I use this code to search and count vowels in the string,

a = "run forest, run";
a = a.split(" ");
var syl = 0;
for (var i = 0; i < a.length - 1; i++) {
    for (var i2 = 0; i2 < a[i].length - 1; i2++) {
        if ('aouie'.search(a[i][i2]) > -1) {
            syl++;
        }
    }
}

alert(syl + " vowels")

Obviously it should alert up 4 vowels, but it returns 3. What's wrong and how you can simplify it?

I use this code to search and count vowels in the string,

a = "run forest, run";
a = a.split(" ");
var syl = 0;
for (var i = 0; i < a.length - 1; i++) {
    for (var i2 = 0; i2 < a[i].length - 1; i2++) {
        if ('aouie'.search(a[i][i2]) > -1) {
            syl++;
        }
    }
}

alert(syl + " vowels")

Obviously it should alert up 4 vowels, but it returns 3. What's wrong and how you can simplify it?

• javascript
• string

• I'm not native English speaking, so I might be wrong, but isn't y sometimes considered a vowel as well? – Christofer Eliasson Commented Feb 16, 2013 at 17:20
• Apparently y can be both an vowel and a consonant in English, depending on its use, which sounds like a tough nut to crack with a regular expression. More about it here. – Christofer Eliasson Commented Feb 16, 2013 at 17:23

6 Answers 6

Try this:

var syl = ("|"+a+"|").split(/[aeiou]/i).length-1;

The | ensures there are no edge cases, such as having a vowel at the start or end of the string.

Regarding your code, your if condition needs no i2

if('aouie'.search(a[i]) > -1){ 

I wonder, why all that use of arrays and nested loops, the below regex could do it better,

var str = "run forest, run";
var matches = str.match(/[aeiou]/gi);
    var count = matches ? matches.length : 0;
    alert(count + " vowel(s)");

Demo

Try:

a = "run forest, run";

var syl = 0;

for(var i=0; i<a.length; i++) {
        if('aouie'.search(a[i]) > -1){ 
            syl++;
        }
}

alert(syl+" vowels")

First, the split is useless since you can already cycle through every character.

Second: you need to use i<a.length, this gets the last character in the string, too.

The simplest way is

s.match(/[aeiou]/gi).length

You can use the .match to pare a string to a regular expression. g is global which will run through the entire string. i makes the string readable as upper and lower case.

function getVowels(str) {
      var m = str.match(/[aeiou]/gi);
      return m === null ? 0 : m.length;
    }

A different approach,

const vowels = (str) => [...str].filter((ch) => ["a", "e", "i", "o", "u"].includes(ch)).length;
console.log(vowels("aaaa"));