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"));

       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。

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);

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 );
}

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

此函数在三种模式下工作：查找字符串中单个字符的频率，查找字符串中相邻子字符串的频率，然后如果它与一个匹配，则会直接向前移动到它后面的下一个，第三个与前一个相似，但它也会计算给定字符串中的交叉子字符串

函数substringFrequency（字符串、子字符串、连接）{let索引允许发生频率=0for（设i=0；i<string.length；i++）{index=string.indexOf（substring，i）如果（索引！=-1）{if（（子字符串长度==1）||连接==true）{i=索引}其他{i=索引+1}发生频率++}其他{打破} }return（发生频率）}console.log（substringFrequency（'vvv'，'v'））console.log（substringFrequency（'vvv'，'vv'））console.log（substringFrequency（'vvv'，'vv'））