Question

52

javascript isNaN() function not working?

rated 0 times [ 56] [ 4] / answers: 1 / hits: 15505 / 8 Years ago, wed, january 27, 2016, 12:00:00

I have a function to test if a prompt input is a number, like so:

function myFunction() 

{

    var person = prompt(Please enter your name, );



    if (person != null) 

    {

        if(isNaN(person))

        {

            document.write(hello  + person + <br><br>);

        }

        else

            document.write(You gave me a number);



    }

    else

    {

        document.write(You didn't answer.<br><br>);

    }

}

but every time I enter a number it keeps outputting hello + the number. I've been googling this function for quite some time and it doesn't make sense to me, it seems like it should work. Why is person returning true?

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2 Years ago

answered 8 Years ago benitoh · Accepted Answer

Your code is the correct way of using the isNaN method. However for anyone else reading this post I have seen a strange anomaly where the documented usage of iNaN hasn't worked properly and I got around the problem by combining the parseInt method with the isNaN method. According to w3schools the isNan('123') should return false and isNan('g12') should return true, but I've seen scenarios where this isn't the case.
If you're having trouble getting the documented methods to work try this code below:

var unitsToAdd = parseInt($('#unitsToAdd').val());

if(isNaN(unitsToAdd)) {

    alert('not a number');

    $('#unitsToAdd').val('1');

    returnVal = false;

}

Alternatively you can try this method which is well tested.

function isNumber(searchValue) {

    var found = searchValue.search(/^(d*.?d*)$/);

    //Change to ^(d*.?d+)$ if you don't want the number to end with a . such as 2.

    //Currently validates .2, 0.2, 2.0 and 2.

    if(found > -1) {

        return true;

    }

    else {

        return false;

    }

}