[javascript] Check if string contains only digits

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 wanna 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?

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

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.

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

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));_x000D_
_x000D_
console.log(digits_only('123')); // true_x000D_
console.log(digits_only('+123')); // false_x000D_
console.log(digits_only('-123')); // false_x000D_
console.log(digits_only('123.')); // false_x000D_
console.log(digits_only('.123')); // false_x000D_
console.log(digits_only('123.0')); // false_x000D_
console.log(digits_only('0.123')); // false_x000D_
console.log(digits_only('Hello, world!')); // false

This is what you want

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

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

If a string contains only digits it will return null

it checks valid number integers or float or double not a string

regex that i used

simple regex

 1. ^[0-9]*[.]?[0-9]*$

advance regex

 2. ^-?[\d.]+(?:e-?\d+)?$    

eg:only numbers

var str='1232323';
var reg=/^[0-9]*[.]?[0-9]*$/;

console.log(reg.test(str))

it allows

eg:123434

eg:.1232323

eg:12.3434434

it does not allow

eg:1122212.efsffasf

eg:2323fdf34434

eg:0.3232rf3333

advance regex

////////////////////////////////////////////////////////////////////

int or float or exponent all types of number in string

^-?[\d.]+(?:e-?\d+)?$

eg:13123123

eg:12344.3232

eg:2323323e4

eg:0.232332

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

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

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

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

in case you need integer and float at same validation

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

Well, you can use the following regex:

^\d+$

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 

If you use jQuery:

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

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