这里有一个和split()和replace方法一样快的方法，它们比regex方法快一点点(在Chrome和Firefox中都是如此)。

let num = 0;
let str = "str1,str2,str3,str4";
//Note: Pre-calculating `.length` is an optimization;
//otherwise, it recalculates it every loop iteration.
let len = str.length;
//Note: Don't use a `for (... of ...)` loop, it's slow!
for (let charIndex = 0; charIndex < len; ++charIndex) {
  if (str[charIndex] === ',') {
    ++num;
  }
}

我的解决方案:

function countOcurrences(str, value){
   var regExp = new RegExp(value, "gi");
   return str.match(regExp) ? str.match(regExp).length : 0;  
}

我知道这可能是一个老问题，但我有一个简单的解决方案为低水平的JavaScript初学者。

作为一个初学者，我只能理解这个问题的一些解决方案，所以我使用了两个嵌套的FOR循环来检查字符串中的每个字符，为每个发现等于该字符的字符增加一个计数变量。

我创建了一个新的空白对象，其中每个属性键是一个字符，值是每个字符在字符串中出现的次数(count)。

例子函数:

function countAllCharacters(str) {
  var obj = {};
  if(str.length!==0){
    for(i=0;i<str.length;i++){
      var count = 0;
      for(j=0;j<str.length;j++){
        if(str[i] === str[j]){
          count++;
        }
      }
      if(!obj.hasOwnProperty(str[i])){
        obj[str[i]] = count;
      }
    }
  }
  return obj;
}

var a = "acvbasbb";
var b= {};
for (let i=0;i<a.length;i++){
    if((a.match(new RegExp(a[i], "g"))).length > 1){
        b[a[i]]=(a.match(new RegExp(a[i], "g"))).length;
    }
}
console.log(b);

在javascript中，你可以使用上面的代码来获得字符串中出现的字符。

更新:这可能是简单的，但它不是最快的。参见下面的基准测试。

令人惊讶的是，13年了，这个答案还没有出现。从直觉上看，它应该是最快的:

const s = "The quick brown fox jumps over the lazy dog.";
const oCount = s.length - s.replaceAll('o', '').length;

如果字符串中只有两种字符，那么这样仍然更快:

const s = "001101001";
const oneCount = s.replaceAll('0', '').length;

基准

const { performance } = require('node:perf_hooks');

const ITERATIONS = 10000000;
const TEST_STRING = "The quick brown fox jumps over the lazy dog.";

console.log(ITERATIONS, "iterations");

let sum = 0; // make sure compiler doesn't optimize code out
let start = performance.now();
for (let i = 0; i < ITERATIONS; ++i) {
  sum += TEST_STRING.length - TEST_STRING.replaceAll('o', '').length;
}
let end = performance.now();
console.log("  replaceAll duration", end - start, `(sum ${sum})`);

sum = 0;
start = performance.now();
for (let i = 0; i < ITERATIONS; ++i) {
  sum += TEST_STRING.split('o').length - 1
}
end = performance.now();
console.log("  split duration", end - start, `(sum ${sum})`);

10000 iterations
  replaceAll duration 2.6167500019073486 (sum 40000)
  split duration 2.0777920186519623 (sum 40000)
100000 iterations
  replaceAll duration 17.563208997249603 (sum 400000)
  split duration 8.087624996900558 (sum 400000)
1000000 iterations
  replaceAll duration 128.71587499976158 (sum 4000000)
  split duration 64.15841698646545 (sum 4000000)
10000000 iterations
  replaceAll duration 1223.3415840268135 (sum 40000000)
  split duration 629.1629169881344 (sum 40000000)

var i = 0; var split_start = new Date().getTime(); while (i < 30000) { "1234,453,123,324".split(",").length -1; i++; } var split_end = new Date().getTime(); var split_time = split_end - split_start; i= 0; var reg_start = new Date().getTime(); while (i < 30000) { ("1234,453,123,324".match(/,/g) || []).length; i++; } var reg_end = new Date().getTime(); var reg_time = reg_end - reg_start; alert ('Split Execution time: ' + split_time + "\n" + 'RegExp Execution time: ' + reg_time + "\n");