Javascript: Counting number of vowels in a string - Stack Overflow

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I am trying to count the number of vowels in a string, but my counter does not seem to be returning more than one. Can someone please tell me what is wrong with my code? Thanks!

var vowelCount = function(str){
  var count = 0;
  for(var i = 0; i < str.length; i++){
    if(str[i] == 'a' || str[i] == 'i' || str[i] == 'o' ||str[i] == 'e' ||str[i] == 'u'){
      count+=1;
    }
  }
  return count;
}
console.log(vowelCount('aide'));

I am trying to count the number of vowels in a string, but my counter does not seem to be returning more than one. Can someone please tell me what is wrong with my code? Thanks!

var vowelCount = function(str){
  var count = 0;
  for(var i = 0; i < str.length; i++){
    if(str[i] == 'a' || str[i] == 'i' || str[i] == 'o' ||str[i] == 'e' ||str[i] == 'u'){
      count+=1;
    }
  }
  return count;
}
console.log(vowelCount('aide'));

• javascript

Share Improve this question Follow edited Jun 20, 2018 at 15:16 giodamelio 5,6051414 gold badges4646 silver badges7272 bronze badges asked Feb 23, 2017 at 5:10 Colin SygielColin Sygiel 94566 gold badges1919 silver badges3131 bronze badges 7

• 9 return count; move out to for loop – Jaydip Jadhav Commented Feb 23, 2017 at 5:12
• 3 try string.match(/[aeiou]/g).length – Rajesh Commented Feb 23, 2017 at 5:13
• 3 @Rajesh I'd use /[aeiou]/ig – Phil Commented Feb 23, 2017 at 5:14
• 1 @JaydipJ That makes sense, thanks! – Colin Sygiel Commented Feb 23, 2017 at 5:15
• 3 Possible duplicate of Counting number of vowels in a string with JavaScript – Rajesh Commented Feb 23, 2017 at 5:18

 |  Show 2 more ments

2 Answers 2

Sorted by: Reset to default 7

return count outside of for loop, or use RegExp /[^aeiou]/ig as first parameter to .replace() with "" as replacement string, get .legnth of string returned by .replace()

vowelLength = "aide".replace(/[^aeiou]/ig, "").length;

console.log(vowelLength);

vowelLength = "gggg".replace(/[^aeiou]/ig, "").length;

console.log(vowelLength);

RegExp description

Character set

[^xyz] A negated or plemented character set. That is, it matches anything that is not enclosed in the brackets.

Flags

i ignore case

g global match; find all matches rather than stopping after the first match

---

Using spread element, Array.prototype.reduce(), String.prototype.indexOf() or String.prototype.contains() where supported

const v = "aeiouAEIOU";

var vowelLength = [..."aide"].reduce((n, c) => v.indexOf(c) > -1 ? ++n : n, 0);

console.log(vowelLength);

var vowelLength = [..."gggg"].reduce((n, c) => v.indexOf(c) > -1 ? ++n : n, 0);

console.log(vowelLength);

---

Alternatively, instead of creating a new string or new array to get .length property or iterate characters of string, you can use for..of loop, RegExp.prototype.test with RegExp /[aeiou]/i to increment a variable initially set to 0 if .test() evaluates to true for the character passed.

var [re, vowelLength] = [/[aeiou]/i, 0]; 

for (let c of "aide") re.test(c) && ++vowelLength;

console.log(vowelLength); 

vowelLength = 0;

for (let c of "gggg") re.test(c) && ++vowelLength;

console.log(vowelLength); 

You need to also do this. use toLowerCase() also

 var vowelCount = function(str){
  var count = 0;
  for(var i = 0; i < str.length; i++){
    if(str[i].toLowerCase() == 'a' || str[i].toLowerCase() == 'i' || str[i].toLowerCase() == 'o' ||str[i].toLowerCase() == 'e' ||str[i].toLowerCase() == 'u'){
      count+=1;
    }
  }
 return count;
}
vowelCount('aide')

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