How to check if a variable is an integer in JavaScript?


How do I check if a variable is an integer in JavaScript, and throw an alert if it isn't? I tried this, but it doesn't work:

<html>
    <head>
        <script type="text/javascript">
            var data = 22;
            alert(NaN(data));
        </script>
    </head>
</html>

Use the === operator (strict equality) as below,

if (data === parseInt(data, 10))
    alert("data is integer")
else
    alert("data is not an integer")

That depends, do you also want to cast strings as potential integers as well?

This will do:

function isInt(value) {
  return !isNaN(value) && 
         parseInt(Number(value)) == value && 
         !isNaN(parseInt(value, 10));
}

With Bitwise operations

Simple parse and check

function isInt(value) {
  var x = parseFloat(value);
  return !isNaN(value) && (x | 0) === x;
}

Short-circuiting, and saving a parse operation:

function isInt(value) {
  if (isNaN(value)) {
    return false;
  }
  var x = parseFloat(value);
  return (x | 0) === x;
}

Or perhaps both in one shot:

function isInt(value) {
  return !isNaN(value) && (function(x) { return (x | 0) === x; })(parseFloat(value))
}

Tests:

isInt(42)        // true
isInt("42")      // true
isInt(4e2)       // true
isInt("4e2")     // true
isInt(" 1 ")     // true
isInt("")        // false
isInt("  ")      // false
isInt(42.1)      // false
isInt("1a")      // false
isInt("4e2a")    // false
isInt(null)      // false
isInt(undefined) // false
isInt(NaN)       // false

Here's the fiddle: http://jsfiddle.net/opfyrqwp/28/

Performance

Testing reveals that the short-circuiting solution has the best performance (ops/sec).

// Short-circuiting, and saving a parse operation
function isInt(value) {
  var x;
  if (isNaN(value)) {
    return false;
  }
  x = parseFloat(value);
  return (x | 0) === x;
}

Here is a benchmark: http://jsben.ch/#/htLVw

If you fancy a shorter, obtuse form of short circuiting:

function isInt(value) {
  var x;
  return isNaN(value) ? !1 : (x = parseFloat(value), (0 | x) === x);
}

Of course, I'd suggest letting the minifier take care of that.


Assuming you don't know anything about the variable in question, you should take this approach:

if(typeof data === 'number') {
    var remainder = (data % 1);
    if(remainder === 0) {
        // yes, it is an integer
    }
    else if(isNaN(remainder)) {
        // no, data is either: NaN, Infinity, or -Infinity
    }
    else {
        // no, it is a float (still a number though)
    }
}
else {
    // no way, it is not even a number
}

To put it simply:

if(typeof data==='number' && (data%1)===0) {
    // data is an integer
}

Number.isInteger() seems to be the way to go.

MDN has also provided the following polyfill for browsers not supporting Number.isInteger(), mainly all versions of IE.

Link to MDN page

Number.isInteger = Number.isInteger || function(value) {
    return typeof value === "number" && 
           isFinite(value) && 
           Math.floor(value) === value;
};

You could check if the number has a remainder:

var data = 22;

if(data % 1 === 0){
   // yes it's an integer.
}

Mind you, if your input could also be text and you want to check first it is not, then you can check the type first:

var data = 22;

if(typeof data === 'number'){
     // yes it is numeric

    if(data % 1 === 0){
       // yes it's an integer.
    }
}

You can use a simple regular expression:

function isInt(value) {
    var er = /^-?[0-9]+$/;
    return er.test(value);
}

First off, NaN is a "number" (yes I know it's weird, just roll with it), and not a "function".

You need to check both if the type of the variable is a number, and to check for integer I would use modulus.

alert(typeof data === 'number' && data%1 == 0);

Be careful while using

num % 1

empty string ('') or boolean (true or false) will return as integer. You might not want to do that

false % 1 // true
'' % 1 //true

Number.isInteger(data)

Number.isInteger(22); //true
Number.isInteger(22.2); //false
Number.isInteger('22'); //false

build in function in the browser. Dosnt support older browsers

Alternatives:

Math.round(num)=== num

However, Math.round() also will fail for empty string and boolean


To check if integer like poster wants:

if (+data===parseInt(data)) {return true} else {return false}

notice + in front of data (converts string to number), and === for exact.

Here are examples:

data=10
+data===parseInt(data)
true

data="10"
+data===parseInt(data)
true

data="10.2"
+data===parseInt(data)
false

if(Number.isInteger(Number(data))){
    //-----
}

The simplest and cleanest pre-ECMAScript-6 solution (which is also sufficiently robust to return false even if a non-numeric value such as a string or null is passed to the function) would be the following:

function isInteger(x) { return (x^0) === x; } 

The following solution would also work, although not as elegant as the one above:

function isInteger(x) { return Math.round(x) === x; }

Note that Math.ceil() or Math.floor() could be used equally well (instead of Math.round()) in the above implementation.

Or alternatively:

function isInteger(x) { return (typeof x === 'number') && (x % 1 === 0); }

One fairly common incorrect solution is the following:

function isInteger(x) { return parseInt(x, 10) === x; }

While this parseInt-based approach will work well for many values of x, once x becomes quite large, it will fail to work properly. The problem is that parseInt() coerces its first parameter to a string before parsing digits. Therefore, once the number becomes sufficiently large, its string representation will be presented in exponential form (e.g., 1e+21). Accordingly, parseInt() will then try to parse 1e+21, but will stop parsing when it reaches the e character and will therefore return a value of 1. Observe:

> String(1000000000000000000000)
'1e+21'

> parseInt(1000000000000000000000, 10)
1

> parseInt(1000000000000000000000, 10) === 1000000000000000000000
false

Why hasnt anyone mentioned Number.isInteger() ?

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/isInteger

Works perfectly for me and solves the issue with the NaN beginning a number.


In ES6 2 new methods are added for Number Object.

In it Number.isInteger() method returns true if the argument is an integer.

Example usage :

Number.isInteger(10);        // returns true
Number.isInteger(10.5);      // returns false

ECMA-262 6.0 (ES6) standard include Number.isInteger function.

In order to add support for old browser I highly recommend using strong and community supported solution from:

https://github.com/paulmillr/es6-shim

which is pure ES6 JS polyfills library.

Note that this lib require es5-shim, just follow README.md.


You could tryNumber.isInteger(Number(value)) if value might be an integer in string form e.g var value = "23" and you want this to evaluate to true. Avoid trying Number.isInteger(parseInt(value)) because this won't always return the correct value. e.g if var value = "23abc" and you use the parseInt implementation, it would still return true.

But if you want strictly integer values then probably Number.isInteger(value) should do the trick.


var x = 1.5;
if(!isNaN(x)){
 console.log('Number');
 if(x % 1 == 0){
   console.log('Integer');
 }
}else {
 console.log('not a number');
}

Check if the variable is equal to that same variable rounded to an integer, like this:

if(Math.round(data) != data) {
    alert("Variable is not an integer!");
}

Besides, Number.isInteger(). Maybe Number.isSafeInteger() is another option here by using the ES6-specified.

To polyfill Number.isSafeInteger(..) in pre-ES6 browsers:

Number.isSafeInteger = Number.isSafeInteger || function(num) {
    return typeof num === "number" && 
           isFinite(num) && 
           Math.floor(num) === num &&
           Math.abs( num ) <= Number.MAX_SAFE_INTEGER;
};

Number.isInteger() is the best way if your browser support it, if not, I think there are so many ways to go:

function isInt1(value){
  return (value^0) === value
}

or:

function isInt2(value){
  return (typeof value === 'number') && (value % 1 === 0); 
}

or:

function isInt3(value){
  return parseInt(value, 10) === value; 
}

or:

function isInt4(value){
  return Math.round(value) === value; 
}

now we can test the results:

var value = 1
isInt1(value)   // return true
isInt2(value)   // return true
isInt3(value)   // return true
isInt4(value)   // return true

var value = 1.1
isInt1(value)   // return false
isInt2(value)   // return false
isInt3(value)   // return false
isInt4(value)   // return false

var value = 1000000000000000000
isInt1(value)   // return false
isInt2(value)   // return true
isInt3(value)   // return false
isInt4(value)   // return true

var value = undefined
isInt1(value)   // return false
isInt2(value)   // return false
isInt3(value)   // return false
isInt4(value)   // return false

var value = '1' //number as string
isInt1(value)   // return false
isInt2(value)   // return false
isInt3(value)   // return false
isInt4(value)   // return false

So, all of these methods are works, but when the number is very big, parseInt and ^ operator would not works well.


You could use this function:

function isInteger(value) {
    return (value == parseInt(value));
}

It will return true even if the value is a string containing an integer value.
So, the results will be:

alert(isInteger(1)); // true
alert(isInteger(1.2)); // false
alert(isInteger("1")); // true
alert(isInteger("1.2")); // false
alert(isInteger("abc")); // false

Just try this:

let number = 5;
if (Number.isInteger(number)) {
    //do something
}

My approach:

a >= 1e+21 Test can only pass for a value which has to be a number and a very large one. This will cover all cases for sure, unlike other solutions which has been provided in this discussion.

a === (a|0) If the given function's argument is exactly (===) the same as the bitwise-transformed value, it means that the argument is an integer.

a|0 will return 0 for any value of a that isn't a number, and if a is indeed a number, it will strip away anything after the decimal point, so 1.0001 will become 1

function isInteger(a){
    return a >= 1e+21 ? true : a === (a|0)
}

/// tests ///////////////////////////
[
  1,                        // true
  1000000000000000000000,   // true
  4e2,                      // true
  Infinity,                 // true
  1.0,                      // true
  1.0000000000001,          // false
  0.1,                      // false
  "0",                      // false
  "1",                      // false
  "1.1",                    // false
  NaN,                      // false
  [],                       // false
  {},                       // false
  true,                     // false
  false,                    // false
  null,                     // false
  undefined                 // false
].forEach( a => console.log(typeof a, a, isInteger(a)) )


You can use regexp for this:

function isInteger(n) {
    return (typeof n == 'number' && /^-?\d+$/.test(n+''));
}

From http://www.toptal.com/javascript/interview-questions:

function isInteger(x) { return (x^0) === x; } 

Found it to be the best way to do this.


Use the | operator:

(5.3 | 0) === 5.3 // => false
(5.0 | 0) === 5.0 // => true

So, a test function might look like this:

var isInteger = function (value) {
  if (typeof value !== 'number') {
    return false;
  }

  if ((value | 0) !== value) {
    return false;
  }

  return true;
};

This will solve one more scenario (121.), a dot at end

function isInt(value) {
        var ind = value.indexOf(".");
        if (ind > -1) { return false; }

        if (isNaN(value)) {
            return false;
        }

        var x = parseFloat(value);
        return (x | 0) === x;

    }

For positive integer values without separators:

return ( data !== '' && data === data.replace(/\D/, '') );

Tests 1. if not empty and 2. if value is equal to the result of a replace of a non-digit char in its value.


Ok got minus, cause didn't describe my example, so more examples:):

I use regular expression and test method:

var isInteger = /^[0-9]\d*$/;

isInteger.test(123); //true
isInteger.test('123'); // true
isInteger.test('sdf'); //false
isInteger.test('123sdf'); //false

// If u want to avoid string value:
typeof testVal !== 'string' && isInteger.test(testValue);

you can also try it this way

var data = 22;
if (Number.isInteger(data)) {
    console.log("integer");
 }else{
     console.log("not an integer");
 }

or

if (data === parseInt(data, 10)){
    console.log("integer");
}else{
    console.log("not an integer");
}

I had to check if a variable (string or number) is an integer and I used this condition:

function isInt(a){
    return !isNaN(a) && parseInt(a) == parseFloat(a);
}

http://jsfiddle.net/e267369d/1/

Some of the other answers have a similar solution (rely on parseFloat combined with isNaN), but mine should be more straight forward and self explaining.


Edit: I found out that my method fails for strings containing comma (like "1,2") and I also realized that in my particular case I want the function to fail if a string is not a valid integer (should fail on any float, even 1.0). So here is my function Mk II:

function isInt(a){
    return !isNaN(a) && parseInt(a) == parseFloat(a) && (typeof a != 'string' || (a.indexOf('.') == -1 && a.indexOf(',') == -1));
}

http://jsfiddle.net/e267369d/3/

Of course in case you actually need the function to accept integer floats (1.0 stuff), you can always remove the dot condition a.indexOf('.') == -1.