In Typescript, How to check if a string is Numeric

Question

In Typescript, this shows an error saying isNaN accepts only numeric values

isNaN('9BX46B6A')

and this returns false because parseFloat('9BX46B6A') evaluates to 9

isNaN(parseFloat('9BX46B6A'))

I can still run with the error showing up in Visual Studio, but I would like to do it the right way.

Currently, I have written this modified function -

static isNaNModified = (inputStr: string) => {
    var numericRepr = parseFloat(inputStr);
    return isNaN(numericRepr) || numericRepr.toString().length != inputStr.length;
}

Answer 1

The way to convert a string to a number is with Number , not parseFloat .

Number('1234') // 1234
Number('9BX9') // NaN

You can also use the unary plus operator if you like shorthand:

+'1234' // 1234
+'9BX9' // NaN

Be careful when checking against NaN (the operator === and !== don't work as expected with NaN ). Use:

 isNaN(+maybeNumber) // returns true if NaN, otherwise false

Answer 2

Update 2

This method is no longer available in rxjs v6

I'm solved it by using the isNumeric operator from rxjs library (importing rxjs/util/isNumeric

Update

import { isNumeric } from 'rxjs/util/isNumeric';

. . .

var val = "5700";
if (isNumeric(val)){
   alert("it is number !");
}

Answer 3

function isNumber(value: string | number): boolean
{
   return ((value != null) &&
           (value !== '') &&
           !isNaN(Number(value.toString())));
}

Answer 4

You can use the Number.isFinite() function:

Number.isFinite(Infinity);  // false
Number.isFinite(NaN);       // false
Number.isFinite(-Infinity); // false
Number.isFinite('0');       // false
Number.isFinite(null);      // false

Number.isFinite(0);         // true
Number.isFinite(2e64);      // true

Note: there's a significant difference between the global function isFinite() and the latter Number.isFinite() . In the case of the former, string coercion is performed - so isFinite('0') === true whilst Number.isFinite('0') === false .

Also, note that this is not available in IE!

Answer 5

Simple answer: (watch for blank & null)

isNaN(+'111') = false;
isNaN(+'111r') = true;
isNaN(+'r') = true;
isNaN(+'') = false;   
isNaN(null) = false;   

https://codepen.io/CQCoder/pen/zYGEjxd?editors=1111

Answer 6

I would choose an existing and already tested solution. For example this from rxjs in typescript:

function isNumeric(val: any): val is number | string {
  // parseFloat NaNs numeric-cast false positives (null|true|false|"")
  // ...but misinterprets leading-number strings, particularly hex literals ("0x...")
  // subtraction forces infinities to NaN
  // adding 1 corrects loss of precision from parseFloat (#15100)
  return !isArray(val) && (val - parseFloat(val) + 1) >= 0;
}

rxjs/isNumeric.ts

Without rxjs isArray() function and with simplefied typings:

function isNumeric(val: any): boolean {
  return !(val instanceof Array) && (val - parseFloat(val) + 1) >= 0;
}

You should always test such functions with your use cases. If you have special value types, this function may not be your solution. You can test the function here.

Results are:

enum         : CardTypes.Debit   : true
decimal      : 10                : true
hexaDecimal  : 0xf10b            : true
binary       : 0b110100          : true
octal        : 0o410             : true
stringNumber : '10'              : true

string       : 'Hello'           : false
undefined    : undefined         : false
null         : null              : false
function     : () => {}          : false
array        : [80, 85, 75]      : false
turple       : ['Kunal', 2018]   : false
object       : {}                : false

As you can see, you have to be careful, if you use this function with enums.

Answer 7

对于 Angular 中的完整数字（非浮点数），您可以使用：

if (Number.isInteger(yourVariable)) { ... }

Answer 8

Most of the time the value that we want to check is string or number, so here is function that i use:

const isNumber = (n: string | number): boolean => 
    !isNaN(parseFloat(String(n))) && isFinite(Number(n));

---

Codesandbox tests .

const willBeTrue = [0.1, '1', '-1', 1, -1, 0, -0, '0', "-0", 2e2, 1e23, 1.1, -0.1, '0.1', '2e2', '1e23', '-0.1', ' 898', '080']

const willBeFalse = ['9BX46B6A', "+''", '', '-0,1', [], '123a', 'a', 'NaN', 1e10000, undefined, null, NaN, Infinity, () => {}]

Answer 9

My simple solution here is:

const isNumeric = (val: string) : boolean => {
   return !isNaN(Number(val));
}

// isNumberic("2") => true
// isNumeric("hi") => false;

Answer 10

滚动浏览每个答案，惊讶地发现这个简单的单行文字还没有出现：

const isNumber = (val: string | number) => !!(val || val === 0) && !isNaN(Number(val.toString()));

Answer 11

For Rxjs version above 6, we need to import the function " isNumeric " like below. Below import worked for me.

import { isNumeric } from 'rxjs/internal-compatibility/index';

Answer 12

My solution:

 function isNumeric(val: unknown): val is string | number {
  return (
    !isNaN(Number(Number.parseFloat(String(val)))) &&
    isFinite(Number(val))
  );
}

// true
isNumeric("-10");
isNumeric("0");
isNumeric("0xff");
isNumeric("0xFF");
isNumeric("8e5");
isNumeric("3.1415");
isNumeric("+10");
isNumeric("144");
isNumeric("5");

// false
isNumeric("-0x42");
isNumeric("7.2acdgs");
isNumeric("");
isNumeric({});
isNumeric(NaN);
isNumeric(null);
isNumeric(true);
isNumeric(Infinity);
isNumeric(undefined);
isNumeric([]);
isNumeric("some string");

Answer 13

 const isNumeric = (value: string): boolean =>
  !new RegExp(/[^\d]/g).test(value.trim());

If you want to allow decimal point

   const isNumeric = (value: string): boolean =>
      !new RegExp(/[^\d.]/g).test(value.trim());

Answer 14

if var sum=0; var x;

then, what about this? sum+=(x|0);

Answer 15

Whether a string can be parsed as a number is a runtime concern. Typescript does not support this use case as it is focused on compile time (not runtime) safety.