如何计算特定字符串在另一个字符串中出现的次数。例如,这就是我试图在Javascript中做的事情:

var temp = "This is a string.";
alert(temp.count("is")); //should output '2'

当前回答

       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。

其他回答

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;
};

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);

你可以试试这个

let count = s.length - s.replace(/is/g, "").length;

ES2020提供了一个新的MatchAll,它可能在这个特定的环境中使用。

这里我们创建了一个新的RegExp,请确保将“g”传递到函数中。

使用Array.from转换结果并计算长度,根据原始请求者所需的输出返回2。

let strToCheck=RegExp('is','g')let matchesReg=“这是一个字符串。”.matchAll(strToCheck)console.log(Array.from(matchesReg).length)//2

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