Check if string contains only digits

Question

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?

Answer 1

how about

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

Answer 2

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

Answer 3

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

Answer 4

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

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

Answer 5

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

Answer 6

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.

Answer 7

This is what you want

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

Answer 8

in case you need integer and float at same validation

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

Answer 9

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

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

Answer 10

Well, you can use the following regex:

^\d+$

Answer 11

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 

Answer 12

If you use jQuery:

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

Answer 13

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

Answer 14

If you want to leave room for . you can try the below regex.

/[^0-9.]/g

Answer 15

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

If a string contains only digits it will return null