Check if character is number?

Question

I need to check whether justPrices[i].substr(commapos+2,1).

The string is something like: \"blabla,120\"

In this case it would check whether \'0

Answer 1

var Is = {
    character: {
        number: (function() {
            // Only computed once
            var zero = "0".charCodeAt(0), nine = "9".charCodeAt(0);

            return function(c) {
                return (c = c.charCodeAt(0)) >= zero && c <= nine;
            }
        })()
    }
};

Answer 2

you can either use parseInt and than check with isNaN

or if you want to work directly on your string you can use regexp like this:

function is_numeric(str){
    return /^\d+$/.test(str);
}

Answer 3

The shortest solution is:

const isCharDigit = n => n < 10;

You can apply these as well:

const isCharDigit = n => Boolean(++n);

const isCharDigit = n => '/' < n && n < ':';

const isCharDigit = n => !!++n;

if you want to check more than 1 chatacter, you might use next variants

Regular Expression:

const isDigit = n => /\d+/.test(n);

Comparison:

const isDigit = n => +n == n;

Check if it is not NaN

const isDigit = n => !isNaN(n);

Answer 4

isNumber = function(obj, strict) {
    var strict = strict === true ? true : false;
    if (strict) {
        return !isNaN(obj) && obj instanceof Number ? true : false;
    } else {
        return !isNaN(obj - parseFloat(obj));
    }
}

output without strict mode:

var num = 14;
var textnum = '14';
var text = 'yo';
var nan = NaN;

isNumber(num);
isNumber(textnum);
isNumber(text);
isNumber(nan);

true
true
false
false

output with strict mode:

var num = 14;
var textnum = '14';
var text = 'yo';
var nan = NaN;

isNumber(num, true);
isNumber(textnum, true);
isNumber(text, true);
isNumber(nan);

true
false
false
false

Answer 5

As far as I know, easiest way is just to multiply by 1:

var character = ... ; // your character
var isDigit = ! isNaN(character * 1);

Multiplication by one makes a number from any numeric string (as you have only one character it will always be an integer from 0 to 9) and a NaN for any other string.

Answer 6

I wonder why nobody has posted a solution like:

var charCodeZero = "0".charCodeAt(0);
var charCodeNine = "9".charCodeAt(0);

function isDigitCode(n) {
   return(n >= charCodeZero && n <= charCodeNine);
}

with an invocation like:

if (isDigitCode(justPrices[i].charCodeAt(commapos+2))) {
    ... // digit
} else {
    ... // not a digit
}