使用JavaScript计算字符串中元音的数量 [英] Counting number of vowels in a string with JavaScript

问题描述

我正在使用基本的JavaScript来计算字符串中元音的数量。下面的代码工作，但我想清理一下。考虑到它是一个字符串，会使用 .includes（）帮助吗？我想使用像 string.includes（a，e，i，o，u）这样的东西，如果可能的话条件陈述。此外，是否需要将输入转换为字符串？

I'm using basic JavaScript to count the number of vowels in a string. The below code works but I would like to have it cleaned up a bit. Would using .includes() help at all considering it is a string? I would like to use something like string.includes("a", "e", "i", "o", "u") if at all possible to clean up the conditional statement. Also, is it needed to convert the input into a string?

function getVowels(str) {
  var vowelsCount = 0;

  //turn the input into a string
  var string = str.toString();

  //loop through the string
  for (var i = 0; i <= string.length - 1; i++) {

  //if a vowel, add to vowel count
    if (string.charAt(i) == "a" || string.charAt(i) == "e" || string.charAt(i) == "i" || string.charAt(i) == "o" || string.charAt(i) == "u") {
      vowelsCount += 1;
    }
  }
  return vowelsCount;
}

推荐答案

你实际上可以用一个小的正则表达式：

You can actually do this with a small regex:

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

这只是与正则表达式匹配（ g 使其搜索整个字符串， i 使其不区分大小写）并返回匹配数。我们检查 null 如果没有匹配（即没有元音），则在这种情况下返回0。

This just matches against the regex (g makes it search the whole string, i makes it case-insensitive) and returns the number of matches. We check for null incase there are no matches (ie no vowels), and return 0 in that case.

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