]); // false
        isNumber([1]); // false
        isNumber([1,2]); // false
        isNumber(''); // false
        isNumber(null); // false
        isNumber({}); // false
        isNumber(true); // false
        isNumber('true'); // false
        isNumber('false'); // false
        isNumber('123asdf'); // false
        isNumber('123.asdf'); // false
        isNumber(undefined); // false
        isNumber(Number.POSITIVE_INFINITY); // false
        isNumber(Number.NEGATIVE_INFINITY); // false
        isNumber('Infinity'); // false
        isNumber('-Infinity'); // false
        isNumber(Number.NaN); // false
        isNumber(new Date('December 17, 1995 03:24:00')); // false
        isNumber(0); // true
        isNumber('0'); // true
        isNumber(123); // true
        isNumber(123.456); // true
        isNumber(-123.456); // true
        isNumber(-.123456); // true
        isNumber('123'); // true
        isNumber('123.456'); // true
        isNumber('.123'); // true
        isNumber(.123); // true
        isNumber(Number.MAX_SAFE_INTEGER); // true
        isNumber(Number.MAX_VALUE); // true
        isNumber(Number.MIN_VALUE); // true
        isNumber(new Number(123)); // true
    */

    return (
        (typeof x === 'number' || x instanceof Number || (!noStr && x && typeof x === 'string' && !isNaN(x))) &&
        isFinite(x)
    ) || false;
};

Answer 30

So, it will depend on the test cases that you want it to handle.

function isNumeric(number) {
  return !isNaN(parseFloat(number)) && !isNaN(+number);
}

What I was looking for was regular types of numbers in javascript. 0, 1 , -1, 1.1 , -1.1 , 1E1 , -1E1 , 1e1 , -1e1, 0.1e10, -0.1.e10 , 0xAF1 , 0o172, Math.PI, Number.NEGATIVE_INFINITY, Number.POSITIVE_INFINITY

And also they're representations as strings:
'0', '1', '-1', '1.1', '-1.1', '1E1', '-1E1', '1e1', '-1e1', '0.1e10', '-0.1.e10', '0xAF1', '0o172'

I did want to leave out and not mark them as numeric '', ' ', [], {}, null, undefined, NaN

As of today, all other answers seemed to failed one of these test cases.

Answer 31

Checking the number in JS:

1. Best way for check if it's a number:

   isFinite(20)
   //True
   

2. Read a value out of a string. CSS *:

   parseInt('2.5rem')
   //2
   parseFloat('2.5rem')
   //2.5  
   

3. For an integer:

   isInteger(23 / 0)
   //False
   

4. If value is NaN:

   isNaN(20)
   //False
   

Answer 32

You can also use a simple parseInt function too... with an if the condition for example

if (parseInt(i)){
    (i in dic) ? dic[i] += 1 : dic[i] = 1
}

Answer 33

Well, I'm using this one I made...

It's been working so far:

function checkNumber(value) {
    return value % 1 == 0;
}

If you spot any problem with it, tell me, please.

Answer 34

Here's an elegant one-liner to check if sNum is a valid numeric value. The code has been tested for a wide variety of inputs as well.

// returns True if sNum is a numeric value    
!!sNum && !isNaN(+sNum.replace(/\s|\$/g, ''));  

Hat tip to @gman for catching the error.

Answer 35

I used this function as a form validation tool, and I didn't want users to be able to write exponential function, so I came up with this function:

<script>

    function isNumber(value, acceptScientificNotation) {

        if(true !== acceptScientificNotation){
            return /^-{0,1}\d+(\.\d+)?$/.test(value);
        }

        if (true === Array.isArray(value)) {
            return false;
        }
        return !isNaN(parseInt(value, 10));
    }


    console.log(isNumber(""));              // false
    console.log(isNumber(false));           // false
    console.log(isNumber(true));            // false
    console.log(isNumber("0"));             // true
    console.log(isNumber("0.1"));           // true
    console.log(isNumber("12"));            // true
    console.log(isNumber("-12"));           // true
    console.log(isNumber(-45));             // true
    console.log(isNumber({jo: "pi"}));      // false
    console.log(isNumber([]));              // false
    console.log(isNumber([78, 79]));        // false
    console.log(isNumber(NaN));             // false
    console.log(isNumber(Infinity));        // false
    console.log(isNumber(undefined));       // false
    console.log(isNumber("0,1"));           // false



    console.log(isNumber("1e-1"));          // false
    console.log(isNumber("1e-1", true));    // true
</script>

Answer 36

This is a possible way to check whether a variable is NOT numerical:

(isNaN(foo) || ((foo !== 0) && (!foo)))

Which means that foo is either falsy but different from 0 or isNaN(foo) is true.

Another way to perform such a check is

!isNaN(parseFloat(foo))

Answer 37

I often employ a method of iterating and checking each character of a string for my checks:

const isInt = aString => aString.length !== 0 && [...aString].every(char => ! Number.isNaN(char));

const isDecimal = aString =>
{
    if (aString.length === 0)
    {
        return false;
    }

    const chars = [...aString];

    return chars.filter(aChar => aChar === ".").length === 1 && chars.filter(aChar => aChar !== ".").every(char => ! Number.isNaN(char));
};

These methods could be further optimized as much as you like. For instance they both require creating a new string array and the decimal check function iterates through all the characters of the string being checked twice. But I find that these get the job done and haven't noticed any bottlenecks from them before.

Answer 38

This function gets a string and returns a boolean as a result :

const isNumber = (str) => typeof str === 'string' && str.length > 0 && !isNaN(Number(str))

Answer 39

If you only want to support positive integers and forbid any other characters, you can use this function:

const isNumeric = value =>
  value.length !== 0 && [...value].every(c => c >= '0' && c <= '9');
isNumeric('')          // false
isNumeric('1234')      // true
isNumeric('1234asdf')  // false

Answer 40

In my application we are only allowing a-z A-Z and 0-9 characters. I found the answer above using " string % 1 === 0" worked unless the string began with 0xnn (like 0x10) and then it would return it as numeric when we didn't want it to. The following simple trap in my numeric check seems to do the trick in our specific cases.

function isStringNumeric(str_input){   
    //concat a temporary 1 during the modulus to keep a beginning hex switch combination from messing us up   
    //very simple and as long as special characters (non a-z A-Z 0-9) are trapped it is fine   
    return '1'.concat(str_input) % 1 === 0;}

Warning : This might be exploiting a longstanding bug in Javascript and Actionscript [Number("1" + the_string) % 1 === 0)], I can't speak for that, but it is exactly what we needed.

Answer 41

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 42

My attempt at a slightly confusing, Pherhaps not the best solution

function isInt(a){
    return a === ""+~~a
}


console.log(isInt('abcd'));         // false
console.log(isInt('123a'));         // false
console.log(isInt('1'));            // true
console.log(isInt('0'));            // true
console.log(isInt('-0'));           // false
console.log(isInt('01'));           // false
console.log(isInt('10'));           // true
console.log(isInt('-1234567890'));  // true
console.log(isInt(1234));           // false
console.log(isInt('123.4'));        // false
console.log(isInt(''));             // false

// other types then string returns false
console.log(isInt(5));              // false
console.log(isInt(undefined));      // false
console.log(isInt(null));           // false
console.log(isInt('0x1'));          // false
console.log(isInt(Infinity));       // false

Answer 43

You could make use of types, like with the flow library, to get static, compile time checking. Of course not terribly useful for user input.

// @flow

function acceptsNumber(value: number) {
  // ...
}

acceptsNumber(42);       // Works!
acceptsNumber(3.14);     // Works!
acceptsNumber(NaN);      // Works!
acceptsNumber(Infinity); // Works!
acceptsNumber("foo");    // Error!

Answer 44

Here is a high-performance (2.5*10^7 iterations/s @3.8GHz Haswell) version of a isNumber implementation. It works for every testcase i could find (including Symbols):

var isNumber = (function () {
  var isIntegerTest = /^\d+$/;
  var isDigitArray = [!0, !0, !0, !0, !0, !0, !0, !0, !0, !0];
  function hasLeading0s (s) {
    return !(typeof s !== 'string' ||
    s.length < 2 ||
    s[0] !== '0' ||
    !isDigitArray[s[1]] ||
    isIntegerTest.test(s));
  }
  var isWhiteSpaceTest = /\s/;
  return function isNumber (s) {
    var t = typeof s;
    var n;
    if (t === 'number') {
      return (s <= 0) || (s > 0);
    } else if (t === 'string') {
      n = +s;
      return !((!(n <= 0) && !(n > 0)) || n === '0' || hasLeading0s(s) || !(n !== 0 || !(s === '' || isWhiteSpaceTest.test(s))));
    } else if (t === 'object') {
      return !(!(s instanceof Number) || ((n = +s), !(n <= 0) && !(n > 0)));
    }
    return false;
  };
})();

Answer 45

Test if a string or number is a number

const isNumeric = stringOrNumber =>
  stringOrNumber == 0 || !!+stringOrNumber;

Or if you want to convert a string or number to a number

const toNumber = stringOrNumber =>
  stringOrNumber == 0 || +stringOrNumber ? +stringOrNumber : NaN;

Answer 46

I used this function in Angular

 isNumeric(value: string): boolean {
             let valueToNumber = Number(value);
             var result = typeof valueToNumber == 'number' ;
             if(valueToNumber.toString() == 'NaN')
             {
                result = false;
             }
             return result;
          }

Answer 47

I didn't like the provided solutions because I only need numbers.

nums = new Set(['1','2','3','4','5','6','7','8','9','0']);
hasOnlyNumbers= str =>[...str].every( x=>nums.has(x) );

https://jsfiddle.net/gaby_de_wilde/05fb7k3g/

If you want to allow minus (ignore minus at start of string)

nums = new Set(['1','2','3','4','5','6','7','8','9','0']);
hasOnlyNumbers=str=>[...str[0]=='-'?str.substring(1):str].every(x=>nums.has(x));

https://jsfiddle.net/gaby_de_wilde/y4fdwvxk/

If you want it more readable:

nums = new Set(['1','2','3','4','5','6','7','8','9','0']);
hasOnlyNumbers = function(str){
  if(str[0]=='-'){ 
    str = str.substring(1) 
  }
  return [...str].every(x=>nums.has(x));
}

https://jsfiddle.net/gaby_de_wilde/6a5qwx2t/

allow a single dot (allow at most 1)

nums = new Set(['.','1','2','3','4','5','6','7','8','9','0']);
hasOnlyNumbers=str=>str.split('.').length<3&&[...str[0]=='-'?str.substring(1):str].every(x=>nums.has(x));

https://jsfiddle.net/gaby_de_wilde/hnx8d6so/

more readable:

nums = new Set(['.','1','2','3','4','5','6','7','8','9','0']);
hasOnlyNumbers = function(str){
  if(str.split('.').length > 2) { 
    return false 
  }
  if(str[0]=='-'){ 
    str = str.substring(1) 
  }
  return [...str].every(x=>nums.has(x));
}

https://jsfiddle.net/gaby_de_wilde/sz7e95mw/

Note: the Set now has a dot too.

Answer 48

Just use isNaN(), this will convert the string to a number and if get a valid number, will return false...

isNaN("Alireza"); //return true
isNaN("123"); //return false

Answer 49

I'm using the following:

const isNumber = s => !isNaN(+s)

Answer 50

If you are looking for a positive number (housenumber for example), just use:

if (mystring > 0) ...

Answer 51

If you like a tricky way and you like getting your coworkers confused, you can use:

const isNumeric = str => parseFloat(str) === parseFloat(str)

Proof:

 const isNumeric = str => parseFloat(str) === parseFloat(str)
 
console.log(isNumeric('10'))
console.log(isNumeric('-10.2'))
console.log(isNumeric('15abc'))
console.log(isNumeric('0.0001'))
console.log(isNumeric('abc'))
console.log(isNumeric('abc123'))

Answer 52

How about the following?

const a = '1'

const isNumber = (a) => Number(a) === +a

Answer 53

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.