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 );
}

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。

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

参数：ustring：超集字符串countChar：子字符串

一个计算JavaScript中子字符串出现次数的函数：

函数subStringCount（ustring，countChar）{var correspCount=0；var corresp=false；变量量=0；var prevChar=空；对于（var i=0；i！=ustring.length；i++）{如果（ustring.charAt（i）==countChar.charAt（0）&&corresp==false）{corresp=真；correspCount+=1；如果（correspCount==countChar.length）{数量+=1；corresp=false；correspCount=0；}prevChar=1；}否则如果（ustring.charAt（i）==countChar.charAt（prevChar）&&corresp==true）{correspCount+=1；如果（correspCount==countChar.length）{数量+=1；corresp=false；correspCount=0；prevChar=空；}其他{prevChar+=1；}}其他{corresp=false；correspCount=0；}} 回报金额；}console.log（subStringCount（“Hello World，Hello World”，“ll”））；

试试看：

function countString(str, search){
    var count=0;
    var index=str.indexOf(search);
    while(index!=-1){
        count++;
        index=str.indexOf(search,index+1);
    }
    return count;
}

subst_count从php转换为Javascript

Locutus（将Php翻译为JS的包）subst_count（官方页面，代码复制如下）

function substr_count (haystack, needle, offset, length) { 
  // eslint-disable-line camelcase
  //  discuss at: https://locutus.io/php/substr_count/
  // original by: Kevin van Zonneveld (https://kvz.io)
  // bugfixed by: Onno Marsman (https://twitter.com/onnomarsman)
  // improved by: Brett Zamir (https://brett-zamir.me)
  // improved by: Thomas
  //   example 1: substr_count('Kevin van Zonneveld', 'e')
  //   returns 1: 3
  //   example 2: substr_count('Kevin van Zonneveld', 'K', 1)
  //   returns 2: 0
  //   example 3: substr_count('Kevin van Zonneveld', 'Z', 0, 10)
  //   returns 3: false

  var cnt = 0

  haystack += ''
  needle += ''
  if (isNaN(offset)) {
    offset = 0
  }
  if (isNaN(length)) {
    length = 0
  }
  if (needle.length === 0) {
    return false
  }
  offset--

  while ((offset = haystack.indexOf(needle, offset + 1)) !== -1) {
    if (length > 0 && (offset + needle.length) > length) {
      return false
    }
    cnt++
  }

  return cnt
}

查看Locutus对Php的subst_count函数的翻译