using an if statement to check if NaN

Question

I'm getting a numeric value from a form. Then I check to see if it's NaN. If it is a number I want to set that value to a variable. The problem is that when I enter a valid number I still get an alert and the number isn't passed to the variable "date". How should I modify my statement so that when it is a valid number I can assign it to the variable date?

var adate = document.getElementById("dueDate").value;    

    if ( adate == NaN || " ") {
    alert("Please enter a due date");
    return;
    }

    else {
    var date = (new Date()).setDate(adate);
    }

    processDate(date);

`document.getElementById("dueDate").value` will allways be a string (maybe `undefined` in some corner cases), so you should first try to convert it to your needed data type before checking for NaN. — Yoshi, Feb 26 '13 at 18:54
I would make it a number first with `parseInt(adate)` and then check for `NaN`. — Bart, Mar 14 '13 at 12:31

score 19 · Answer 1 · edited May 23 '17 at 12:25

19

Use Javascript's isNaN() function.

Checking equality with NaN is always false, as per IEEE's standards. Stephen Canon, a member of the IEEE-754 committee that decided this, has an excellent answer explaining this here.

edited May 23 '17 at 12:25

Community

1
1

answered Feb 26 '13 at 18:51

Bryan Herbst

66,602
10
133
120

Isn't isNAN() ECMA6 only? – geoidesic Oct 17 '15 at 16:10
`Number.isNaN` is ES6, `isNaN` is a builtin function present since 1st standard definition. – Andre Figueiredo Feb 26 '18 at 19:05

benekastah · Answer 2 · 2013-02-26T19:20:40.747

10

As strange as it seems, NaN !== NaN.

if (adate !== adate || adate !== " ") {
  //...
}

The isNaN function would work in a lot of cases. There is a good case to be made that it is broken, though.

One nice way of getting around this is:

MyNamespace.isNaN = function (x) {
  return x !== x;
}

edited Feb 26 '13 at 19:20

answered Feb 26 '13 at 18:51

benekastah

5,651
1
35
50

score 5 · Answer 3 · edited Sep 22 '14 at 07:55

5

you could Use if( isNaN(adate))

good luck

edited Sep 22 '14 at 07:55

benka

4,732
35
47
58

answered Sep 22 '14 at 07:15

Sandun

51
1
1

score 4 · Answer 4 · answered Feb 26 '13 at 18:57

You have two problems here. The result is that the conditional will always pass. This is what it does:

adate == NaN // first, test if adate == NaN (this always returns false)
||           // if the first test fails (i.e. always), carry on checking
" "          // test if the string " " is truthy (this always returns true)

The || does two separate checks. It does not test to see if adate is "either NaN or " "", which seems to be what you expect.

Your code might as well say

if ( true ) {

You would be able to sort this out, however, if you tried two comparisons:

if ( (adate == NaN) || (adate === " ")) {

As other people have said, however, this doesn't work, because NaN !== NaN. So the solution is to use isNaN:

if (isNaN(adate) || (adate === " ")) {

thanks for clearing me up on isNaN vs NaN and the correct meaning of || . I ended up using if (isNaN(adate) ||(0 == adate.length) || adate == 0) . Which I'm not sure is the best way to go about it yet. I may end up doing what @Yoshi was talking about and try to parse the values first before running this check — user2084813, Feb 26 '13 at 19:14

score 1 · Answer 5 · edited Jun 28 '22 at 23:09

By using isNaN method we can verify if the given input is number or not.

let num1 = parseInt(prompt('Enter your number-1'));
let num2 = parseInt(prompt('Enter your number-2'));
alert(num1 + " is of type " + typeof num1 + " & " + num2 + " is of type " + typeof num2);
if (isNaN(num1) || isNaN(num2)) {
  alert("Can not add incompatible types");
} else {
  let sum = num1 + num2;
  alert("Sum is " + sum);
}

using an if statement to check if NaN

5 Answers5