你可以在没有正则表达式的情况下使用reduce():

(str, n) => {
  return str.split('').reduce(
    (acc, rec, index) => {
      return ((index % n) || !(index)) ? acc.concat(rec) : acc.concat(',', rec)
    },
    ''
  ).split(',')
}

包括左版本和右版本的预分配。 对于小块，这和RegExp impl一样快，但是随着块大小的增加，速度会更快。它的内存效率很高。

function chunkLeft (str, size = 3) {
  if (typeof str === 'string') {
    const length = str.length
    const chunks = Array(Math.ceil(length / size))
    for (let i = 0, index = 0; index < length; i++) {
      chunks[i] = str.slice(index, index += size)
    }
    return chunks
  }
}

function chunkRight (str, size = 3) {
  if (typeof str === 'string') {
    const length = str.length
    const chunks = Array(Math.ceil(length / size))
    if (length) {
      chunks[0] = str.slice(0, length % size || size)
      for (let i = 1, index = chunks[0].length; index < length; i++) {
        chunks[i] = str.slice(index, index += size)
      }
    }
    return chunks
  }
}

console.log(chunkRight())  // undefined
console.log(chunkRight(''))  // []
console.log(chunkRight('1'))  // ["1"]
console.log(chunkRight('123'))  // ["123"]
console.log(chunkRight('1234'))  // ["1", "234"]
console.log(chunkRight('12345'))  // ["12", "345"]
console.log(chunkRight('123456'))  // ["123", "456"]
console.log(chunkRight('1234567'))  // ["1", "234", "567"]

    window.format = function(b, a) {
        if (!b || isNaN(+a)) return a;
        var a = b.charAt(0) == "-" ? -a : +a,
            j = a < 0 ? a = -a : 0,
            e = b.match(/[^\d\-\+#]/g),
            h = e && e[e.length - 1] || ".",
            e = e && e[1] && e[0] || ",",
            b = b.split(h),
            a = a.toFixed(b[1] && b[1].length),
            a = +a + "",
            d = b[1] && b[1].lastIndexOf("0"),
            c = a.split(".");
        if (!c[1] || c[1] && c[1].length <= d) a = (+a).toFixed(d + 1);
        d = b[0].split(e);
        b[0] = d.join("");
        var f = b[0] && b[0].indexOf("0");
        if (f > -1)
            for (; c[0].length < b[0].length - f;) c[0] = "0" + c[0];
        else +c[0] == 0 && (c[0] = "");
        a = a.split(".");
        a[0] = c[0];
        if (c = d[1] && d[d.length -
                1].length) {
            for (var d = a[0], f = "", k = d.length % c, g = 0, i = d.length; g < i; g++) f += d.charAt(g), !((g - k + 1) % c) && g < i - c && (f += e);
            a[0] = f
        }
        a[1] = b[1] && a[1] ? h + a[1] : "";
        return (j ? "-" : "") + a[0] + a[1]
    };

var str="1234567890";
var formatstr=format( "##,###.", str);
alert(formatstr);


This will split the string in reverse order with comma separated after 3 char's. If you want you can change the position.

你绝对可以做一些

let pieces = "1234567890 ".split(/(.{2})/).filter(x => x.length == 2);

要得到这个:

[ '12', '34', '56', '78', '90' ]

如果你想动态输入/调整数据块的大小，使数据块的大小为n，你可以这样做:

n = 2;
let pieces = "1234567890 ".split(new RegExp("(.{"+n.toString()+"})")).filter(x => x.length == n);

要在原始字符串中找到所有大小为n的块，尝试这样做:

let subs = new Set();
let n = 2;
let str = "1234567890 ";
let regex = new RegExp("(.{"+n.toString()+"})");     //set up regex expression dynamically encoded with n

for (let i = 0; i < n; i++){               //starting from all possible offsets from position 0 in the string
    let pieces = str.split(regex).filter(x => x.length == n);    //divide the string into chunks of size n...
    for (let p of pieces)                 //...and add the chunks to the set
        subs.add(p);
    str = str.substr(1);    //shift the string reading frame
}

你应该得到:

[ '12', '23', '34', '45', '56', '67', '78', '89', '90', '0 ' ]

var l = str.length, lc = 0, chunks = [], c = 0, chunkSize = 2;
for (; lc < l; c++) {
  chunks[c] = str.slice(lc, lc += chunkSize);
}

var str = "123456789";
var chunks = [];
var chunkSize = 2;

while (str) {
    if (str.length < chunkSize) {
        chunks.push(str);
        break;
    }
    else {
        chunks.push(str.substr(0, chunkSize));
        str = str.substr(chunkSize);
    }
}

alert(chunks); // chunks == 12,34,56,78,9