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); ｝ ｝

填充字符串已在新的javascript版本中实现。

str.padStart(目标，pad弦)

https://developer.mozilla.org/es/docs/Web/JavaScript/Referencia/Objetos_globales/String/padStart

如果你想要自己的函数，检查这个例子:

const myString = 'Welcome to my house';
String.prototype.padLeft = function(times = 0, str = ' ') {
    return (Array(times).join(str) + this);
}
console.log(myString.padLeft(12, ':'));
//:::::::::::Welcome to my house

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"

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。

这里有一个简单的答案，基本上只有一行代码。

var value = 35 // the numerical value
var x = 5 // the minimum length of the string

var padded = ("00000" + value).substr(-x);

确保你填充的字符数量，这里的0，至少和你预期的最小长度一样多。因此，实际上，把它放在一行中，在这种情况下，得到“00035”的结果是:

var padded = ("00000" + 35).substr(-5);

下面是一个JavaScript函数，它使用自定义符号添加指定数量的填充。该函数接受三个参数。

padMe --> string or number to left pad
pads  --> number of pads
padSymble --> custom symbol, default is "0"

function leftPad(padMe, pads, padSymble) {
    if(typeof padMe === "undefined") {
        padMe = "";
    }
    if (typeof pads === "undefined") {
        pads = 0;
    }
    if (typeof padSymble === "undefined") {
        padSymble = "0";
    }

    var symble = "";
    var result = [];
    for(var i=0; i < pads; i++) {
       symble += padSymble;
    }
    var length = symble.length - padMe.toString().length;
    result = symble.substring(0, length);
    return result.concat(padMe.toString());
}

以下是一些结果:

> leftPad(1)
"1"

> leftPad(1, 4)
"0001"

> leftPad(1, 4, "0")
"0001"

> leftPad(1, 4, "@")
"@@@1"