还有一种结合了几个解决方案的方法:

/**
 * pad string on left
 * @param {number} number of digits to pad, default is 2
 * @param {string} string to use for padding, default is '0' *
 * @returns {string} padded string
 */
String.prototype.paddingLeft = function (b, c) {
    if (this.length > (b||2))
        return this + '';
  return (this || c || 0) + '', b = new Array((++b || 3) - this.length).join(c || 0), b + this
};

/**
 * pad string on right
 * @param {number} number of digits to pad, default is 2
 * @param {string} string to use for padding, default is '0' *
 * @returns {string} padded string
 */
String.prototype.paddingRight = function (b, c) {
  if (this.length > (b||2))
        return this + '';
  return (this||c||0) + '', b = new Array((++b || 3) - this.length).join(c || 0), this + b
};

继承塞缪尔的想法，在上面这里。记住一个旧的SQL脚本，我尝试了这个:

a=1234;
'0000'.slice(a.toString().length)+a;

它适用于我能想到的所有情况:

a=     1 result  0001
a=    12 result  0012
a=   123 result  0123
a=  1234 result  1234
a= 12345 result 12345
a=  '12' result  0012

这是一个递归的方法。

function pad(width, string, padding) { 
  return (width <= string.length) ? string : pad(width, padding + string, padding)
}

一个例子……

pad(5, 'hi', '0')
=> "000hi"

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); ｝ ｝

基于这个问题的最佳答案，我为String做了一个名为padLeft的原型(就像我们在c#中所做的一样):

String.prototype.padLeft = function (paddingChar, totalWidth) {
    if (this.toString().length >= totalWidth)
        return this.toString();

    var array = new Array(totalWidth); 

    for (i = 0; i < array.length; i++)
        array[i] = paddingChar;

    return (array.join("") + this.toString()).slice(-array.length);
}

用法:

var str = "12345";
console.log(str.padLeft("0", 10)); //Result is: "0000012345"

小提琴

如果你想要一些紧凑的东西，只需一行:

String.prototype.pad = function(len, chr){ 返回((((新数组(len)) .fill(杆)). join (" ") +) .substring (this.length)); ｝