我的解决方案：

var temp=“这是一个字符串。”；函数countOccurrences（str，value）{var regExp=新regExp（值，“gi”）；return（str.match（regExp）| |[]）.length；}console.log（countOccurrences（temp，'is'））；

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;
}

无正则表达式的简单版本：

var temp=“这是一个字符串。”；var计数=（临时拆分（'is'）.长度-1）；警报（计数）；

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>

/** Function that count occurrences of a substring in a string;
 * @param {String} string               The string
 * @param {String} subString            The sub string to search for
 * @param {Boolean} [allowOverlapping]  Optional. (Default:false)
 *
 * @author Vitim.us https://gist.github.com/victornpb/7736865
 * @see Unit Test https://jsfiddle.net/Victornpb/5axuh96u/
 * @see https://stackoverflow.com/a/7924240/938822
 */
function occurrences(string, subString, allowOverlapping) {

    string += "";
    subString += "";
    if (subString.length <= 0) return (string.length + 1);

    var n = 0,
        pos = 0,
        step = allowOverlapping ? 1 : subString.length;

    while (true) {
        pos = string.indexOf(subString, pos);
        if (pos >= 0) {
            ++n;
            pos += step;
        } else break;
    }
    return n;
}

用法

occurrences("foofoofoo", "bar"); //0

occurrences("foofoofoo", "foo"); //3

occurrences("foofoofoo", "foofoo"); //1

允许重叠

occurrences("foofoofoo", "foofoo", true); //2

比赛：

  foofoofoo
1 `----´
2    `----´

单元测试

https://jsfiddle.net/Victornpb/5axuh96u/

基准

我做了一个基准测试，我的功能超过了10倍比gumbo发布的regexp匹配函数更快。在我的测试中字符串长度为25个字符。字符“o”出现2次。我在Safari中执行了1000 000次。Safari 5.1基准>执行总时间：5617ms（正则表达式）基准测试>执行总时间：881毫秒（我的功能快6.4倍）Firefox 4基准>执行总时间：8547毫秒（Rexep）基准测试>总执行时间：634毫秒（我的功能更快13.5倍）编辑：我所做的更改缓存的子字符串长度为字符串添加了类型转换。添加了可选的“allowOverlapping”参数修复了“”空子字符串大小写的正确输出。

Gist

https://gist.github.com/victornpb/7736865

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;
};