太老了，但我今天需要做这样的事情，只想事后检查一下。对我来说工作很快。

String.prototype.count = function(substr,start,overlap) {
    overlap = overlap || false;
    start = start || 0;

    var count = 0, 
        offset = overlap ? 1 : substr.length;

    while((start = this.indexOf(substr, start) + offset) !== (offset - 1))
        ++count;
    return count;
};

我认为regex的目的与indexOf有很大不同。indexOf只需查找某个字符串的出现，而在正则表达式中，您可以使用[a-Z]之类的通配符，这意味着它将查找单词中的任何大写字符，而无需说明实际字符。

例子：

var index=“This is a string”.indexOf（“is”）；console.log（索引）；var length=“This is a string”.match（/[a-z]/g）.length；//其中[a-z]是正则表达式通配符，这就是为什么其速度较慢的原因console.log（长度）；

只需编码打高尔夫球丽贝卡·切尔诺夫的解决方案：-）

alert(("This is a string.".match(/is/g) || []).length);

var countInstances=函数（主体，目标）{var全局计数器=0；var concatstring=“”；for（var i=0，j=target.length；i<body.length；i++）{concatstring=body.substring（i-1，j）；if（concatstring===目标）{全局计数器+=1；concatstring=''；}}返回全局计数器；};console.log（countInstance（'abcabc'，'abc'））；//==>2.console.log（countInstance（'ababa'，'aba'））；//==>2.console.log（countInstance（'aaabbb'，'ab'））；//==>1.

const getLetterMatchCount = (guessedWord, secretWord) => {
  const secretLetters = secretWord.split('');
  const guessedLetterSet = new Set(guessedWord);
  return secretLetters.filter(letter => guessedLetterSet.has(letter)).length;
};
const str = "rahul";
const str1 = "rajendra";

getLetterMatchCount(str, str1)

基于@Vittim.us的上述回答。我喜欢他的方法给我的控制，使其易于扩展，但我需要添加不区分大小写的功能，并将匹配限制在支持标点符号的整个单词中。（例如，“洗澡”是指“洗澡”，而不是“洗澡”）

标点正则表达式来自：https://stackoverflow.com/a/25575009/497745（如何使用正则表达式从JavaScript字符串中删除所有标点符号？）

function keywordOccurrences(string, subString, allowOverlapping, caseInsensitive, wholeWord)
{

    string += "";
    subString += "";
    if (subString.length <= 0) return (string.length + 1); //deal with empty strings

    if(caseInsensitive)
    {            
        string = string.toLowerCase();
        subString = subString.toLowerCase();
    }

    var n = 0,
        pos = 0,
        step = allowOverlapping ? 1 : subString.length,
        stringLength = string.length,
        subStringLength = subString.length;

    while (true)
    {
        pos = string.indexOf(subString, pos);
        if (pos >= 0)
        {
            var matchPos = pos;
            pos += step; //slide forward the position pointer no matter what

            if(wholeWord) //only whole word matches are desired
            {
                if(matchPos > 0) //if the string is not at the very beginning we need to check if the previous character is whitespace
                {                        
                    if(!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchPos - 1])) //ignore punctuation
                    {
                        continue; //then this is not a match
                    }
                }

                var matchEnd = matchPos + subStringLength;
                if(matchEnd < stringLength - 1)
                {                        
                    if (!/[\s\u2000-\u206F\u2E00-\u2E7F\\'!"#$%&\(\)*+,\-.\/:;<=>?@\[\]^_`{|}~]/.test(string[matchEnd])) //ignore punctuation
                    {
                        continue; //then this is not a match
                    }
                }
            }

            ++n;                
        } else break;
    }
    return n;
}

如果发现错误或改进，请随时修改和重构此答案。