我最喜欢的答案是gouder hicham的。但我稍微修改了一下，让它更有意义。

let myString = "Able was I ere I saw elba";

let splitString = [];
for (let i = 0; i < myString.length; i = i + 3) {
    splitString.push(myString.slice(i, i + 3));
}

console.log(splitString);

下面是该代码的一个功能化版本。

function stringSplitter(myString, chunkSize) {
    let splitString = [];
    for (let i = 0; i < myString.length; i = i + chunkSize) {
        splitString.push(myString.slice(i, i + chunkSize));
    }
    return splitString;
}

以及函数的用法:

let myString = "Able was I ere I saw elba";
let mySplitString = stringSplitter(myString, 3);
console.log(mySplitString);

结果是:

>(9) ['Abl', 'e w', 'as ', 'I e', 're ', 'I s', 'aw ', 'elb', 'a']

试试这个简单的代码，它会像魔法一样工作!

let letters = "abcabcabcabcabc";
// we defined our variable or the name whatever
let a = -3;
let finalArray = [];
for (let i = 0; i <= letters.length; i += 3) {
    finalArray.push(letters.slice(a, i));
  a += 3;
}
// we did the shift method cause the first element in the array will be just a string "" so we removed it
finalArray.shift();
// here the final result
console.log(finalArray);

我的解决方案(ES6语法):

const source = "8d7f66a9273fc766cd66d1d";
const target = [];
for (
    const array = Array.from(source);
    array.length;
    target.push(array.splice(0,2).join(''), 2));

我们甚至可以这样创建一个函数:

function splitStringBySegmentLength(source, segmentLength) {
    if (!segmentLength || segmentLength < 1) throw Error('Segment length must be defined and greater than/equal to 1');
    const target = [];
    for (
        const array = Array.from(source);
        array.length;
        target.push(array.splice(0,segmentLength).join('')));
    return target;
}

然后你可以以一种可重用的方式轻松地调用函数:

const source = "8d7f66a9273fc766cd66d1d";
const target = splitStringBySegmentLength(source, 2);

干杯

这里我们每隔n个字符就在一个字符串中穿插另一个字符串:

export const intersperseString = (n: number, intersperseWith: string, str: string): string => {

  let ret = str.slice(0,n), remaining = str;

  while (remaining) {
    let v = remaining.slice(0, n);
    remaining = remaining.slice(v.length);
    ret += intersperseWith + v;
  }

  return ret;

};

如果我们像这样使用上面的语句:

console.log(splitString(3,'|', 'aagaegeage'));

我们得到:

亚美大陆煤层气有限公司|亚美大陆煤层气有限公司| aeg |坚毅不屈| e

这里我们做同样的事情，但是push到一个数组:

export const sperseString = (n: number, str: string): Array<string> => {

  let ret = [], remaining = str;

  while (remaining) {
    let v = remaining.slice(0, n);
    remaining = remaining.slice(v.length);
    ret.push(v);
  }

  return ret;

};

然后运行它:

console.log(sperseString(5, 'foobarbaztruck'));

我们得到:

[“fooba”、“rbazt”、“ruck”]

如果有人知道简化上述代码的方法，请使用lmk，但它应该适用于字符串。

如果你不想使用正则表达式…

var chunks = [];

for (var i = 0, charsLength = str.length; i < charsLength; i += 3) {
    chunks.push(str.substring(i, i + 3));
}

jsFiddle。

.．.否则正则表达式的解决方案是相当好的:)

稍后再讨论，但这里有一个变化，比子字符串+数组push 1快一点。

// substring + array push + end precalc
var chunks = [];

for (var i = 0, e = 3, charsLength = str.length; i < charsLength; i += 3, e += 3) {
    chunks.push(str.substring(i, e));
}

作为for循环的一部分，预先计算结束值比在子字符串中进行内联计算要快。我已经在Firefox和Chrome上测试过了，它们都显示了加速。

你可以在这里试试