JavaScript check if the value is number, but allow no leading zeros

Question

I am currently working on a project where I need to check if the user input a number value. The problem comes when he input a number starting with a decimal point (.20). I've been looking for a solution, but now I just give up. Neither isNaN or isNumeric works. Also tried to check like: if(value > 0), but this only works when the user input a leading zero. Any help would be greatly appreciated! Thanks in advance!

SOLVED: isNaN doesn't work for me because I use parseInt() instead of parseFloat(). Now everything works just fine. I want the administrators to excuse me for the dumb question I asked! This thread could now be deleted.

I don't get it, why wouldn't isNaN work -> http://jsfiddle.net/6NrZA/1/ — adeneo, Jul 20 '13 at 08:52
*"isNaN doesn't work for me because I use parseInt() instead of parseFloat()"*: Uh? You don't have to use `parseInt` or `parseFloat` at all, just pass the value to `isNaN`. It will be converted to a number, just like if you used the unary plus operator. *"This thread could now be deleted."* That's not how it works. You mark the answer the solved the problem as accepted. — Felix Kling, Jul 20 '13 at 09:25

Tushar Gupta - curioustushar · Answer 1 · 2013-07-20T09:03:49.953

0

Working Demohttp://jsfiddle.net/sPt7q/1

use

if(parseFloat(n) > 0){}

js in fiddle

var n = '.2';
if (n > 0) {
    alert(n);
}
if(parseFloat(n) > 0){
    alert(parseFloat(n));
}

edited Jul 20 '13 at 09:03

answered Jul 20 '13 at 08:54

Tushar Gupta - curioustushar

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What's the point of using `parseFloat` if `n` *is* already a number? – Felix Kling Jul 20 '13 at 09:02
oh you are correct i updated the code pls check now it's working fine. – Tushar Gupta - curioustushar Jul 20 '13 at 09:04

score 0 · Answer 2 · edited May 23 '17 at 11:49

0

Try

function isInt(n) {
   return typeof n === 'number' && n % 1 == 0;
}

from: How do I check that a number is float or integer?

this does fail on 0.0 though.

edited May 23 '17 at 11:49

Community

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answered Jul 20 '13 at 08:56

CBIII

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Rajesh · Answer 3 · 2013-07-20T09:37:34.280

0

A better and fool-proof way is to use regular expression:

var reg = new RegExp('^[1-9]\d*(\.\d+)?$');

or

var reg = new RegExp('^[1-9]\d*(\.\d+)?$');

The user input will be a string (from a form field). Check that string against this regex and that should work.

example:

 if(reg.test(userInput)) {
  //returns true
} else {
  //returns false
}

edited Jul 20 '13 at 09:37

answered Jul 20 '13 at 09:04

Rajesh

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*Why* is it better than using `isNaN`? Do you have any facts to support your claim? Btw, your expression would not handle `'.2'` correctly. – Felix Kling Jul 20 '13 at 09:26
updated my regex. This will handle cases like 2, 0.2, .2 (without leadng zero), 2.222 etc. This regex checks for a number which can be a decimal also. – Rajesh Jul 20 '13 at 09:38
Still doesn't explain why it is "better". – Felix Kling Jul 20 '13 at 09:42
I would say better because it checks the exact pattern while eliminating use of parseInt(), isNaN and other number-manipulating functions. – Rajesh Jul 20 '13 at 09:47

score 0 · Answer 4 · answered Jul 20 '13 at 09:05

0

Try this:

var val = .21;
if (!isNaN(val) && val.toString().indexOf('.') == -1) {
    console.log('valid number:' + val);
} else {
    console.log('invalid number: ' + val);
}

jsFiddle

answered Jul 20 '13 at 09:05

Manoj Yadav

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JavaScript check if the value is number, but allow no leading zeros

4 Answers4