Using a regular expression to validate whether input has any non digits in it

Question

function validInteger(theNumber){
    var anyNonDigits = new  RegExp('\D','g');
    if(parseInt(theNumber)&&!anyNonDigits.test(theNumber)){
        return true;
    }else{
        return false;
    }
}

Above is a function I've written to validate some input. I want all positive integers. The problem I'm facing is with the RegExp object. This seems like it should be super simple, but for some reason it's not working.

For example if I pass 'f5' I get true, but if I pass '5f' I get false. I'm also having problems when passing negative numbers. -3 doesn't get caught even if I stringify the variable before passing it into the RegExp. I can fix this by adding '&&parseInt(theNumber)>0' in my if statement, but I feel like the RegExp should catch that too. Thanks in advance!

The regular expression doesn't work because you need to escape the `\ ` character since `\ ` also has meaning in string literal notation. `new RegExp('\\D','g');` — , Jun 07 '12 at 16:57
...also, you wouldn't really need to `g` modifier, since it sounds as though even one non-digit should be enough to flag the input. — , Jun 07 '12 at 17:17
@amnotiam thanks for the info on why what I was doing wasn't working. Regular expressions are something I'm still new to. Hopefully I'll change that soon. — Frank B, Jun 07 '12 at 18:24

gdoron · Accepted Answer · 2019-01-05T19:25:04.647

11

Simply:

function validInteger(theNumber){    
    return theNumber.match(/^\d+$/) && parseInt(theNumber) > 0;
}

Live DEMO

Or even simpler with regex only as suggested by @Eric:

return /^[0-9]\d*$/.test(theNumber);

Live DEMO

Update:

~~An excellent cheat sheet.~~ The link died after 5 years, sorry.

edited Jan 05 '19 at 19:25

answered Jun 07 '12 at 16:55

gdoron

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No need for `parseInt(theNumber) > 0`: `/^[1-9]\d*$/` – Eric Jun 07 '12 at 16:58
@Eric. Thanks for the suggestion, I added it with your name on it. – gdoron Jun 07 '12 at 17:02
i think that regex would call an empty string valid since `*` is "zero or more", you may want to use `+` instead "one or more" – jbabey Jun 07 '12 at 17:11
@jbabey. It's o.k. as it is because of the `[0-9]` before of that. – gdoron Jun 07 '12 at 17:12
@gdoron oh right, the `*` is only applying to the `\d`. got it. – jbabey Jun 07 '12 at 17:13
Thanks for the help guys. Regular expressions have always been a problem for me, can someone point me in the right direction of a good tutorial? – Frank B Jun 07 '12 at 18:23
@FrankB. I added a cheat sheet, for tutorials, just google it up. – gdoron Jun 07 '12 at 18:25
Cheat Sheet link was so useful – abhihello123 Aug 18 '13 at 18:34
Cheat Sheet link is broken :/ – Josiel Faleiros Jan 05 '19 at 16:16

score 1 · Answer 2 · answered Jun 07 '12 at 17:01

1

If it's okay don't use RegExp, you can have:

function validInteger(theNumber){
    var number = +theNumber;

    return number > -1 && number % 1 === 0;
}

Assuming that you consider 0 as positive integer, and you don't want to make a distinction between +0 and -0.

Notice that this function will accept any value for theNumber that can be converted in a Number, so not just "string", and you can pass Number as well of course.

answered Jun 07 '12 at 17:01

ZER0

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Probably you mean `-1 < number`? – Jun 07 '12 at 17:03
@amnotiam. Is there any difference (I know they do different things) between `if(str.match())` and `if (/.../.test(str))`? – gdoron Jun 07 '12 at 17:05
@gdoron: Should give the same boolean coerced result. I personally prefer `.test()`. – Jun 07 '12 at 17:06

score 0 · Answer 3 · answered Jun 07 '12 at 17:06

0

Be simple!

function validate(num){
    return (num | 0) > 0;
};

This function will return "true" only for positive integers.

answered Jun 07 '12 at 17:06

sgmonda

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1

Depends on how you want to evaluate hexadecimal notation. `('0xff' | 0) > 0; // true` – Jun 07 '12 at 17:09
1

`("+123" | 0) > 0` => `true` => broken... `:)` – gdoron Jun 07 '12 at 17:11

Using a regular expression to validate whether input has any non digits in it

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