I'm hoping there's something in the same conceptual space as the old VB6 IsNumeric() function?
This question is related to
javascript
validation
numeric
The answer is
You could make use of types, like with the flow library, to get static, compile time checking. Of course not terribly useful for user input.
// @flow
function acceptsNumber(value: number) {
// ...
}
acceptsNumber(42); // Works!
acceptsNumber(3.14); // Works!
acceptsNumber(NaN); // Works!
acceptsNumber(Infinity); // Works!
acceptsNumber("foo"); // Error!
Just use isNaN(), this will convert the string to a number and if get a valid number, will return false...
isNaN("Alireza"); //return true
isNaN("123"); //return false
I recently wrote an article about ways to ensure a variable is a valid number: https://github.com/jehugaleahsa/artifacts/blob/master/2018/typescript_num_hack.md The article explains how to ensure floating point or integer, if that's important (+x vs ~~x).
The article assumes the variable is a string or a number to begin with and trim is available/polyfilled. It wouldn't be hard to extend it to handle other types, as well. Here's the meat of it:
// Check for a valid float
if (x == null
|| ("" + x).trim() === ""
|| isNaN(+x)) {
return false; // not a float
}
// Check for a valid integer
if (x == null
|| ("" + x).trim() === ""
|| ~~x !== +x) {
return false; // not an integer
}
The accepted answer for this question has quite a few flaws (as highlighted by couple of other users). This is one of the easiest & proven way to approach it in javascript:
function isNumeric(n) {
return !isNaN(parseFloat(n)) && isFinite(n);
}
Below are some good test cases:
console.log(isNumeric(12345678912345678912)); // true
console.log(isNumeric('2 ')); // true
console.log(isNumeric('-32.2 ')); // true
console.log(isNumeric(-32.2)); // true
console.log(isNumeric(undefined)); // false
// the accepted answer fails at these tests:
console.log(isNumeric('')); // false
console.log(isNumeric(null)); // false
console.log(isNumeric([])); // false
If you're just trying to check if a string is a whole number (no decimal places), regex is a good way to go. Other methods such as isNaN are too complicated for something so simple.
function isNumeric(value) {
return /^-?\d+$/.test(value);
}
console.log(isNumeric('abcd')); // false
console.log(isNumeric('123a')); // false
console.log(isNumeric('1')); // true
console.log(isNumeric('1234567890')); // true
console.log(isNumeric('-23')); // true
console.log(isNumeric(1234)); // true
console.log(isNumeric('123.4')); // false
console.log(isNumeric('')); // false
console.log(isNumeric(undefined)); // false
console.log(isNumeric(null)); // false
To only allow positive whole numbers use this:
function isNumeric(value) {
return /^\d+$/.test(value);
}
console.log(isNumeric('123')); // true
console.log(isNumeric('-23')); // false
2019: Practical and tight numerical validity check
Often, a 'valid number' means a Javascript number excluding NaN and Infinity, ie a 'finite number'.
To check the numerical validity of a value (from an external source for example), you can define in ESlint Airbnb style :
/**
* Returns true if 'candidate' is a finite number or a string referring (not just 'including') a finite number
* To keep in mind:
* Number(true) = 1
* Number('') = 0
* Number(" 10 ") = 10
* !isNaN(true) = true
* parseFloat('10 a') = 10
*
* @param {?} candidate
* @return {boolean}
*/
function isReferringFiniteNumber(candidate) {
if (typeof (candidate) === 'number') return Number.isFinite(candidate);
if (typeof (candidate) === 'string') {
return (candidate.trim() !== '') && Number.isFinite(Number(candidate));
}
return false;
}
and use it this way:
if (isReferringFiniteNumber(theirValue)) {
myCheckedValue = Number(theirValue);
} else {
console.warn('The provided value doesn\'t refer to a finite number');
}
Just use isNaN(), this will convert the string to a number and if get a valid number, will return false...
isNaN("Alireza"); //return true
isNaN("123"); //return false
If you're just trying to check if a string is a whole number (no decimal places), regex is a good way to go. Other methods such as isNaN are too complicated for something so simple.
function isNumeric(value) {
return /^-?\d+$/.test(value);
}
console.log(isNumeric('abcd')); // false
console.log(isNumeric('123a')); // false
console.log(isNumeric('1')); // true
console.log(isNumeric('1234567890')); // true
console.log(isNumeric('-23')); // true
console.log(isNumeric(1234)); // true
console.log(isNumeric('123.4')); // false
console.log(isNumeric('')); // false
console.log(isNumeric(undefined)); // false
console.log(isNumeric(null)); // false
To only allow positive whole numbers use this:
function isNumeric(value) {
return /^\d+$/.test(value);
}
console.log(isNumeric('123')); // true
console.log(isNumeric('-23')); // false
It is not valid for TypeScript as:
declare function isNaN(number: number): boolean;
For TypeScript you can use:
/^\d+$/.test(key)
Try the isNan function:
The isNaN() function determines whether a value is an illegal number (Not-a-Number).
This function returns true if the value equates to NaN. Otherwise it returns false.
This function is different from the Number specific Number.isNaN() method.
The global isNaN() function, converts the tested value to a Number, then tests it.
Number.isNan() does not convert the values to a Number, and will not return true for any value that is not of the type Number...
My solution:
// returns true for positive ints;
// no scientific notation, hexadecimals or floating point dots
var isPositiveInt = function(str) {
var result = true, chr;
for (var i = 0, n = str.length; i < n; i++) {
chr = str.charAt(i);
if ((chr < "0" || chr > "9") && chr != ",") { //not digit or thousands separator
result = false;
break;
};
if (i == 0 && (chr == "0" || chr == ",")) { //should not start with 0 or ,
result = false;
break;
};
};
return result;
};
You can add additional conditions inside the loop, to fit you particular needs.
And you could go the RegExp-way:
var num = "987238";
if(num.match(/^-?\d+$/)){
//valid integer (positive or negative)
}else if(num.match(/^\d+\.\d+$/)){
//valid float
}else{
//not valid number
}
Here is a high-performance (2.5*10^7 iterations/s @3.8GHz Haswell) version of a isNumber implementation. It works for every testcase i could find (including Symbols):
var isNumber = (function () {
var isIntegerTest = /^\d+$/;
var isDigitArray = [!0, !0, !0, !0, !0, !0, !0, !0, !0, !0];
function hasLeading0s (s) {
return !(typeof s !== 'string' ||
s.length < 2 ||
s[0] !== '0' ||
!isDigitArray[s[1]] ||
isIntegerTest.test(s));
}
var isWhiteSpaceTest = /\s/;
return function isNumber (s) {
var t = typeof s;
var n;
if (t === 'number') {
return (s <= 0) || (s > 0);
} else if (t === 'string') {
n = +s;
return !((!(n <= 0) && !(n > 0)) || n === '0' || hasLeading0s(s) || !(n !== 0 || !(s === '' || isWhiteSpaceTest.test(s))));
} else if (t === 'object') {
return !(!(s instanceof Number) || ((n = +s), !(n <= 0) && !(n > 0)));
}
return false;
};
})();
Old question, but there are several points missing in the given answers.
Scientific notation.
!isNaN('1e+30') is true, however in most of the cases when people ask for numbers, they do not want to match things like 1e+30.
Large floating numbers may behave weird
Observe (using Node.js):
> var s = Array(16 + 1).join('9')
undefined
> s.length
16
> s
'9999999999999999'
> !isNaN(s)
true
> Number(s)
10000000000000000
> String(Number(s)) === s
false
>
On the other hand:
> var s = Array(16 + 1).join('1')
undefined
> String(Number(s)) === s
true
> var s = Array(15 + 1).join('9')
undefined
> String(Number(s)) === s
true
>
So, if one expects String(Number(s)) === s, then better limit your strings to 15 digits at most (after omitting leading zeros).
Infinity
> typeof Infinity
'number'
> !isNaN('Infinity')
true
> isFinite('Infinity')
false
>
Given all that, checking that the given string is a number satisfying all of the following:
- non scientific notation
- predictable conversion to
Numberand back toString - finite
is not such an easy task. Here is a simple version:
function isNonScientificNumberString(o) {
if (!o || typeof o !== 'string') {
// Should not be given anything but strings.
return false;
}
return o.length <= 15 && o.indexOf('e+') < 0 && o.indexOf('E+') < 0 && !isNaN(o) && isFinite(o);
}
However, even this one is far from complete. Leading zeros are not handled here, but they do screw the length test.
Quote:
isNaN(num) // returns true if the variable does NOT contain a valid number
is not entirely true if you need to check for leading/trailing spaces - for example when a certain quantity of digits is required, and you need to get, say, '1111' and not ' 111' or '111 ' for perhaps a PIN input.
Better to use:
var num = /^\d+$/.test(num)
Here's a one-liner to check if sNum is a valid numeric value; it has been tested for a wide variety of inputs:
!isNaN(+s.replace(/\s|\$/g, '')); // returns True if numeric value
And you could go the RegExp-way:
var num = "987238";
if(num.match(/^-?\d+$/)){
//valid integer (positive or negative)
}else if(num.match(/^\d+\.\d+$/)){
//valid float
}else{
//not valid number
}
If anyone ever gets this far down, I spent some time hacking on this trying to patch moment.js (https://github.com/moment/moment). Here's something that I took away from it:
function isNumeric(val) {
var _val = +val;
return (val !== val + 1) //infinity check
&& (_val === +val) //Cute coercion check
&& (typeof val !== 'object') //Array/object check
}
Handles the following cases:
True! :
isNumeric("1"))
isNumeric(1e10))
isNumeric(1E10))
isNumeric(+"6e4"))
isNumeric("1.2222"))
isNumeric("-1.2222"))
isNumeric("-1.222200000000000000"))
isNumeric("1.222200000000000000"))
isNumeric(1))
isNumeric(0))
isNumeric(-0))
isNumeric(1010010293029))
isNumeric(1.100393830000))
isNumeric(Math.LN2))
isNumeric(Math.PI))
isNumeric(5e10))
False! :
isNumeric(NaN))
isNumeric(Infinity))
isNumeric(-Infinity))
isNumeric())
isNumeric(undefined))
isNumeric('[1,2,3]'))
isNumeric({a:1,b:2}))
isNumeric(null))
isNumeric([1]))
isNumeric(new Date()))
Ironically, the one I am struggling with the most:
isNumeric(new Number(1)) => false
Any suggestions welcome. :]
Maybe there are one or two people coming across this question who need a much stricter check than usual (like I did). In that case, this might be useful:
if(str === String(Number(str))) {
// it's a "perfectly formatted" number
}
Beware! This will reject strings like .1, 40.000, 080, 00.1. It's very picky - the string must match the "most minimal perfect form" of the number for this test to pass.
It uses the String and Number constructor to cast the string to a number and back again and thus checks if the JavaScript engine's "perfect minimal form" (the one it got converted to with the initial Number constructor) matches the original string.
I have tested and Michael's solution is best. Vote for his answer above (search this page for "If you really want to make sure that a string" to find it). In essence, his answer is this:
function isNumeric(num){
num = "" + num; //coerce num to be a string
return !isNaN(num) && !isNaN(parseFloat(num));
}
It works for every test case, which I documented here: https://jsfiddle.net/wggehvp9/5/
Many of the other solutions fail for these edge cases: ' ', null, "", true, and []. In theory, you could use them, with proper error handling, for example:
return !isNaN(num);
or
return (+num === +num);
with special handling for /\s/, null, "", true, false, [] (and others?)
When guarding against empty strings and null
// Base cases that are handled properly
Number.isNaN(Number('1')); // => false
Number.isNaN(Number('-1')); // => false
Number.isNaN(Number('1.1')); // => false
Number.isNaN(Number('-1.1')); // => false
Number.isNaN(Number('asdf')); // => true
Number.isNaN(Number(undefined)); // => true
// Special notation cases that are handled properly
Number.isNaN(Number('1e1')); // => false
Number.isNaN(Number('1e-1')); // => false
Number.isNaN(Number('-1e1')); // => false
Number.isNaN(Number('-1e-1')); // => false
Number.isNaN(Number('0b1')); // => false
Number.isNaN(Number('0o1')); // => false
Number.isNaN(Number('0xa')); // => false
// Edge cases that will FAIL if not guarded against
Number.isNaN(Number('')); // => false
Number.isNaN(Number(' ')); // => false
Number.isNaN(Number(null)); // => false
// Edge cases that are debatable
Number.isNaN(Number('-0b1')); // => true
Number.isNaN(Number('-0o1')); // => true
Number.isNaN(Number('-0xa')); // => true
Number.isNaN(Number('Infinity')); // => false
Number.isNaN(Number('INFINITY')); // => true
Number.isNaN(Number('-Infinity')); // => false
Number.isNaN(Number('-INFINITY')); // => true
When NOT guarding against empty strings and null
Using parseInt:
// Base cases that are handled properly
Number.isNaN(parseInt('1')); // => false
Number.isNaN(parseInt('-1')); // => false
Number.isNaN(parseInt('1.1')); // => false
Number.isNaN(parseInt('-1.1')); // => false
Number.isNaN(parseInt('asdf')); // => true
Number.isNaN(parseInt(undefined)); // => true
Number.isNaN(parseInt('')); // => true
Number.isNaN(parseInt(' ')); // => true
Number.isNaN(parseInt(null)); // => true
// Special notation cases that are handled properly
Number.isNaN(parseInt('1e1')); // => false
Number.isNaN(parseInt('1e-1')); // => false
Number.isNaN(parseInt('-1e1')); // => false
Number.isNaN(parseInt('-1e-1')); // => false
Number.isNaN(parseInt('0b1')); // => false
Number.isNaN(parseInt('0o1')); // => false
Number.isNaN(parseInt('0xa')); // => false
// Edge cases that are debatable
Number.isNaN(parseInt('-0b1')); // => false
Number.isNaN(parseInt('-0o1')); // => false
Number.isNaN(parseInt('-0xa')); // => false
Number.isNaN(parseInt('Infinity')); // => true
Number.isNaN(parseInt('INFINITY')); // => true
Number.isNaN(parseInt('-Infinity')); // => true
Number.isNaN(parseInt('-INFINITY')); // => true
Using parseFloat:
// Base cases that are handled properly
Number.isNaN(parseFloat('1')); // => false
Number.isNaN(parseFloat('-1')); // => false
Number.isNaN(parseFloat('1.1')); // => false
Number.isNaN(parseFloat('-1.1')); // => false
Number.isNaN(parseFloat('asdf')); // => true
Number.isNaN(parseFloat(undefined)); // => true
Number.isNaN(parseFloat('')); // => true
Number.isNaN(parseFloat(' ')); // => true
Number.isNaN(parseFloat(null)); // => true
// Special notation cases that are handled properly
Number.isNaN(parseFloat('1e1')); // => false
Number.isNaN(parseFloat('1e-1')); // => false
Number.isNaN(parseFloat('-1e1')); // => false
Number.isNaN(parseFloat('-1e-1')); // => false
Number.isNaN(parseFloat('0b1')); // => false
Number.isNaN(parseFloat('0o1')); // => false
Number.isNaN(parseFloat('0xa')); // => false
// Edge cases that are debatable
Number.isNaN(parseFloat('-0b1')); // => false
Number.isNaN(parseFloat('-0o1')); // => false
Number.isNaN(parseFloat('-0xa')); // => false
Number.isNaN(parseFloat('Infinity')); // => false
Number.isNaN(parseFloat('INFINITY')); // => true
Number.isNaN(parseFloat('-Infinity')); // => false
Number.isNaN(parseFloat('-INFINITY')); // => true
Notes:
- Only string, empty, and uninitialized values are considered in keeping with addressing the original question. Additional edge cases exist if arrays and objects are the values being considered.
- Characters in binary, octal, hexadecimal, and exponential notation are not case-sensitive (ie: '0xFF', '0XFF', '0xfF' etc. will all yield the same result in the test cases shown above).
- Unlike with
Infinity(case-sensitive) in some cases, constants from theNumberandMathobjects passed as test cases in string format to any of the methods above will be determined to not be numbers. - See here for an explanation of how arguments are converted to a
Numberand why the edge cases fornulland empty strings exist.
This is built on some of the previous answers and comments. The following covers all the edge cases and fairly concise as well:
const isNumRegEx = /^-?(\d*\.)?\d+$/;
function isNumeric(n, allowScientificNotation = false) {
return allowScientificNotation ?
!Number.isNaN(parseFloat(n)) && Number.isFinite(n) :
isNumRegEx.test(n);
}
Save yourself the headache of trying to find a "built-in" solution.
There isn't a good answer, and the hugely upvoted answer in this thread is wrong.
npm install is-number
In JavaScript, it's not always as straightforward as it should be to reliably check if a value is a number. It's common for devs to use +, -, or Number() to cast a string value to a number (for example, when values are returned from user input, regex matches, parsers, etc). But there are many non-intuitive edge cases that yield unexpected results:
console.log(+[]); //=> 0
console.log(+''); //=> 0
console.log(+' '); //=> 0
console.log(typeof NaN); //=> 'number'
I used this function as a form validation tool, and I didn't want users to be able to write exponential function, so I came up with this function:
<script>
function isNumber(value, acceptScientificNotation) {
if(true !== acceptScientificNotation){
return /^-{0,1}\d+(\.\d+)?$/.test(value);
}
if (true === Array.isArray(value)) {
return false;
}
return !isNaN(parseInt(value, 10));
}
console.log(isNumber("")); // false
console.log(isNumber(false)); // false
console.log(isNumber(true)); // false
console.log(isNumber("0")); // true
console.log(isNumber("0.1")); // true
console.log(isNumber("12")); // true
console.log(isNumber("-12")); // true
console.log(isNumber(-45)); // true
console.log(isNumber({jo: "pi"})); // false
console.log(isNumber([])); // false
console.log(isNumber([78, 79])); // false
console.log(isNumber(NaN)); // false
console.log(isNumber(Infinity)); // false
console.log(isNumber(undefined)); // false
console.log(isNumber("0,1")); // false
console.log(isNumber("1e-1")); // false
console.log(isNumber("1e-1", true)); // true
</script>
My attempt at a slightly confusing, Pherhaps not the best solution
function isInt(a){
return a === ""+~~a
}
console.log(isInt('abcd')); // false
console.log(isInt('123a')); // false
console.log(isInt('1')); // true
console.log(isInt('0')); // true
console.log(isInt('-0')); // false
console.log(isInt('01')); // false
console.log(isInt('10')); // true
console.log(isInt('-1234567890')); // true
console.log(isInt(1234)); // false
console.log(isInt('123.4')); // false
console.log(isInt('')); // false
// other types then string returns false
console.log(isInt(5)); // false
console.log(isInt(undefined)); // false
console.log(isInt(null)); // false
console.log(isInt('0x1')); // false
console.log(isInt(Infinity)); // false
function isNumberCandidate(s) {
const str = (''+ s).trim();
if (str.length === 0) return false;
return !isNaN(+str);
}
console.log(isNumberCandidate('1')); // true
console.log(isNumberCandidate('a')); // false
console.log(isNumberCandidate('000')); // true
console.log(isNumberCandidate('1a')); // false
console.log(isNumberCandidate('1e')); // false
console.log(isNumberCandidate('1e-1')); // true
console.log(isNumberCandidate('123.3')); // true
console.log(isNumberCandidate('')); // false
console.log(isNumberCandidate(' ')); // false
console.log(isNumberCandidate(1)); // true
console.log(isNumberCandidate(0)); // true
console.log(isNumberCandidate(NaN)); // false
console.log(isNumberCandidate(undefined)); // false
console.log(isNumberCandidate(null)); // false
console.log(isNumberCandidate(-1)); // true
console.log(isNumberCandidate('-1')); // true
console.log(isNumberCandidate('-1.2')); // true
console.log(isNumberCandidate(0.0000001)); // true
console.log(isNumberCandidate('0.0000001')); // true
console.log(isNumberCandidate(Infinity)); // true
console.log(isNumberCandidate(-Infinity)); // true
console.log(isNumberCandidate('Infinity')); // true
if (isNumberCandidate(s)) {
// use +s as a number
+s ...
}
parseInt(), but be aware that this function is a bit different in the sense that it for example returns 100 for parseInt("100px").
Maybe there are one or two people coming across this question who need a much stricter check than usual (like I did). In that case, this might be useful:
if(str === String(Number(str))) {
// it's a "perfectly formatted" number
}
Beware! This will reject strings like .1, 40.000, 080, 00.1. It's very picky - the string must match the "most minimal perfect form" of the number for this test to pass.
It uses the String and Number constructor to cast the string to a number and back again and thus checks if the JavaScript engine's "perfect minimal form" (the one it got converted to with the initial Number constructor) matches the original string.
PFB the working solution:
function(check){
check = check + "";
var isNumber = check.trim().length>0? !isNaN(check):false;
return isNumber;
}
Try the isNan function:
The isNaN() function determines whether a value is an illegal number (Not-a-Number).
This function returns true if the value equates to NaN. Otherwise it returns false.
This function is different from the Number specific Number.isNaN() method.
The global isNaN() function, converts the tested value to a Number, then tests it.
Number.isNan() does not convert the values to a Number, and will not return true for any value that is not of the type Number...
So, it will depend on the test cases that you want it to handle.
function isNumeric(number) {
return !isNaN(parseFloat(number)) && !isNaN(+number);
}
What I was looking for was regular types of numbers in javascript.
0, 1 , -1, 1.1 , -1.1 , 1E1 , -1E1 , 1e1 , -1e1, 0.1e10, -0.1.e10 , 0xAF1 , 0o172, Math.PI, Number.NEGATIVE_INFINITY, Number.POSITIVE_INFINITY
And also they're representations as strings:
'0', '1', '-1', '1.1', '-1.1', '1E1', '-1E1', '1e1', '-1e1', '0.1e10', '-0.1.e10', '0xAF1', '0o172'
I did want to leave out and not mark them as numeric
'', ' ', [], {}, null, undefined, NaN
As of today, all other answers seemed to failed one of these test cases.
And you could go the RegExp-way:
var num = "987238";
if(num.match(/^-?\d+$/)){
//valid integer (positive or negative)
}else if(num.match(/^\d+\.\d+$/)){
//valid float
}else{
//not valid number
}
I'm using the following:
const isNumber = s => !isNaN(+s)
This appears to catch the seemingly infinite number of edge cases:
function isNumber(x, noStr) {
/*
- Returns true if x is either a finite number type or a string containing only a number
- If empty string supplied, fall back to explicit false
- Pass true for noStr to return false when typeof x is "string", off by default
isNumber(); // false
isNumber([]); // false
isNumber([1]); // false
isNumber([1,2]); // false
isNumber(''); // false
isNumber(null); // false
isNumber({}); // false
isNumber(true); // false
isNumber('true'); // false
isNumber('false'); // false
isNumber('123asdf'); // false
isNumber('123.asdf'); // false
isNumber(undefined); // false
isNumber(Number.POSITIVE_INFINITY); // false
isNumber(Number.NEGATIVE_INFINITY); // false
isNumber('Infinity'); // false
isNumber('-Infinity'); // false
isNumber(Number.NaN); // false
isNumber(new Date('December 17, 1995 03:24:00')); // false
isNumber(0); // true
isNumber('0'); // true
isNumber(123); // true
isNumber(123.456); // true
isNumber(-123.456); // true
isNumber(-.123456); // true
isNumber('123'); // true
isNumber('123.456'); // true
isNumber('.123'); // true
isNumber(.123); // true
isNumber(Number.MAX_SAFE_INTEGER); // true
isNumber(Number.MAX_VALUE); // true
isNumber(Number.MIN_VALUE); // true
isNumber(new Number(123)); // true
*/
return (
(typeof x === 'number' || x instanceof Number || (!noStr && x && typeof x === 'string' && !isNaN(x))) &&
isFinite(x)
) || false;
};
Save yourself the headache of trying to find a "built-in" solution.
There isn't a good answer, and the hugely upvoted answer in this thread is wrong.
npm install is-number
In JavaScript, it's not always as straightforward as it should be to reliably check if a value is a number. It's common for devs to use +, -, or Number() to cast a string value to a number (for example, when values are returned from user input, regex matches, parsers, etc). But there are many non-intuitive edge cases that yield unexpected results:
console.log(+[]); //=> 0
console.log(+''); //=> 0
console.log(+' '); //=> 0
console.log(typeof NaN); //=> 'number'
Try the isNan function:
The isNaN() function determines whether a value is an illegal number (Not-a-Number).
This function returns true if the value equates to NaN. Otherwise it returns false.
This function is different from the Number specific Number.isNaN() method.
The global isNaN() function, converts the tested value to a Number, then tests it.
Number.isNan() does not convert the values to a Number, and will not return true for any value that is not of the type Number...
In my application we are only allowing a-z A-Z and 0-9 characters. I found the answer above using " string % 1 === 0" worked unless the string began with 0xnn (like 0x10) and then it would return it as numeric when we didn't want it to. The following simple trap in my numeric check seems to do the trick in our specific cases.
function isStringNumeric(str_input){
//concat a temporary 1 during the modulus to keep a beginning hex switch combination from messing us up
//very simple and as long as special characters (non a-z A-Z 0-9) are trapped it is fine
return '1'.concat(str_input) % 1 === 0;}
Warning : This might be exploiting a longstanding bug in Javascript and Actionscript [Number("1" + the_string) % 1 === 0)], I can't speak for that, but it is exactly what we needed.
parseInt(), but be aware that this function is a bit different in the sense that it for example returns 100 for parseInt("100px").
I like the simplicity of this.
Number.isNaN(Number(value))
The above is regular Javascript, but I'm using this in conjunction with a typescript typeguard for smart type checking. This is very useful for the typescript compiler to give you correct intellisense, and no type errors.
Typescript typeguards
isNotNumber(value: string | number): value is string {
return Number.isNaN(Number(this.smartImageWidth));
}
isNumber(value: string | number): value is number {
return Number.isNaN(Number(this.smartImageWidth)) === false;
}
Let's say you have a property width which is number | string. You may want to do logic based on whether or not it's a string.
var width: number|string;
width = "100vw";
if (isNotNumber(width))
{
// the compiler knows that width here must be a string
if (width.endsWith('vw'))
{
// we have a 'width' such as 100vw
}
}
else
{
// the compiler is smart and knows width here must be number
var doubleWidth = width * 2;
}
The typeguard is smart enough to constrain the type of width within the if statement to be ONLY string. This permits the compiler to allow width.endsWith(...) which it wouldn't allow if the type was string | number.
You can call the typeguard whatever you want isNotNumber, isNumber, isString, isNotString but I think isString is kind of ambiguous and harder to read.
parseInt(), but be aware that this function is a bit different in the sense that it for example returns 100 for parseInt("100px").
2019: Including ES3, ES6 and TypeScript Examples
Maybe this has been rehashed too many times, however I fought with this one today too and wanted to post my answer, as I didn't see any other answer that does it as simply or thoroughly:
ES3
var isNumeric = function(num){
return (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
}
ES6
const isNumeric = (num) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
Typescript
const isNumeric = (num: any) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num as number);
This seems quite simple and covers all the bases I saw on the many other posts and thought up myself:
// Positive Cases
console.log(0, isNumeric(0) === true);
console.log(1, isNumeric(1) === true);
console.log(1234567890, isNumeric(1234567890) === true);
console.log('1234567890', isNumeric('1234567890') === true);
console.log('0', isNumeric('0') === true);
console.log('1', isNumeric('1') === true);
console.log('1.1', isNumeric('1.1') === true);
console.log('-1', isNumeric('-1') === true);
console.log('-1.2354', isNumeric('-1.2354') === true);
console.log('-1234567890', isNumeric('-1234567890') === true);
console.log(-1, isNumeric(-1) === true);
console.log(-32.1, isNumeric(-32.1) === true);
console.log('0x1', isNumeric('0x1') === true); // Valid number in hex
// Negative Cases
console.log(true, isNumeric(true) === false);
console.log(false, isNumeric(false) === false);
console.log('1..1', isNumeric('1..1') === false);
console.log('1,1', isNumeric('1,1') === false);
console.log('-32.1.12', isNumeric('-32.1.12') === false);
console.log('[blank]', isNumeric('') === false);
console.log('[spaces]', isNumeric(' ') === false);
console.log('null', isNumeric(null) === false);
console.log('undefined', isNumeric(undefined) === false);
console.log([], isNumeric([]) === false);
console.log('NaN', isNumeric(NaN) === false);
You can also try your own isNumeric function and just past in these use cases and scan for "true" for all of them.
Or, to see the values that each return:
So, it will depend on the test cases that you want it to handle.
function isNumeric(number) {
return !isNaN(parseFloat(number)) && !isNaN(+number);
}
What I was looking for was regular types of numbers in javascript.
0, 1 , -1, 1.1 , -1.1 , 1E1 , -1E1 , 1e1 , -1e1, 0.1e10, -0.1.e10 , 0xAF1 , 0o172, Math.PI, Number.NEGATIVE_INFINITY, Number.POSITIVE_INFINITY
And also they're representations as strings:
'0', '1', '-1', '1.1', '-1.1', '1E1', '-1E1', '1e1', '-1e1', '0.1e10', '-0.1.e10', '0xAF1', '0o172'
I did want to leave out and not mark them as numeric
'', ' ', [], {}, null, undefined, NaN
As of today, all other answers seemed to failed one of these test cases.
function isNumberCandidate(s) {
const str = (''+ s).trim();
if (str.length === 0) return false;
return !isNaN(+str);
}
console.log(isNumberCandidate('1')); // true
console.log(isNumberCandidate('a')); // false
console.log(isNumberCandidate('000')); // true
console.log(isNumberCandidate('1a')); // false
console.log(isNumberCandidate('1e')); // false
console.log(isNumberCandidate('1e-1')); // true
console.log(isNumberCandidate('123.3')); // true
console.log(isNumberCandidate('')); // false
console.log(isNumberCandidate(' ')); // false
console.log(isNumberCandidate(1)); // true
console.log(isNumberCandidate(0)); // true
console.log(isNumberCandidate(NaN)); // false
console.log(isNumberCandidate(undefined)); // false
console.log(isNumberCandidate(null)); // false
console.log(isNumberCandidate(-1)); // true
console.log(isNumberCandidate('-1')); // true
console.log(isNumberCandidate('-1.2')); // true
console.log(isNumberCandidate(0.0000001)); // true
console.log(isNumberCandidate('0.0000001')); // true
console.log(isNumberCandidate(Infinity)); // true
console.log(isNumberCandidate(-Infinity)); // true
console.log(isNumberCandidate('Infinity')); // true
if (isNumberCandidate(s)) {
// use +s as a number
+s ...
}
If you really want to make sure that a string contains only a number, any number (integer or floating point), and exactly a number, you cannot use parseInt()/ parseFloat(), Number(), or !isNaN() by themselves. Note that !isNaN() is actually returning true when Number() would return a number, and false when it would return NaN, so I will exclude it from the rest of the discussion.
The problem with parseFloat() is that it will return a number if the string contains any number, even if the string doesn't contain only and exactly a number:
parseFloat("2016-12-31") // returns 2016
parseFloat("1-1") // return 1
parseFloat("1.2.3") // returns 1.2
The problem with Number() is that it will return a number in cases where the passed value is not a number at all!
Number("") // returns 0
Number(" ") // returns 0
Number(" \u00A0 \t\n\r") // returns 0
The problem with rolling your own regex is that unless you create the exact regex for matching a floating point number as Javascript recognizes it you are going to miss cases or recognize cases where you shouldn't. And even if you can roll your own regex, why? There are simpler built-in ways to do it.
However, it turns out that Number() (and isNaN()) does the right thing for every case where parseFloat() returns a number when it shouldn't, and vice versa. So to find out if a string is really exactly and only a number, call both functions and see if they both return true:
function isNumber(str) {
if (typeof str != "string") return false // we only process strings!
// could also coerce to string: str = ""+str
return !isNaN(str) && !isNaN(parseFloat(str))
}
I do it like this:
function isString(value)
{
return value.length !== undefined;
}
function isNumber(value)
{
return value.NaN !== undefined;
}
Of course isString() will be tripped up here if you pass some other object that has 'length' defined.
This is built on some of the previous answers and comments. The following covers all the edge cases and fairly concise as well:
const isNumRegEx = /^-?(\d*\.)?\d+$/;
function isNumeric(n, allowScientificNotation = false) {
return allowScientificNotation ?
!Number.isNaN(parseFloat(n)) && Number.isFinite(n) :
isNumRegEx.test(n);
}
I do it like this:
function isString(value)
{
return value.length !== undefined;
}
function isNumber(value)
{
return value.NaN !== undefined;
}
Of course isString() will be tripped up here if you pass some other object that has 'length' defined.
It is not valid for TypeScript as:
declare function isNaN(number: number): boolean;
For TypeScript you can use:
/^\d+$/.test(key)
Well, I'm using this one I made...
It's been working so far:
function checkNumber(value) {
return value % 1 == 0;
}
If you spot any problem with it, tell me, please.
The accepted answer for this question has quite a few flaws (as highlighted by couple of other users). This is one of the easiest & proven way to approach it in javascript:
function isNumeric(n) {
return !isNaN(parseFloat(n)) && isFinite(n);
}
Below are some good test cases:
console.log(isNumeric(12345678912345678912)); // true
console.log(isNumeric('2 ')); // true
console.log(isNumeric('-32.2 ')); // true
console.log(isNumeric(-32.2)); // true
console.log(isNumeric(undefined)); // false
// the accepted answer fails at these tests:
console.log(isNumeric('')); // false
console.log(isNumeric(null)); // false
console.log(isNumeric([])); // false
And you could go the RegExp-way:
var num = "987238";
if(num.match(/^-?\d+$/)){
//valid integer (positive or negative)
}else if(num.match(/^\d+\.\d+$/)){
//valid float
}else{
//not valid number
}
Here's a one-liner to check if sNum is a valid numeric value; it has been tested for a wide variety of inputs:
!isNaN(+s.replace(/\s|\$/g, '')); // returns True if numeric value
I used this function as a form validation tool, and I didn't want users to be able to write exponential function, so I came up with this function:
<script>
function isNumber(value, acceptScientificNotation) {
if(true !== acceptScientificNotation){
return /^-{0,1}\d+(\.\d+)?$/.test(value);
}
if (true === Array.isArray(value)) {
return false;
}
return !isNaN(parseInt(value, 10));
}
console.log(isNumber("")); // false
console.log(isNumber(false)); // false
console.log(isNumber(true)); // false
console.log(isNumber("0")); // true
console.log(isNumber("0.1")); // true
console.log(isNumber("12")); // true
console.log(isNumber("-12")); // true
console.log(isNumber(-45)); // true
console.log(isNumber({jo: "pi"})); // false
console.log(isNumber([])); // false
console.log(isNumber([78, 79])); // false
console.log(isNumber(NaN)); // false
console.log(isNumber(Infinity)); // false
console.log(isNumber(undefined)); // false
console.log(isNumber("0,1")); // false
console.log(isNumber("1e-1")); // false
console.log(isNumber("1e-1", true)); // true
</script>
Well, I'm using this one I made...
It's been working so far:
function checkNumber(value) {
return value % 1 == 0;
}
If you spot any problem with it, tell me, please.
parseInt(), but be aware that this function is a bit different in the sense that it for example returns 100 for parseInt("100px").
If you really want to make sure that a string contains only a number, any number (integer or floating point), and exactly a number, you cannot use parseInt()/ parseFloat(), Number(), or !isNaN() by themselves. Note that !isNaN() is actually returning true when Number() would return a number, and false when it would return NaN, so I will exclude it from the rest of the discussion.
The problem with parseFloat() is that it will return a number if the string contains any number, even if the string doesn't contain only and exactly a number:
parseFloat("2016-12-31") // returns 2016
parseFloat("1-1") // return 1
parseFloat("1.2.3") // returns 1.2
The problem with Number() is that it will return a number in cases where the passed value is not a number at all!
Number("") // returns 0
Number(" ") // returns 0
Number(" \u00A0 \t\n\r") // returns 0
The problem with rolling your own regex is that unless you create the exact regex for matching a floating point number as Javascript recognizes it you are going to miss cases or recognize cases where you shouldn't. And even if you can roll your own regex, why? There are simpler built-in ways to do it.
However, it turns out that Number() (and isNaN()) does the right thing for every case where parseFloat() returns a number when it shouldn't, and vice versa. So to find out if a string is really exactly and only a number, call both functions and see if they both return true:
function isNumber(str) {
if (typeof str != "string") return false // we only process strings!
// could also coerce to string: str = ""+str
return !isNaN(str) && !isNaN(parseFloat(str))
}
You could make use of types, like with the flow library, to get static, compile time checking. Of course not terribly useful for user input.
// @flow
function acceptsNumber(value: number) {
// ...
}
acceptsNumber(42); // Works!
acceptsNumber(3.14); // Works!
acceptsNumber(NaN); // Works!
acceptsNumber(Infinity); // Works!
acceptsNumber("foo"); // Error!
Quote:
isNaN(num) // returns true if the variable does NOT contain a valid number
is not entirely true if you need to check for leading/trailing spaces - for example when a certain quantity of digits is required, and you need to get, say, '1111' and not ' 111' or '111 ' for perhaps a PIN input.
Better to use:
var num = /^\d+$/.test(num)
Old question, but there are several points missing in the given answers.
Scientific notation.
!isNaN('1e+30') is true, however in most of the cases when people ask for numbers, they do not want to match things like 1e+30.
Large floating numbers may behave weird
Observe (using Node.js):
> var s = Array(16 + 1).join('9')
undefined
> s.length
16
> s
'9999999999999999'
> !isNaN(s)
true
> Number(s)
10000000000000000
> String(Number(s)) === s
false
>
On the other hand:
> var s = Array(16 + 1).join('1')
undefined
> String(Number(s)) === s
true
> var s = Array(15 + 1).join('9')
undefined
> String(Number(s)) === s
true
>
So, if one expects String(Number(s)) === s, then better limit your strings to 15 digits at most (after omitting leading zeros).
Infinity
> typeof Infinity
'number'
> !isNaN('Infinity')
true
> isFinite('Infinity')
false
>
Given all that, checking that the given string is a number satisfying all of the following:
- non scientific notation
- predictable conversion to
Numberand back toString - finite
is not such an easy task. Here is a simple version:
function isNonScientificNumberString(o) {
if (!o || typeof o !== 'string') {
// Should not be given anything but strings.
return false;
}
return o.length <= 15 && o.indexOf('e+') < 0 && o.indexOf('E+') < 0 && !isNaN(o) && isFinite(o);
}
However, even this one is far from complete. Leading zeros are not handled here, but they do screw the length test.
Why is jQuery's implementation not good enough?
function isNumeric(a) {
var b = a && a.toString();
return !$.isArray(a) && b - parseFloat(b) + 1 >= 0;
};
Michael suggested something like this (although I've stolen "user1691651 - John"'s altered version here):
function isNumeric(num){
num = "" + num; //coerce num to be a string
return !isNaN(num) && !isNaN(parseFloat(num));
}
The following is a solution with most likely bad performance, but solid results. It is a contraption made from the jQuery 1.12.4 implementation and Michael's answer, with an extra check for leading/trailing spaces (because Michael's version returns true for numerics with leading/trailing spaces):
function isNumeric(a) {
var str = a + "";
var b = a && a.toString();
return !$.isArray(a) && b - parseFloat(b) + 1 >= 0 &&
!/^\s+|\s+$/g.test(str) &&
!isNaN(str) && !isNaN(parseFloat(str));
};
The latter version has two new variables, though. One could get around one of those, by doing:
function isNumeric(a) {
if ($.isArray(a)) return false;
var b = a && a.toString();
a = a + "";
return b - parseFloat(b) + 1 >= 0 &&
!/^\s+|\s+$/g.test(a) &&
!isNaN(a) && !isNaN(parseFloat(a));
};
I haven't tested any of these very much, by other means than manually testing the few use-cases I'll be hitting with my current predicament, which is all very standard stuff. This is a "standing-on-the-shoulders-of-giants" situation.
In my application we are only allowing a-z A-Z and 0-9 characters. I found the answer above using " string % 1 === 0" worked unless the string began with 0xnn (like 0x10) and then it would return it as numeric when we didn't want it to. The following simple trap in my numeric check seems to do the trick in our specific cases.
function isStringNumeric(str_input){
//concat a temporary 1 during the modulus to keep a beginning hex switch combination from messing us up
//very simple and as long as special characters (non a-z A-Z 0-9) are trapped it is fine
return '1'.concat(str_input) % 1 === 0;}
Warning : This might be exploiting a longstanding bug in Javascript and Actionscript [Number("1" + the_string) % 1 === 0)], I can't speak for that, but it is exactly what we needed.
My solution:
// returns true for positive ints;
// no scientific notation, hexadecimals or floating point dots
var isPositiveInt = function(str) {
var result = true, chr;
for (var i = 0, n = str.length; i < n; i++) {
chr = str.charAt(i);
if ((chr < "0" || chr > "9") && chr != ",") { //not digit or thousands separator
result = false;
break;
};
if (i == 0 && (chr == "0" || chr == ",")) { //should not start with 0 or ,
result = false;
break;
};
};
return result;
};
You can add additional conditions inside the loop, to fit you particular needs.
You can use the result of Number when passing an argument to its constructor.
If the argument (a string) cannot be converted into a number, it returns NaN, so you can determinate if the string provided was a valid number or not.
Notes: Note when passing empty string or '\t\t' and '\n\t' as Number will return 0; Passing true will return 1 and false returns 0.
Number('34.00') // 34
Number('-34') // -34
Number('123e5') // 12300000
Number('123e-5') // 0.00123
Number('999999999999') // 999999999999
Number('9999999999999999') // 10000000000000000 (integer accuracy up to 15 digit)
Number('0xFF') // 255
Number('Infinity') // Infinity
Number('34px') // NaN
Number('xyz') // NaN
Number('true') // NaN
Number('false') // NaN
// cavets
Number(' ') // 0
Number('\t\t') // 0
Number('\n\t') // 0
If anyone ever gets this far down, I spent some time hacking on this trying to patch moment.js (https://github.com/moment/moment). Here's something that I took away from it:
function isNumeric(val) {
var _val = +val;
return (val !== val + 1) //infinity check
&& (_val === +val) //Cute coercion check
&& (typeof val !== 'object') //Array/object check
}
Handles the following cases:
True! :
isNumeric("1"))
isNumeric(1e10))
isNumeric(1E10))
isNumeric(+"6e4"))
isNumeric("1.2222"))
isNumeric("-1.2222"))
isNumeric("-1.222200000000000000"))
isNumeric("1.222200000000000000"))
isNumeric(1))
isNumeric(0))
isNumeric(-0))
isNumeric(1010010293029))
isNumeric(1.100393830000))
isNumeric(Math.LN2))
isNumeric(Math.PI))
isNumeric(5e10))
False! :
isNumeric(NaN))
isNumeric(Infinity))
isNumeric(-Infinity))
isNumeric())
isNumeric(undefined))
isNumeric('[1,2,3]'))
isNumeric({a:1,b:2}))
isNumeric(null))
isNumeric([1]))
isNumeric(new Date()))
Ironically, the one I am struggling with the most:
isNumeric(new Number(1)) => false
Any suggestions welcome. :]
You can use the result of Number when passing an argument to its constructor.
If the argument (a string) cannot be converted into a number, it returns NaN, so you can determinate if the string provided was a valid number or not.
Notes: Note when passing empty string or '\t\t' and '\n\t' as Number will return 0; Passing true will return 1 and false returns 0.
Number('34.00') // 34
Number('-34') // -34
Number('123e5') // 12300000
Number('123e-5') // 0.00123
Number('999999999999') // 999999999999
Number('9999999999999999') // 10000000000000000 (integer accuracy up to 15 digit)
Number('0xFF') // 255
Number('Infinity') // Infinity
Number('34px') // NaN
Number('xyz') // NaN
Number('true') // NaN
Number('false') // NaN
// cavets
Number(' ') // 0
Number('\t\t') // 0
Number('\n\t') // 0
Here is a high-performance (2.5*10^7 iterations/s @3.8GHz Haswell) version of a isNumber implementation. It works for every testcase i could find (including Symbols):
var isNumber = (function () {
var isIntegerTest = /^\d+$/;
var isDigitArray = [!0, !0, !0, !0, !0, !0, !0, !0, !0, !0];
function hasLeading0s (s) {
return !(typeof s !== 'string' ||
s.length < 2 ||
s[0] !== '0' ||
!isDigitArray[s[1]] ||
isIntegerTest.test(s));
}
var isWhiteSpaceTest = /\s/;
return function isNumber (s) {
var t = typeof s;
var n;
if (t === 'number') {
return (s <= 0) || (s > 0);
} else if (t === 'string') {
n = +s;
return !((!(n <= 0) && !(n > 0)) || n === '0' || hasLeading0s(s) || !(n !== 0 || !(s === '' || isWhiteSpaceTest.test(s))));
} else if (t === 'object') {
return !(!(s instanceof Number) || ((n = +s), !(n <= 0) && !(n > 0)));
}
return false;
};
})();
This appears to catch the seemingly infinite number of edge cases:
function isNumber(x, noStr) {
/*
- Returns true if x is either a finite number type or a string containing only a number
- If empty string supplied, fall back to explicit false
- Pass true for noStr to return false when typeof x is "string", off by default
isNumber(); // false
isNumber([]); // false
isNumber([1]); // false
isNumber([1,2]); // false
isNumber(''); // false
isNumber(null); // false
isNumber({}); // false
isNumber(true); // false
isNumber('true'); // false
isNumber('false'); // false
isNumber('123asdf'); // false
isNumber('123.asdf'); // false
isNumber(undefined); // false
isNumber(Number.POSITIVE_INFINITY); // false
isNumber(Number.NEGATIVE_INFINITY); // false
isNumber('Infinity'); // false
isNumber('-Infinity'); // false
isNumber(Number.NaN); // false
isNumber(new Date('December 17, 1995 03:24:00')); // false
isNumber(0); // true
isNumber('0'); // true
isNumber(123); // true
isNumber(123.456); // true
isNumber(-123.456); // true
isNumber(-.123456); // true
isNumber('123'); // true
isNumber('123.456'); // true
isNumber('.123'); // true
isNumber(.123); // true
isNumber(Number.MAX_SAFE_INTEGER); // true
isNumber(Number.MAX_VALUE); // true
isNumber(Number.MIN_VALUE); // true
isNumber(new Number(123)); // true
*/
return (
(typeof x === 'number' || x instanceof Number || (!noStr && x && typeof x === 'string' && !isNaN(x))) &&
isFinite(x)
) || false;
};
My attempt at a slightly confusing, Pherhaps not the best solution
function isInt(a){
return a === ""+~~a
}
console.log(isInt('abcd')); // false
console.log(isInt('123a')); // false
console.log(isInt('1')); // true
console.log(isInt('0')); // true
console.log(isInt('-0')); // false
console.log(isInt('01')); // false
console.log(isInt('10')); // true
console.log(isInt('-1234567890')); // true
console.log(isInt(1234)); // false
console.log(isInt('123.4')); // false
console.log(isInt('')); // false
// other types then string returns false
console.log(isInt(5)); // false
console.log(isInt(undefined)); // false
console.log(isInt(null)); // false
console.log(isInt('0x1')); // false
console.log(isInt(Infinity)); // false
2019: Including ES3, ES6 and TypeScript Examples
Maybe this has been rehashed too many times, however I fought with this one today too and wanted to post my answer, as I didn't see any other answer that does it as simply or thoroughly:
ES3
var isNumeric = function(num){
return (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
}
ES6
const isNumeric = (num) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num);
Typescript
const isNumeric = (num: any) => (typeof(num) === 'number' || typeof(num) === "string" && num.trim() !== '') && !isNaN(num as number);
This seems quite simple and covers all the bases I saw on the many other posts and thought up myself:
// Positive Cases
console.log(0, isNumeric(0) === true);
console.log(1, isNumeric(1) === true);
console.log(1234567890, isNumeric(1234567890) === true);
console.log('1234567890', isNumeric('1234567890') === true);
console.log('0', isNumeric('0') === true);
console.log('1', isNumeric('1') === true);
console.log('1.1', isNumeric('1.1') === true);
console.log('-1', isNumeric('-1') === true);
console.log('-1.2354', isNumeric('-1.2354') === true);
console.log('-1234567890', isNumeric('-1234567890') === true);
console.log(-1, isNumeric(-1) === true);
console.log(-32.1, isNumeric(-32.1) === true);
console.log('0x1', isNumeric('0x1') === true); // Valid number in hex
// Negative Cases
console.log(true, isNumeric(true) === false);
console.log(false, isNumeric(false) === false);
console.log('1..1', isNumeric('1..1') === false);
console.log('1,1', isNumeric('1,1') === false);
console.log('-32.1.12', isNumeric('-32.1.12') === false);
console.log('[blank]', isNumeric('') === false);
console.log('[spaces]', isNumeric(' ') === false);
console.log('null', isNumeric(null) === false);
console.log('undefined', isNumeric(undefined) === false);
console.log([], isNumeric([]) === false);
console.log('NaN', isNumeric(NaN) === false);
You can also try your own isNumeric function and just past in these use cases and scan for "true" for all of them.
Or, to see the values that each return:
2019: Practical and tight numerical validity check
Often, a 'valid number' means a Javascript number excluding NaN and Infinity, ie a 'finite number'.
To check the numerical validity of a value (from an external source for example), you can define in ESlint Airbnb style :
/**
* Returns true if 'candidate' is a finite number or a string referring (not just 'including') a finite number
* To keep in mind:
* Number(true) = 1
* Number('') = 0
* Number(" 10 ") = 10
* !isNaN(true) = true
* parseFloat('10 a') = 10
*
* @param {?} candidate
* @return {boolean}
*/
function isReferringFiniteNumber(candidate) {
if (typeof (candidate) === 'number') return Number.isFinite(candidate);
if (typeof (candidate) === 'string') {
return (candidate.trim() !== '') && Number.isFinite(Number(candidate));
}
return false;
}
and use it this way:
if (isReferringFiniteNumber(theirValue)) {
myCheckedValue = Number(theirValue);
} else {
console.warn('The provided value doesn\'t refer to a finite number');
}
I recently wrote an article about ways to ensure a variable is a valid number: https://github.com/jehugaleahsa/artifacts/blob/master/2018/typescript_num_hack.md The article explains how to ensure floating point or integer, if that's important (+x vs ~~x).
The article assumes the variable is a string or a number to begin with and trim is available/polyfilled. It wouldn't be hard to extend it to handle other types, as well. Here's the meat of it:
// Check for a valid float
if (x == null
|| ("" + x).trim() === ""
|| isNaN(+x)) {
return false; // not a float
}
// Check for a valid integer
if (x == null
|| ("" + x).trim() === ""
|| ~~x !== +x) {
return false; // not an integer
}
I'm using the following:
const isNumber = s => !isNaN(+s)
PFB the working solution:
function(check){
check = check + "";
var isNumber = check.trim().length>0? !isNaN(check):false;
return isNumber;
}
Why is jQuery's implementation not good enough?
function isNumeric(a) {
var b = a && a.toString();
return !$.isArray(a) && b - parseFloat(b) + 1 >= 0;
};
Michael suggested something like this (although I've stolen "user1691651 - John"'s altered version here):
function isNumeric(num){
num = "" + num; //coerce num to be a string
return !isNaN(num) && !isNaN(parseFloat(num));
}
The following is a solution with most likely bad performance, but solid results. It is a contraption made from the jQuery 1.12.4 implementation and Michael's answer, with an extra check for leading/trailing spaces (because Michael's version returns true for numerics with leading/trailing spaces):
function isNumeric(a) {
var str = a + "";
var b = a && a.toString();
return !$.isArray(a) && b - parseFloat(b) + 1 >= 0 &&
!/^\s+|\s+$/g.test(str) &&
!isNaN(str) && !isNaN(parseFloat(str));
};
The latter version has two new variables, though. One could get around one of those, by doing:
function isNumeric(a) {
if ($.isArray(a)) return false;
var b = a && a.toString();
a = a + "";
return b - parseFloat(b) + 1 >= 0 &&
!/^\s+|\s+$/g.test(a) &&
!isNaN(a) && !isNaN(parseFloat(a));
};
I haven't tested any of these very much, by other means than manually testing the few use-cases I'll be hitting with my current predicament, which is all very standard stuff. This is a "standing-on-the-shoulders-of-giants" situation.
I have tested and Michael's solution is best. Vote for his answer above (search this page for "If you really want to make sure that a string" to find it). In essence, his answer is this:
function isNumeric(num){
num = "" + num; //coerce num to be a string
return !isNaN(num) && !isNaN(parseFloat(num));
}
It works for every test case, which I documented here: https://jsfiddle.net/wggehvp9/5/
Many of the other solutions fail for these edge cases: ' ', null, "", true, and []. In theory, you could use them, with proper error handling, for example:
return !isNaN(num);
or
return (+num === +num);
with special handling for /\s/, null, "", true, false, [] (and others?)
Source: Stackoverflow.com