一种简单的方法是将字符串拆分为所需单词，即我们要计算出现次数的单词，然后从部分数中减去1：

function checkOccurences(string, word) {
      return string.split(word).length - 1;
}
const text="Let us see. see above, see below, see forward, see backward, see left, see right until we will be right"; 
const count=countOccurences(text,"see "); // 2

 function substrCount( str, x ) {
   let count = -1, pos = 0;
   do {
     pos = str.indexOf( x, pos ) + 1;
     count++;
   } while( pos > 0 );
   return count;
 }

看到这篇帖子。

let str = 'As sly as a fox, as strong as an ox';

let target = 'as'; // let's look for it

let pos = 0;
while (true) {
  let foundPos = str.indexOf(target, pos);
  if (foundPos == -1) break;

  alert( `Found at ${foundPos}` );
  pos = foundPos + 1; // continue the search from the next position
}

相同的算法可以被布置得更短：

let str = "As sly as a fox, as strong as an ox";
let target = "as";

let pos = -1;
while ((pos = str.indexOf(target, pos + 1)) != -1) {
  alert( pos );
}

没有人会看到这一点，但偶尔带回递归和箭头函数是很好的（双关语的意思很好）

String.prototype.occurrencesOf = function(s, i) {
 return (n => (n === -1) ? 0 : 1 + this.occurrencesOf(s, n + 1))(this.indexOf(s, (i || 0)));
};

var temp=“这是一个字符串。”；console.log（（temp.match（新RegExp（“is”，“g”））||[]）.length）；

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。