那么下面这一小段代码呢:

function splitME(str, size) {
    let subStr = new RegExp('.{1,' + size + '}', 'g');
    return str.match(subStr);
};

使用npm库"chkchars" 但是请记住，要确保给定的字符串长度完全除以“number”参数。

const phrase = "1110010111010011100101110100010000011100101110100111001011101001011101001110010111010001000001110010111010011100101110100"
const number = 7

chkchars.splitToChunks(phrase, number)

// result => ['1110010', '1110100','1110010', '1110100','0100000', '1110010','1110100', '1110010','1110100', '1011101','0011100', '1011101','0001000','0011100','1011101', '0011100','1011101']

// perf => 0.287ms

以原型函数的形式:

String.prototype.lsplit = function(){
    return this.match(new RegExp('.{1,'+ ((arguments.length==1)?(isFinite(String(arguments[0]).trim())?arguments[0]:false):1) +'}', 'g'));
}

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

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 ' ]

你可以这样做:

"1234567890".match(/.{1,2}/g);
// Results in:
["12", "34", "56", "78", "90"]

如果字符串的大小不是chunk-size的倍数，该方法仍然有效:

"123456789".match(/.{1,2}/g);
// Results in:
["12", "34", "56", "78", "9"]

一般来说，对于任何你想要提取最多n个子字符串的字符串，你可以这样做:

str.match(/.{1,n}/g); // Replace n with the size of the substring

如果你的字符串可以包含换行符或回车，你会这样做:

str.match(/(.|[\r\n]){1,n}/g); // Replace n with the size of the substring

至于性能，我用了大约10k个字符，在Chrome上花了一秒钟多一点的时间。YMMV。

这也可以用在可重用函数中:

function chunkString(str, length) {
  return str.match(new RegExp('.{1,' + length + '}', 'g'));
}