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
<script type="text/javascript">
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"));
</script>
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<cnt;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
