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
};

函数 var _padLeft = function(paddingString, width, replacementChar) { paddingString返回。长度>=宽度?paddingString: _padLeft(replacementChar + paddingString, width, replacementChar || ' '); }; 字符串的原型 String.prototype.padLeft = function(width, replacementChar) { 返回。长度>=宽度?this. tostring ():(replacementChar + this)。padLeft(width, replacementChar || ' '); }; 片 ('00000' + paddingString).slice(-5)

与简单的字符串连接相比，数组操作真的很慢。当然，是用例的基准测试。

function(string, length, pad_char, append) {
    string = string.toString();
    length = parseInt(length) || 1;
    pad_char = pad_char || ' ';

    while (string.length < length) {
        string = append ? string+pad_char : pad_char+string;
    }
    return string;
};

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。

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)); ｝

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