这是我的解决方案。我希望这会对某人有所帮助

const countOccurence = (string, char) => {
const chars = string.match(new RegExp(char, 'g')).length
return chars;
}

Leandro Batista的答案：只是正则表达式有问题。

“使用严格”；var dataFromDB=“testal”；$（'input[name=“tbInput”]'）.on（“change”，function（）{var charToTest=$（this）.val（）；var howManyChars=charToTest.length；var nrMatches=0；如果（howManyChars！==0）{charToTest=charToTest.charAt（0）；var regexp=新regexp（charToTest，'gi'）；var arrMatches=dataFromDB.match（正则表达式）；nrMatches=arrMatches？arrMatches.length:0；}$（'#result'）.html（nrMatches.toString（））；});<script src=“https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js“></script><div class=“main”>你想数什么<input type=“text”name=“tbInput”value=“”><br/>出现次数=<span id=“result”>0</span></div>

//Try this code

const countSubStr = (str, search) => {
    let arrStr = str.split('');
    let i = 0, count = 0;

    while(i < arrStr.length){
        let subStr = i + search.length + 1 <= arrStr.length ?
                  arrStr.slice(i, i+search.length).join('') :
                  arrStr.slice(i).join('');
        if(subStr === search){
            count++;
            arrStr.splice(i, search.length);
        }else{
            i++;
        }
    }
    return count;
  }

       var myString = "This is a string.";
        var foundAtPosition = 0;
        var Count = 0;
        while (foundAtPosition != -1)
        {
            foundAtPosition = myString.indexOf("is",foundAtPosition);
            if (foundAtPosition != -1)
            {
                Count++;
                foundAtPosition++;
            }
        }
        document.write("There are " + Count + " occurrences of the word IS");

请参阅：-count字符串中出现的子字符串，以了解分步说明。

第二次迭代次数较少（仅当子字符串的第一个字母匹配时），但循环仍使用2：

   function findSubstringOccurrences(str, word) {
        let occurrences = 0;
        for(let i=0; i<str.length; i++){
            if(word[0] === str[i]){ // to make it faster and iterate less
                for(let j=0; j<word.length; j++){
                    if(str[i+j] !== word[j]) break;
                    if(j === word.length - 1) occurrences++;
                }
            }
        }
        return occurrences;
    }
    
    console.log(findSubstringOccurrences("jdlfkfomgkdjfomglo", "omg"));

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)));
};