Check if character is number?
JavascriptJavascript Problem Overview
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' is a number. How can this be done?
Javascript Solutions
Solution 1 - Javascript
You could use comparison operators to see if it is in the range of digit characters:
var c = justPrices[i].substr(commapos+2,1);
if (c >= '0' && c <= '9') {
// it is a number
} else {
// it isn't
}
Solution 2 - Javascript
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);
}
Solution 3 - Javascript
EDIT: Blender's updated answer is the right answer here if you're just checking a single character (namely !isNaN(parseInt(c, 10))). My answer below is a good solution if you want to test whole strings.
Here is jQuery's isNumeric implementation (in pure JavaScript), which works against full strings:
function isNumeric(s) {
return !isNaN(s - parseFloat(s));
}
The comment for this function reads:
> // parseFloat NaNs numeric-cast false positives (null|true|false|"")
> // ...but misinterprets leading-number strings, particularly hex literals ("0x...")
> // subtraction forces infinities to NaN
I think we can trust that these chaps have spent quite a bit of time on this!
Solution 4 - Javascript
Simple function
function isCharNumber(c) {
return c >= '0' && c <= '9';
}
Solution 5 - Javascript
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
}
Solution 6 - Javascript
You can use this:
function isDigit(n) {
return Boolean([true, true, true, true, true, true, true, true, true, true][n]);
}
Here, I compared it to the accepted method: http://jsperf.com/isdigittest/5 . I didn't expect much, so I was pretty suprised, when I found out that accepted method was much slower.
Interesting thing is, that while accepted method is faster correct input (eg. '5') and slower for incorrect (eg. 'a'), my method is exact opposite (fast for incorrect and slower for correct).
Still, in worst case, my method is 2 times faster than accepted solution for correct input and over 5 times faster for incorrect input.
Solution 7 - Javascript
I think it's very fun to come up with ways to solve this. Below are some.
(All functions below assume argument is a single character. Change to n[0] to enforce it)
##Method 1:
function isCharDigit(n){
return !!n.trim() && n > -1;
}
##Method 2:
function isCharDigit(n){
return !!n.trim() && n*0==0;
}
##Method 3: function isCharDigit(n){ return !!n.trim() && !!Number(n+.1); // "+.1' to make it work with "." and "0" Chars }
##Method 4:
var isCharDigit = (function(){
var a = [1,1,1,1,1,1,1,1,1,1];
return function(n){
return !!a[n] // check if a Array has anything in index 'n'. Cast result to boolean
}
})();
##Method 5: function isCharDigit(n){ return !!n.trim() && !isNaN(+n); }
Test string:
var str = ' 90ABcd#?:.+', char;
for( char of str )
console.log( char, isCharDigit(char) );
Solution 8 - Javascript
I suggest a simple regex.
If you're looking for just the last character in the string:
/^.*?[0-9]$/.test("blabla,120"); // true
/^.*?[0-9]$/.test("blabla,120a"); // false
/^.*?[0-9]$/.test("120"); // true
/^.*?[0-9]$/.test(120); // true
/^.*?[0-9]$/.test(undefined); // false
/^.*?[0-9]$/.test(-1); // true
/^.*?[0-9]$/.test("-1"); // true
/^.*?[0-9]$/.test(false); // false
/^.*?[0-9]$/.test(true); // false
And the regex is even simpler if you are just checking a single char as an input:
var char = "0";
/^[0-9]$/.test(char); // true
Solution 9 - Javascript
The shortest solution is:
const isCharDigit = n => n < 10;
You can apply these as well:
const isCharDigit = n => Boolean(++n);
const isCharDigit = n => '/' < n && n < ':';
const isCharDigit = n => !!++n;
if you want to check more than 1 chatacter, you might use next variants
Regular Expression:
const isDigit = n => /\d+/.test(n);
Comparison:
const isDigit = n => +n == n;
Check if it is not NaN
const isDigit = n => !isNaN(n);
Solution 10 - Javascript
If you are testing single characters, then:
var isDigit = (function() {
var re = /^\d$/;
return function(c) {
return re.test(c);
}
}());
will return true or false depending on whether c is a digit or not.
Solution 11 - Javascript
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;
}
})()
}
};
Solution 12 - Javascript
Similar to one of the answers above, I used
var sum = 0; //some value
let num = parseInt(val); //or just Number.parseInt
if(!isNaN(num)) {
sum += num;
}
This blogpost sheds some more light on this check if a string is numeric in Javascript | Typescript & ES6
Solution 13 - Javascript
Here is a simple function that does it.
function is_number(char) {{
return !isNaN(parseInt(char));
}}
Returns: true, false
Solution 14 - Javascript
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
Solution 15 - Javascript
Try:
function is_numeric(str){
try {
return isFinite(str)
}
catch(err) {
return false
}
}
Solution 16 - Javascript
A simple solution by leveraging language's dynamic type checking:
function isNumber (string) {
//it has whitespace
if(string.trim() === ''){
return false
}
return string - 0 === string * 1
}
see test cases below
function isNumber (str) {
if(str.trim() === ''){
return false
}
return str - 0 === str * 1
}
console.log('-1' + ' → ' + isNumber ('-1'))
console.log('-1.5' + ' → ' + isNumber ('-1.5'))
console.log('0' + ' → ' + isNumber ('0'))
console.log(', ,' + ' → ' + isNumber (', ,'))
console.log('0.42' + ' → ' + isNumber ('0.42'))
console.log('.42' + ' → ' + isNumber ('.42'))
console.log('#abcdef' + ' → ' + isNumber ('#abcdef'))
console.log('1.2.3' + ' → ' + isNumber ('1.2.3'))
console.log('' + ' → ' + isNumber (''))
console.log('blah' + ' → ' + isNumber ('blah'))
Solution 17 - Javascript
Use combination of isNaN and parseInt functions:
var character = ... ; // your character
var isDigit = ! isNaN( parseInt(character) );
Another notable way - multiplication by one (like character * 1 instead of parseInt(character)) - makes a number not only from any numeric string, but also a 0 from empty string and a string containing only spaces so it is not suitable here.
Solution 18 - Javascript
I am using this:
const isNumber = (str) => (
str.length === str.trim().length
&& str.length > 0
&& Number(str) >= 0
)
It works for strings or single characters.
Solution 19 - Javascript
modifying this answer to be little more convenient and limiting to chars(and not string):
const charCodeZero = "0".charCodeAt(0);
const charCodeNine = "9".charCodeAt(0);
function isDigit(s:string) {
return s.length==1&& s.charCodeAt(0) >= charCodeZero && s.charCodeAt(0) <= charCodeNine;
}
console.log(isDigit('4')) //true
console.log(isDigit('4s')) //false
console.log(isDigit('s')) //false
Solution 20 - Javascript
Use simple regex solution to check only one char:
function isDigit(chr) {
return chr.match(/[0-9]/i);
}
Solution 21 - Javascript
function is_numeric(mixed_var) {
return (typeof(mixed_var) === 'number' || typeof(mixed_var) === 'string') &&
mixed_var !== '' && !isNaN(mixed_var);
}
Solution 22 - Javascript
square = function(a) {
if ((a * 0) == 0) {
return a*a;
} else {
return "Enter a valid number.";
}
}
Solution 23 - Javascript
You can try this (worked in my case)
If you want to test if the first char of a string is an int:
if (parseInt(YOUR_STRING.slice(0, 1))) {
alert("first char is int")
} else {
alert("first char is not int")
}
If you want to test if the char is a int:
if (parseInt(YOUR_CHAR)) {
alert("first char is int")
} else {
alert("first char is not int")
}
Solution 24 - Javascript
This seems to work:
Static binding:
String.isNumeric = function (value) {
return !isNaN(String(value) * 1);
};
Prototype binding:
String.prototype.isNumeric = function () {
return !isNaN(this.valueOf() * 1);
};
It will check single characters, as well as whole strings to see if they are numeric.
Solution 25 - Javascript
This function works for all test cases that i could find. It's also faster than:
function isNumeric (n) {
if (!isNaN(parseFloat(n)) && isFinite(n) && !hasLeading0s(n)) {
return true;
}
var _n = +n;
return _n === Infinity || _n === -Infinity;
}
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/;
function fIsNaN(n) {
return !(n <= 0) && !(n > 0);
}
function isNumber(s) {
var t = typeof s;
if (t === 'number') {
return (s <= 0) || (s > 0);
} else if (t === 'string') {
var n = +s;
return !(fIsNaN(n) || hasLeading0s(s) || !(n !== 0 || !(s === '' || isWhiteSpaceTest.test(s))));
} else if (t === 'object') {
return !(!(s instanceof Number) || fIsNaN(+s));
}
return false;
}
function testRunner(IsNumeric) {
var total = 0;
var passed = 0;
var failedTests = [];
function test(value, result) {
total++;
if (IsNumeric(value) === result) {
passed++;
} else {
failedTests.push({
value: value,
expected: result
});
}
}
// true
test(0, true);
test(1, true);
test(-1, true);
test(Infinity, true);
test('Infinity', true);
test(-Infinity, true);
test('-Infinity', true);
test(1.1, true);
test(-0.12e-34, true);
test(8e5, true);
test('1', true);
test('0', true);
test('-1', true);
test('1.1', true);
test('11.112', true);
test('.1', true);
test('.12e34', true);
test('-.12e34', true);
test('.12e-34', true);
test('-.12e-34', true);
test('8e5', true);
test('0x89f', true);
test('00', true);
test('01', true);
test('10', true);
test('0e1', true);
test('0e01', true);
test('.0', true);
test('0.', true);
test('.0e1', true);
test('0.e1', true);
test('0.e00', true);
test('0xf', true);
test('0Xf', true);
test(Date.now(), true);
test(new Number(0), true);
test(new Number(1e3), true);
test(new Number(0.1234), true);
test(new Number(Infinity), true);
test(new Number(-Infinity), true);
// false
test('', false);
test(' ', false);
test(false, false);
test('false', false);
test(true, false);
test('true', false);
test('99,999', false);
test('#abcdef', false);
test('1.2.3', false);
test('blah', false);
test('\t\t', false);
test('\n\r', false);
test('\r', false);
test(NaN, false);
test('NaN', false);
test(null, false);
test('null', false);
test(new Date(), false);
test({}, false);
test([], false);
test(new Int8Array(), false);
test(new Uint8Array(), false);
test(new Uint8ClampedArray(), false);
test(new Int16Array(), false);
test(new Uint16Array(), false);
test(new Int32Array(), false);
test(new Uint32Array(), false);
test(new BigInt64Array(), false);
test(new BigUint64Array(), false);
test(new Float32Array(), false);
test(new Float64Array(), false);
test('.e0', false);
test('.', false);
test('00e1', false);
test('01e1', false);
test('00.0', false);
test('01.05', false);
test('00x0', false);
test(new Number(NaN), false);
test(new Number('abc'), false);
console.log('Passed ' + passed + ' of ' + total + ' tests.');
if (failedTests.length > 0) console.log({
failedTests: failedTests
});
}
testRunner(isNumber)
Solution 26 - Javascript
Just use isFinite
const number = "1";
if (isFinite(number)) {
// do something
}