CODEWITHSUNDEEP
javascriptif-statement
user2084813
user2084813

Reputation: 185

using an if statement to check if NaN

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);

Upvotes: 9

Views: 41192

Answers (5)

Shubham Kamble
Shubham Kamble

Reputation: 11

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);
}

Upvotes: 1

Sandun
Sandun

Reputation: 51

you could Use if( isNaN(adate))

good luck

Upvotes: 5

benekastah
benekastah

Reputation: 5711

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;
}

Upvotes: 10

Bryan Herbst
Bryan Herbst

Reputation: 67189

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.

Upvotes: 19

lonesomeday
lonesomeday

Reputation: 237847

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 === " ")) {

Upvotes: 4

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