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
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;
};
其他回答
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)));
};
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>
const getLetterMatchCount = (guessedWord, secretWord) => {
const secretLetters = secretWord.split('');
const guessedLetterSet = new Set(guessedWord);
return secretLetters.filter(letter => guessedLetterSet.has(letter)).length;
};
const str = "rahul";
const str1 = "rajendra";
getLetterMatchCount(str, str1)
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
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;
};
