ECMAScript 2017在String原型中增加了一个padStart方法。这个方法将用空格填充字符串到给定的长度。此方法还接受一个可选字符串，该字符串将用于代替空格作为填充。

'abc'.padStart(10);         // "       abc"
'abc'.padStart(10, "foo");  // "foofoofabc"
'abc'.padStart(6,"123465"); // "123abc"
'abc'.padStart(8, "0");     // "00000abc"
'abc'.padStart(1);          // "abc"

还添加了以相同方式工作的padEnd方法。

关于浏览器兼容性(和一个有用的填充)，请参见此链接。

这两种解决方案的关键技巧是创建具有给定大小(比所需长度大一个)的数组实例，然后立即调用join()方法来生成字符串。join()方法被传递填充字符串(可能是空格)。由于数组是空的，在将数组连接到一个结果字符串的过程中，空单元格将被呈现为空字符串，只有填充将保留。这是一个很好的技巧。

我将上述解决方案的组合添加到我自己的，总是不断发展的版本:)

//in preperation for ES6
String.prototype.lpad || (String.prototype.lpad = function( length, charOptional )
{
    if (length <= this.length) return this;
    return ( new Array((length||0)+1).join(String(charOptional)||' ') + (this||'') ).slice( -(length||0) );
});


'abc'.lpad(5,'.') == '..abc'
String(5679).lpad(10,0) == '0000005679'
String().lpad(4,'-') == '----' // repeat string

http://www.webtoolkit.info/javascript_pad.html

/**
*
*  JavaScript string pad
*  http://www.webtoolkit.info/
*
**/

var STR_PAD_LEFT = 1;
var STR_PAD_RIGHT = 2;
var STR_PAD_BOTH = 3;

function pad(str, len, pad, dir) {

    if (typeof(len) == "undefined") { var len = 0; }
    if (typeof(pad) == "undefined") { var pad = ' '; }
    if (typeof(dir) == "undefined") { var dir = STR_PAD_RIGHT; }

    if (len + 1 >= str.length) {

        switch (dir){

            case STR_PAD_LEFT:
                str = Array(len + 1 - str.length).join(pad) + str;
            break;

            case STR_PAD_BOTH:
                var padlen = len - str.length;
                var right = Math.ceil( padlen / 2 );
                var left = padlen - right;
                str = Array(left+1).join(pad) + str + Array(right+1).join(pad);
            break;

            default:
                str = str + Array(len + 1 - str.length).join(pad);
            break;

        } // switch

    }

    return str;
}

可读性更强。

ES7现在只是草案和建议，但如果你想跟踪与规范的兼容性，你的pad功能需要:

多字符pad支持。 不要截断输入字符串 Pad默认为空格

从我的填充库，但应用你自己的尽职调查的原型扩展。

// Tests
'hello'.lpad(4) === 'hello'
'hello'.rpad(4) === 'hello'
'hello'.lpad(10) === '     hello'
'hello'.rpad(10) === 'hello     '
'hello'.lpad(10, '1234') === '41234hello'
'hello'.rpad(10, '1234') === 'hello12341'

String.prototype.lpad || (String.prototype.lpad = function(length, pad)
{
    if(length < this.length)
        return this;

    pad = pad || ' ';
    let str = this;

    while(str.length < length)
    {
        str = pad + str;
    }

    return str.substr( -length );
});

String.prototype.rpad || (String.prototype.rpad = function(length, pad)
{
    if(length < this.length)
        return this;

    pad = pad || ' ';
    let str = this;

    while(str.length < length)
    {
        str += pad;
    }

    return str.substr(0, length);
});

Never insert data somewhere (especially not at beginning, like str = pad + str;), since the data will be reallocated everytime. Append always at end! Don't pad your string in the loop. Leave it alone and build your pad string first. In the end concatenate it with your main string. Don't assign padding string each time (like str += pad;). It is much faster to append the padding string to itself and extract first x-chars (the parser can do this efficiently if you extract from first char). This is exponential growth, which means that it wastes some memory temporarily (you should not do this with extremely huge texts).

if (!String.prototype.lpad) { String.prototype.lpad =函数(pad, len) { 而(pad。长度< len) { Pad += Pad; ｝ 返回垫。Substr (0, lens -this.length) + this; ｝ ｝ if (!String.prototype.rpad) { String.prototype.rpad = function(pad, len) { 而(pad。长度< len) { Pad += Pad; ｝ 返回这个+ pad。substr (0, len-this.length); ｝ ｝