方法一、自定义函数实现，不方便自定义调用

function FillZero(p) {

return new Array(3 - (p + '').length + 1).join('0') + p;

}

FillZero(6); //输出006

方法二、String方法一扩展(C#中PadLeft、PadRight)

String.prototype.PadLeft = function (len, charStr) {

var s = this + '';

return new Array(len - s.length + 1).join(charStr || '') + s;

}

String.prototype.PadRight = function (len, charStr) {

var s = this + '';

return s + new Array(len - s.length + 1).join(charStr || '');

}

var p = 6;

p.toString().PadLeft(3, '0'); //输出006

p.toString().PadRight(3, '0'); //输出600

方法三、原理同方法二

String.prototype.pad= function (pos,len,padStr) {

var padStrs = new Array(len).join(padStr,'');

return pos ? (padStrs + this.toString()).substr(0-len):(this.toString() + padStrs).substr(0,len);

}

String.prototype.lpad = function(len,padStr){

return this.pad(1,len,padStr);

}

String.prototype.rpad = function(len,padStr){

return this.pad(0,len,padStr);

}

String.prototype.padLeft = String.prototype.lpad;

String.prototype.padRight = String.prototype.rpad;

var str="6";

console.log(str.lpad(2,"0"));

console.log(str.padLeft(2,"0"));

代码四、

String.prototype.padLeft = function (padChar, width) {

var ret = this;

while (ret.length < width) {

if (ret.length + padChar.length < width) {

ret = padChar + ret;

}

else {

ret = padChar.substring(0, width - ret.length) + ret;

}

}

return ret;

};

String.prototype.padRight = function (padChar, width) {

var ret = this;

while (ret.length < width) {

if (ret.length + padChar.length < width) {

ret += padChar;

}

else {

ret += padChar.substring(0, width - ret.length);

}

}

return ret;

};

下面是其他网友的补充

javascript-leftpad方法

原由是微博上出的一则leftpad方法被吐槽的事。

原方法是通过循环挨个拼接字符串的方式，所以效率差了点(会产生N多string对象，导致对象回收慢)

类似

for(i=0;i

str=pad+str;

}

改进如下：

//cache

var padding1="00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000";

var padding2=" ";

function leftpad(str,length,useSpace){

var padding = padding1;

if(useSpace){

padding = padding2;

}

var cnt = length-str.length;

if(cnt<=0) return str;

if(cnt<=200) return padding.substring(0,cnt)+str;

str=padding+str;

return leftpad(str,length,ch);

}

思路：

1.减少string对象的生成，所以尽量减少字符串拼接的次数

2.降低时间复杂度

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