How to count the vowels in a string and return an integer(with consecutive vowels counted twice!)?

Question

So I came up with this, though the counter is not working properly.

const stringVowel = ('hottentottententententoonstellingsbedrijfsacademie')

const letters = stringVowel.split('');

const numberOfVowels = arr => arr.map(num => {
    let counter = 0;
    if (num === 'a' || num === 'o' || num === 'u' || num === 'e' || num === 'i') {
    counter++
    }
    console.log(counter)
})

numberOfVowels(letters)

You're redeclaring `counter` on every iteration, so it will never be more than 1. You'll need to declare it outside the `map` call, but you should be using `forEach` and not `map` as it is since you're not using the returned array. see: [JavaScript: Difference between .forEach() and .map()](https://stackoverflow.com/questions/34426458/javascript-difference-between-foreach-and-map) — pilchard, Nov 08 '21 at 21:42

Nick · Answer 1 · 2021-11-08T21:47:02.667

0

One option is to use the string match method and provide a regular expression with all the vowels:

const withVowels = 'hottentottententententoonstellingsbedrijfsacademie';

const noVowels = 'wqdftzsqq';

function numberOfVowels(str) {
  return str.match(/[aeiou]/ig)?.length ?? 0;
}

console.log(numberOfVowels(withVowels));

console.log(numberOfVowels(noVowels));

Edit: Updated to a more robust solution based on comments

edited Nov 08 '21 at 21:47

answered Nov 08 '21 at 21:42

Nick

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Note that this will get an error if there are no vowels, because `match()` returns `null`. – Barmar Nov 08 '21 at 21:44
Consider `/[aeiou]/i`, and don't forget to check for null `?.length ?? 0` – CertainPerformance Nov 08 '21 at 21:44

How to count the vowels in a string and return an integer(with consecutive vowels counted twice!)?

1 Answers1