ECMAScript 2017 (ES8)增加了字符串。padStart(连同String.padEnd)来实现这个目的:

"Jonas".padStart(10); // Default pad string is a space
"42".padStart(6, "0"); // Pad with "0"
"*".padStart(8, "-/|\\"); // produces '-/|\\-/|*'

如果没有出现在JavaScript主机中，则字符串。padStart可以作为polyfill添加。

ES8的

我在这里找到了这个解，对我来说简单得多:

var n = 123

String("00000" + n).slice(-5); // returns 00123
("00000" + n).slice(-5); // returns 00123
("     " + n).slice(-5); // returns "  123" (with two spaces)

这里我对string对象做了一个扩展:

String.prototype.paddingLeft = function (paddingValue) {
   return String(paddingValue + this).slice(-paddingValue.length);
};

使用它的例子:

function getFormattedTime(date) {
  var hours = date.getHours();
  var minutes = date.getMinutes();

  hours = hours.toString().paddingLeft("00");
  minutes = minutes.toString().paddingLeft("00");

  return "{0}:{1}".format(hours, minutes);
};

String.prototype.format = function () {
    var args = arguments;
    return this.replace(/{(\d+)}/g, function (match, number) {
        return typeof args[number] != 'undefined' ? args[number] : match;
    });
};

这将返回格式为“15:30”的时间。

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"

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
};

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。

包括所有选项

function padding(stringToBePadded, paddingCharacter, totalLength, padLeftElseRight){
    //will pad any string provided in first argument, with padding character provide in 2nd argument and truncate to lenght provided in third argument, padding left if 4th argument true or undefined, right if false. 
    // i.e. padding("lode","x","10")  -->  "xxxxxxlode"
    // i.e. padding("lode","x","10",true)  -->  "xxxxxxlode"
    // i.e. padding("lode","x","10",false)  -->  "lodexxxxxx"
    // i.e. padding("12","0","5")  -->  "00012"
    {
        padLeftElseRight = typeof padLeftElseRight !== 'undefined' ? padLeftElseRight : true;
    }
    if (stringToBePadded.length > totalLength){
        // console.log("string too long to be padded");
        return stringToBePadded;
    }
    var paddingString = paddingCharacter.repeat(totalLength);//make long string of padding characters
    if ( padLeftElseRight){
        return String(paddingString+stringToBePadded).slice(-totalLength);
    }else{ 
        return String(stringToBePadded+paddingString).slice(0,totalLength); 
    }
}

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