How do you check if a string contains only numbers javascript?

I want to check if a string contains only digits. I used this:

var isANumber = isNaN(theValue) === false; if (isANumber){ .. }

But realized that it also allows + and -. Basically, I want to make sure an input contains ONLY digits and no other characters. Since +100 and -5 are both numbers, isNaN() is not the right way to go. Perhaps a regexp is what I need? Any tips?

Jason Aller

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asked Nov 22, 2009 at 15:24

0

how about

let isnum = /^\d+$/.test(val);

Syntle

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answered Nov 22, 2009 at 15:26

Scott EverndenScott Evernden

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9

string.match(/^[0-9]+$/) != null;

answered Nov 22, 2009 at 15:26

Jason SJason S

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String.prototype.isNumber = function(){return /^\d+$/.test(this);} console.log("123123".isNumber()); // outputs true console.log("+12".isNumber()); // outputs false

answered Nov 22, 2009 at 21:25

baluptonbalupton

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If you want to even support for float values (Dot separated values) then you can use this expression :

var isNumber = /^\d+\.\d+$/.test(value);

answered Jul 24, 2017 at 10:06

Adithya SaiAdithya Sai

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Here's another interesting, readable way to check if a string contains only digits.

This method works by splitting the string into an array using the spread operator, and then uses the every() method to test whether all elements (characters) in the array are included in the string of digits '0123456789':

const digits_only = string => [...string].every(c => '0123456789'.includes(c)); console.log(digits_only('123')); // true console.log(digits_only('+123')); // false console.log(digits_only('-123')); // false console.log(digits_only('123.')); // false console.log(digits_only('.123')); // false console.log(digits_only('123.0')); // false console.log(digits_only('0.123')); // false console.log(digits_only('Hello, world!')); // false

answered Nov 21, 2018 at 23:48

Grant MillerGrant Miller

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Here is a solution without using regular expressions:

function onlyDigits(s) { for (let i = s.length - 1; i >= 0; i--) { const d = s.charCodeAt(i); if (d < 48 || d > 57) return false } return true }

where 48 and 57 are the char codes for "0" and "9", respectively.

answered Jun 28, 2018 at 14:11

SC1000SC1000

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This is what you want

function isANumber(str){ return !/\D/.test(str); }

answered Dec 18, 2016 at 7:11

1

function isNumeric(x) { return parseFloat(x).toString() === x.toString(); }

Though this will return false on strings with leading or trailing zeroes.

answered Jul 18, 2018 at 20:32

in case you need integer and float at same validation

/^\d+\.\d+$|^\d+$/.test(val)

answered Aug 5, 2020 at 13:50

1

Well, you can use the following regex:

^\d+$

answered Nov 22, 2009 at 15:26

JoeyJoey

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if you want to include float values also you can use the following code

theValue=$('#balanceinput').val(); var isnum1 = /^\d*\.?\d+$/.test(theValue); var isnum2 = /^\d*\.?\d+$/.test(theValue.split("").reverse().join("")); alert(isnum1+' '+isnum2);

this will test for only digits and digits separated with '.' the first test will cover values such as 0.1 and 0 but also .1 , it will not allow 0. so the solution that I propose is to reverse theValue so .1 will be 1. then the same regular expression will not allow it .

example :

theValue=3.4; //isnum1=true , isnum2=true theValue=.4; //isnum1=true , isnum2=false theValue=3.; //isnum1=flase , isnum2=true

answered Jun 9, 2020 at 15:25

1

Here's a Solution without using regex

const isdigit=(value)=>{ const val=Number(value)?true:false console.log(val); return val } isdigit("10")//true isdigit("any String")//false

answered Oct 24, 2020 at 8:57

3

If you use jQuery:

$.isNumeric('1234'); // true $.isNumeric('1ab4'); // false

answered Jul 3, 2020 at 11:44

JacobJacob

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c="123".match(/\D/) == null #true c="a12".match(/\D/) == null #false

If a string contains only digits it will return null

answered Jul 25, 2019 at 9:04

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