How do you easily check whether a Variable is number or not in JavaScript?

Question

I am not a JavaScript developer, but I just want to know the right way to check a variable contain a number or not. After a couple analysis I reached below solution. Is it correct way?

 function calculation(n1,n2 , ...numbers) { let validateNumber = (num) => { if(num !== num) return 0; else return typeof num === 'number' ? num : 0; } let sum =0; for(n of numbers){ sum += validateNumber(n); } console.log(sum); } calculation(5,6,7,'gg','',null, NaN, undefined, null,8,9,5.4,10);

Please check the 'validateNumber' arrow function.

Answer 1

First version =>
-- simply use isNaN() : sum+(isNaN(val)?0:Number(val)

Second version (asked in comment here) =>
-- just use a strict comparison. to get only numeric type values.

 const calculation =(...numbers)=> // (n1,n2 , ...numbers)=> numbers.reduce((sum,val)=> sum + (Number(val)===val ? val : 0) // sum+(isNaN(val)?0:Number(val)) , 0); console.log( calculation(5,6,7,'gg','',null, NaN, undefined, null,8,9,5.4,'10') )

For more clarity, here it is a table showing the execution of the test according to the different possible types

 const testFunction = val => Number(val)===val , values = [ { v: 123, display: `123`, expected: true } , { v: '123', display: `"123"`, expected: false } , { v: NaN, display: `NaN`, expected: false } , { v: undefined, display: `undefined`, expected: false } , { v: null, display: `null`, expected: false } ] , tBody = document.querySelector('table > tbody') ; let score = 0 ; for (let {v, display, expected} of values) { let TR = tBody.insertRow() , res = testFunction(v) ; score += expected===res ? 1 : 0 TR.insertCell().textContent = display TR.insertCell().textContent = expected TR.insertCell().textContent = res TR.insertCell().className = expected===res ? 'ok' : 'bad' } document.querySelector('table >tfoot td:nth-of-type(2)') .textContent = `${score} / ${values.length}`

 table { font-family: Arial, Helvetica, sans-serif;border-collapse: collapse;margin: 0 1em; } td { padding: .2em .8em;border: 1px solid darkblue; } thead,tfoot { background-color: #84a4ce;text-transform: capitalize; } caption { padding: .3em;caption-side: bottom;font-size: .8em; } .ok:before,.bad:before { font-weight: bold;font-size: 1.2em; } .ok:before { color: green;content: '\\2713'; } .bad:before { color: red; content: '\\2718'; }

 <table> <caption>( Number(val)===val ) ?</caption> <thead> <tr> <td>val</td><td>expected</td><td colspan="2">result</td> </tr> </thead> <tbody></tbody> <tfoot> <tr> <td colspan="3">score</td><td>0/0</td> </tr> </tfoot> </table>

Answer 2

Yes, there is built in function in JavaScript to check if a certain value is number or not which is isNaN()

If you want to get the sum of all the numbers in the array then you can try using reduce() like the following way:

 var data = [5,6,7,'gg','',null, NaN, undefined, null,8,9,5.4,10]; var sum = data.reduce((a,c) =>{ if(!isNaN(c)){ a = a + Number(c); //convert the '' and null to 0 with Number() } return a; },0); console.log(sum)

Answer 3

Yes it is a one liner also

export const isNum = n => Number(n) === Number(n);

First parse to number what Eva param you get them compare to itself. One of them what the hell's that comes with JavaScript is that NaN !== NaN so if the value you get after forcing it to be a number become sa Nan it will be different from itself

Answer 4

You can filter out the elements that aren't numbers by checking to see if they're not null and also a number, and then using reduce to sum up the elements.

 function calculation(n1, n2, ...elements) { return elements // `filter` out the elements that are numbers .filter(el => el && Number(el)) // Add them all up .reduce((acc, c) => acc + c, 0); } console.log(calculation(5,6,7,'gg','',null, NaN, undefined, null,8,9,5.4,10));