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'));
}

我已经写了一个扩展函数，所以块长度也可以是一个数字数组，比如[1,3]

String.prototype.chunkString = function(len) {
    var _ret;
    if (this.length < 1) {
        return [];
    }
    if (typeof len === 'number' && len > 0) {
        var _size = Math.ceil(this.length / len), _offset = 0;
        _ret = new Array(_size);
        for (var _i = 0; _i < _size; _i++) {
            _ret[_i] = this.substring(_offset, _offset = _offset + len);
        }
    }
    else if (typeof len === 'object' && len.length) {
        var n = 0, l = this.length, chunk, that = this;
        _ret = [];
        do {
            len.forEach(function(o) {
                chunk = that.substring(n, n + o);
                if (chunk !== '') {
                    _ret.push(chunk);
                    n += chunk.length;
                }
            });
            if (n === 0) {
                return undefined; // prevent an endless loop when len = [0]
            }
        } while (n < l);
    }
    return _ret;
};

的代码

"1234567890123".chunkString([1,3])

将返回:

[ '1', '234', '5', '678', '9', '012', '3' ]

我对上述解决方案的问题是，不管在句子中的位置如何，它都将字符串划分为正式的大小块。

我认为下面的方法比较好;虽然它需要一些性能调整:

 static chunkString(str, length, size,delimiter='\n' ) {
        const result = [];
        for (let i = 0; i < str.length; i++) {
            const lastIndex = _.lastIndexOf(str, delimiter,size + i);
            result.push(str.substr(i, lastIndex - i));
            i = lastIndex;
        }
        return result;
    }

比较match, slice, substr和substring 不同块大小的匹配和切片的比较 小块大小的匹配和切片的比较

底线:

match非常低效，slice更好，在Firefox上substr/substring更好 匹配对于短字符串来说效率更低(即使使用缓存的regex -可能是因为regex解析设置时间) 对于大块大小的匹配效率更低(可能是由于无法“跳跃”) 对于更长的字符串和非常小的块大小，match在旧的IE上优于slice，但在所有其他系统上仍然失败 jsperf岩石

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 ' ]

    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.