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

一种更快的方法

If you are doing this repeatedly, for example to pad values in an array, and performance is a factor, the following approach can give you nearly a 100x advantage in speed (jsPerf) over other solution that are currently discussed on the inter webs. The basic idea is that you are providing the pad function with a fully padded empty string to use as a buffer. The pad function just appends to string to be added to this pre-padded string (one string concat) and then slices or trims the result to the desired length.

function pad(pad, str, padLeft) {
  if (typeof str === 'undefined') 
    return pad;
  if (padLeft) {
    return (pad + str).slice(-pad.length);
  } else {
    return (str + pad).substring(0, pad.length);
  }
}

例如，要将一个数字零填充为10位，

pad('0000000000',123,true);

要用空格填充字符串，使整个字符串为255个字符，

var padding = Array(256).join(' '), // make a string of 255 spaces
pad(padding,123,true);

性能测试

请在这里查看jsPerf测试。

这比ES6字符串快。重复2倍，正如这里修改后的JsPerf所示

请注意，jsPerf不再联机

请注意，我们最初用来对各种方法进行基准测试的jsPerf站点已不再在线。不幸的是，这意味着我们无法得到那些测试结果。虽然悲伤，但事实如此。

像PHP:

const STR_PAD_RIGHT = 1;
const STR_PAD_LEFT = 0;
const STR_PAD_BOTH = 2;

/**
 * @see http://php.net/str_pad
 * @param mixed input 
 * @param integer length 
 * @param string string 
 * @param integer type 
 * @return string
 */
function str_pad(input, length, string, type) {
    if (type === undefined || (type !== STR_PAD_LEFT && type !== STR_PAD_BOTH)) {
        type = STR_PAD_RIGHT
    }

    if (input.toString().length >= length) {
         return input;
    } else {
        if (type === STR_PAD_BOTH) {
            input = (string + input + string);
        } else if (type == STR_PAD_LEFT) {
            input = (string + input);
        } else {
            input = (input + string);
        }

        return str_pad(input.toString(), length, string, type);
    }
}

试试这个:

function leftPad(number) {
    return (number < 9) ? '0' + number : number;
}

// Call it like this
var month = 3;
month = leftPad(month); // Output: month=04

使用ECMAScript 6方法String#repeat和Arrow函数，一个pad函数就像这样简单:

var leftPad = (s, c, n) => c.repeat(n - s.length) + s;
leftPad("foo", "0", 5); //returns "00foo"

斯菲德尔

编辑: 评论中的建议:

const leftPad = (s, c, n) => n - s.length > 0 ? c.repeat(n - s.length) + s : s;

这样，当s.lengthis大于n时，它就不会抛出错误

edit2: 评论中的建议:

const leftPad = (s, c, n) =>{ s = s.toString(); c = c.toString(); return s.length > n ? s : c.repeat(n - s.length) + s; }

通过这种方式，可以将该函数用于字符串和非字符串。

使用重复，会更简单。

    var padLeft=function(str, pad, fw){
        return fw>str.length ? pad.repeat(fw-str.length)+str : str;
    }

你可以这样使用它: padeLeft('origin-str'， '0'， 20)