Typescript : check a string for number

Question

I'm new to web development, and in my function want to check if a given string value is a number. In case the string isn't a valid number I want to return null.

The following works for all cases except when the string is "0" in which case it returns null.

parseInt(columnSortSettings[0]) || null;

How do I prevent this from happening. Apparantly parseInt doesn't consider 0 as an integer!

I haven't posted this link in a while, but it's always worth a read for anyone learning/using javascript — musefan
– musefan, Commented Sep 24, 2015 at 13:35

Pranav C Balan · Accepted Answer · 2015-09-24 13:02:26Z

9

Since 0 is act as false , so you can use isNaN() in this case

var res = parseInt(columnSortSettings[0], 10);
return isNaN(res) ? null : res;

edited Sep 24, 2015 at 13:02

answered Sep 24, 2015 at 13:00

Pranav C Balan

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1 Comment

Andy Over a year ago

Don't forget the radix.

Eric · Accepted Answer · 2015-09-24 13:04:02Z

4

It's because you are basically testing 0 which is also false. You can do

var n = columnSortSettings[0];
if(parseInt(n, 10) || n === '0'){
    //...
}

You can also test instead if it's a number

if(typeof(parseInt(n, 10)) === 'number'){
  //...
}

But beware cause

typeof Infinity === 'number';
typeof NaN === 'number';

edited Sep 24, 2015 at 13:04

answered Sep 24, 2015 at 12:59

Eric

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3 Comments

Andy Over a year ago

Don't forget the radix.

Eric Over a year ago

@Andy, yup totally forgot :)

Rahul Misra Over a year ago

Thanks @Eric . As SO lets me accept only one answer, I've upvoted your answer.

Richard Lee · Accepted Answer · 2017-03-10 18:18:38Z

1

You can use the isNumeric operator from rxjs library (importing rxjs/util/isNumeric

answered Mar 10, 2017 at 18:18

Richard Lee

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