returning strings instead of numbers

Question

I'm trying to write a function (onlyOddNumbers) that accepts an array of numbers and returns a new array with only the odd numbers, it's working but the problem is I'm getting strings in the new array instead of numbers, why this happen ?

let oddNumbersOnly=[]
const filter = function (numbers) {
  for (number in numbers){
    if(number %2 !==0){
      oddNumbersOnly.push(number)
    }
  } return oddNumbersOnly;
};

Maybe use this `oddNumbersOnly.push(+number)` ? – Hassan Imam Aug 23 '20 at 12:28 — Hassan Imam, Aug 23 '20 at 12:28

score 1 · Accepted Answer · answered Aug 23 '20 at 12:12

1

use for of instead of for in then convert your num to string

const filter = function(numbers) {
  let oddNumbersOnly = []
  for (let number of numbers) {
    if (number % 2 !== 0) {
      oddNumbersOnly.push(number.toString())
    }
  }
  return oddNumbersOnly;
};
const arr = [1, 2, 3, 4, 5, 6];
const result = filter(arr)
console.log(result)

answered Aug 23 '20 at 12:12

Eugen Sunic

12,218
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1

It seems, @AlaaKhalila wants to get numbers: "...the problem is I'm getting strings in the new array instead of numbers". – Alexander Aug 23 '20 at 12:20
problem solved , I just used ( for of ) instead of ( for in ) – Alaa Khalila Aug 23 '20 at 12:32

score 0 · Answer 2 · answered Aug 23 '20 at 12:33

0

let num = [0, 1, 2, 3, 4, 5, 6];
let oddNumbersOnly = num.filter(numb=>numb%2 !==0);
console.log(oddNumbersOnly);

javaScript has an inbuilt es6 filter function. It is shorter and retain the datatype. You may try this.

answered Aug 23 '20 at 12:33

atomty

187
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returning strings instead of numbers

2 Answers2