Filter array of strings, keeping only ones starting with vowels

Question

I realise I've massively overengineered this, but as I'm just starting out with JS, I can't think of how to condense this into something not entirely ridiculous. I know I'm probably going to kick myself here, but can someone refactor this for me?

The aim was to create a new array from a provided one, one that only contained strings starting with vowels. It also needed to be case insensitive.

let results = []

for (let i = 0; i < strings.length; i++) {
  if ((strings[i].startsWith('a')) || (strings[i].startsWith('A')) || (strings[i].startsWith('e')) || (strings[i].startsWith('E')) || (strings[i].startsWith('i')) || (strings[i].startsWith('I')) || (strings[i].startsWith('o')) || (strings[i].startsWith('O')) || (strings[i].startsWith('u')) || (strings[i].startsWith('U'))) {
    results.push(strings[i])
  }
}

return results

Answer 1

You can use a single RegExp and Array.prototype.filter() for that:

console.log([
  'Foo',
  'Bar',
  'Abc',
  'Lorem',
  'Ipsum'
].filter(str => /^[aeiou]/i.test(str)));

Array.prototype.filter() returns a new array with all the elements that pass (return a truthy value) the predicate.

RegExp.prototype.test() returns true if the RegExp finds a match on the string you pass in.

Then, /^[aeiou]/i means:

• ^ matches the start of the string.
• [aeiou] matches any of the characters inside the square brackets, a single time.
• i is a case-insensitive modifier.