blocks|key|3170317|text|function+addLeadingZeros+(n,+length)
{
++++var+str+=+(n+>+0+?+n+:+-n)+%2B+"";
++++var+zeros+=+"";
++++for+(var+i+=+length+-+str.length;+i+>+0;+i--)
++++++++zeros+%2B=+"0";
++++zeros+%2B=+str;
++++return+n+>=+0+?+zeros+:+"-"+%2B+zeros;
}

//addLeadingZeros+(1,+3)+=+++"001"
//addLeadingZeros+(12,+3)+=++"012"
//addLeadingZeros+(123,+3)+=+"123"|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3170318|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>function addLeadingZeros (n, length)
{
 var str = (n &gt; 0 ? n : -n) + "";
 var zeros = "";
 for (var i = length - str.length; i &gt; 0; i--)
 zeros += "0";
 zeros += str;
 return n &gt;= 0 ? zeros : "-" + zeros;
}

//addLeadingZeros (1, 3) = "001"
//addLeadingZeros (12, 3) = "012"
//addLeadingZeros (123, 3) = "123"
</code></pre>

blocks|key|3170326|text|如果在空格上拆分，可以使用如下的简单函数添加前导零：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3170327|function+addZeros(n)+{
++return+(n+<+10)?+'00'+%2B+n+:+(n+<+100)?+'0'+%2B+n+:+''+%2B+n;
}|code-block|syntax|javascript|3170328|所以你可以测试字符串的长度，如果它小于6，在空格上拆分，在数字上加零，然后再把它连接起来。|3170329|或正则表达式：|3170330|function+addZeros(s)+{
++return+s.replace(/+(\d$)/,'+00$1').replace(/+(\d\d)$/,'+0$1');
}|3170331|我相信有人可以用一个替换，而不是两个替换。|3170332|编辑-示例|3170333|alert(addZeros('MR+3'));++++//+MR+003
alert(addZeros('MR+23'));+++//+MR+023
alert(addZeros('MR+123'));++//+MR+123
alert(addZeros('foo+bar+23'));++//+foo+bar+023|3170334|它会将一个数字前面的一个或两个零放在字符串的末尾，前面有一个空格。它并不关心空格之前的位是什么。|3170335|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

If you split on the space, you can add leading zeros using a simple function like:

<pre><code>function addZeros(n) {
 return (n &lt; 10)? '00' + n : (n &lt; 100)? '0' + n : '' + n;
}
</code></pre>

So you can test the length of the string and if it's less than 6, split on the space, add zeros to the number, then join it back together.

Or as a regular expression:

<pre><code>function addZeros(s) {
 return s.replace(/ (\d$)/,' 00$1').replace(/ (\d\d)$/,' 0$1');
}
</code></pre>

I'm sure someone can do it with one replace, not two.

<h3>Edit - examples</h3>

<pre><code>alert(addZeros('MR 3')); // MR 003
alert(addZeros('MR 23')); // MR 023
alert(addZeros('MR 123')); // MR 123
alert(addZeros('foo bar 23')); // foo bar 023
</code></pre>

It will put one or two zeros infront of a number at the end of a string with a space in front of it. It doesn't care what bit before the space is.

blocks|key|3170461|text|假设您将这些值存储在一些字符串中，请尝试执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3170462|function+pad+(str,+max)+{
++str+=+str.toString();
++return+str.length+<+max+?+pad("0"+%2B+str,+max)+:+str;
}

pad("3",+3);++++//+=>+"003"
pad("123",+3);++//+=>+"123"
pad("1234",+3);+//+=>+"1234"

var+test+=+"MR+2";
var+parts+=+test.split("+");
parts[1]+=+pad(parts[1],+3);
parts.join("+");+//+=>+"MR+002"|code-block|syntax|javascript|3170463|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Assuming you have those values stored in some strings, try this:

<pre><code>function pad (str, max) {
 str = str.toString();
 return str.length &lt; max ? pad("0" + str, max) : str;
}

pad("3", 3); // =&gt; "003"
pad("123", 3); // =&gt; "123"
pad("1234", 3); // =&gt; "1234"

var test = "MR 2";
var parts = test.split(" ");
parts[1] = pad(parts[1], 3);
parts.join(" "); // =&gt; "MR 002"
</code></pre>

blocks|key|261371|text|这是我通常在代码中用来在数字或字符串前面加上零的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|261372|输入是字符串或数字(str)，以及所需的输出长度(len)。|261373|var+PrependZeros+=+function+(str,+len)+{
++++if(typeof+str+===+'number'+%7C%7C+Number(str)){
++++str+=+str.toString();
++++return+(len+-+str.length+>+0)+?+new+Array(len+%2B+1+-+str.length).join('0')+%2B+str:+str;
}
else{
++++for(var+i+=+0,spl+=+str.split('+');+i+<+spl.length;+spl[i]+=+(Number(spl[i])&&+spl[i].length+<+len)?PrependZeros(spl[i],len):spl[i],str+=+(i+==+spl.length+-1)?spl.join('+'):str,i%2B%2B);
++++return+str;
}|code-block|syntax|javascript|261374|}；|261375|示例：|261376|PrependZeros('MR+3',3);++++//+MR+003
PrependZeros('MR+23',3);+++//+MR+023
PrependZeros('MR+123',3);++//+MR+123
PrependZeros('foo+bar+23',3);++//+foo+bar+023|261377|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This is the function that I generally use in my code to prepend zeros to a number or string.

The inputs are the string or number (str), and the desired length of the output (len).

<pre><code>var PrependZeros = function (str, len) {
 if(typeof str === 'number' || Number(str)){
 str = str.toString();
 return (len - str.length &gt; 0) ? new Array(len + 1 - str.length).join('0') + str: str;
}
else{
 for(var i = 0,spl = str.split(' '); i &lt; spl.length; spl[i] = (Number(spl[i])&amp;&amp; spl[i].length &lt; len)?PrependZeros(spl[i],len):spl[i],str = (i == spl.length -1)?spl.join(' '):str,i++);
 return str;
}
</code></pre>

};

Examples:

<pre><code>PrependZeros('MR 3',3); // MR 003
PrependZeros('MR 23',3); // MR 023
PrependZeros('MR 123',3); // MR 123
PrependZeros('foo bar 23',3); // foo bar 023
</code></pre>

blocks|key|3170572|text|我有一个潜在的解决方案，我想这是相关的，我在这里发布了它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3170573|https://www.facebook.com/antimatterstudios/posts/10150752380719364|offset|length|3170574|基本上，你想要的最小长度是2或3，你可以调整你在这段代码中放了多少个0|3170575|var+d+=+new+Date();
var+h+=+("0"%2Bd.getHours()).slice(-2);
var+m+=+("0"%2Bd.getMinutes()).slice(-2);
var+s+=+("0"%2Bd.getSeconds()).slice(-2);|code-block|syntax|javascript|3170576|我知道我总是会得到一个整数作为最小值(因为小时1，小时2)等等，但是如果你不能确定得到的只是一个空字符串，你可以只做"000"%2Bd.getHours()来确保得到最小值。|3170577|那你想要3个数字？在我的代码中使用-3而不是-2，我之所以写这个是因为我想以一种非常简单的方式构造一个24小时时钟。|3170578|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1U|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@$D|Z|E|10|1|11]]|A|$]]|$1|F|3|G|5|6|7|12|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|13|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|14|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|15|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|C]]]]

I have a potential solution which I guess is relevent, I posted about it here: 

<a href="https://www.facebook.com/antimatterstudios/posts/10150752380719364" rel="noreferrer">https://www.facebook.com/antimatterstudios/posts/10150752380719364</a>

basically, you want a minimum length of 2 or 3, you can adjust how many 0's you put in this piece of code

<pre><code>var d = new Date();
var h = ("0"+d.getHours()).slice(-2);
var m = ("0"+d.getMinutes()).slice(-2);
var s = ("0"+d.getSeconds()).slice(-2);
</code></pre>

I knew I would always get a single integer as a minimum (cause hour 1, hour 2) etc, but if you can't be sure of getting anything but an empty string, you can just do "000"+d.getHours() to make sure you get the minimum.

then you want 3 numbers? just use -3 instead of -2 in my code, I'm just writing this because I wanted to construct a 24 hour clock in a super easy fashion.

blocks|key|266959|text|简单地说，我们可以这样写：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|266960|for(var+i=1;i<=31;i%2B%2B)
++++i=(i<10)+?+'0'%2Bi+:+i;|code-block|syntax|javascript|266961|//因为大多数情况下，我们需要它来表示日、月或金额。|266962|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

In simple terms we can written as follows,

<pre><code>for(var i=1;i&lt;=31;i++)
 i=(i&lt;10) ? '0'+i : i;
</code></pre>

//Because most of the time we need this for day, month or amount matters.

blocks|key|3170641|text|搞笑，搞笑做起来又长又脏的方式...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3170642|(注意:我没有使用过这个，我也不建议使用这个。)|3170643|function+pad(str,+new_length)+{
++++('00000000000000000000000000000000000000000000000000'+%2B+str).
++++substr((50+%2B+str.toString().length)+-+new_length,+new_length)
}|code-block|syntax|javascript|3170644|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Just for a laugh do it the long nasty way....: 
(NOTE: ive not used this, and i would not advise using this.!)

<pre><code>function pad(str, new_length) {
 ('00000000000000000000000000000000000000000000000000' + str).
 substr((50 + str.toString().length) - new_length, new_length)
}
</code></pre>

blocks|key|2956697|text|前几天我自己也需要这样的东西，Pud而不是总是0，我希望能够告诉它我想要填充在前面的东西。下面是我想出的代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2956698|function+lpad(n,+e,+d)+{
++var+o+=+'';+if(typeof(d)+===+'undefined'){+d='0';+}+if(typeof(e)+===+'undefined'){+e=2;+}
++if(n.length+<+e){+for(var+r=0;+r+<+e+-+n.length;+r%2B%2B){+o+%2B=+d;+}+o+%2B=+n;+}+else+{+o=n;+}
++return+o;+}|code-block|syntax|javascript|2956699|其中，n是您想要填充的内容，e是您想要填充的幂(长度应该是多少个字符)，d是您想要填充的内容。似乎能很好地满足我的需要，但如果"d“在某些情况下超过一个字符长度，它就会失败。|2956700|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I needed something like this myself the other day, Pud instead of always a 0, I wanted to be able to tell it what I wanted padded ing the front. Here's what I came up with for code:

<pre><code>function lpad(n, e, d) {
 var o = ''; if(typeof(d) === 'undefined'){ d='0'; } if(typeof(e) === 'undefined'){ e=2; }
 if(n.length &lt; e){ for(var r=0; r &lt; e - n.length; r++){ o += d; } o += n; } else { o=n; }
 return o; }
</code></pre>

Where n is what you want padded, e is the power you want it padded to (number of characters long it should be), and d is what you want it to be padded with. Seems to work well for what I needed it for, but it would fail if "d" was more than one character long is some cases.

blocks|key|267057|text|我知道这是一个古老的帖子，但这里有另一个简短而有效的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|267058|📷|atomic|offset|length|267059|​|267060|编辑:+dur.如果num不是字符串，则应添加：|267061|len+-=+String(num).length;|style|CODE|267062|除此之外，一切都很好|267063|function+addLeadingZeros(sNum,+len)+{
++++len+-=+sNum.length;
++++while+(len--)+sNum+=+'0'+%2B+sNum;
++++return+sNum;
}|code-block|syntax|javascript|267064|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/ef50b54408ac3e4375ce633b51bf193e.jpeg|imageAlt^0|0|0|1|0|0|0|0|0|Q|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@$E|16|F|17|1|18]]|A|$]]|$1|G|3|H|5|6|7|19|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|1A|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1B|8|@$E|1C|F|1D|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1F|8|@]|9|@]|A|$T|U]]|$1|V|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12|13|-4]]]]

Know this is an old post, but here's another short, effective way:
<img src="https://i.stack.imgur.com/Au3Bf.jpg" alt="zeros" />
<h1>edit: dur. if num isn't string, you'd add:</h1>
<code>len -= String(num).length;</code>
else, it's all good
<pre><code>function addLeadingZeros(sNum, len) {
 len -= sNum.length;
 while (len--) sNum = '0' + sNum;
 return sNum;
}
</code></pre>

blocks|key|263871|text|var+str+=+"43215";+
console.log("Before+:+\n+string+:"%2Bstr%2B"\n+Length+:"%2Bstr.length);
var+max+=+9;
while(str.length+<+max+){
++++++++++++++++++++++++++++++++str+=+"0"+%2B+str;

++++++++++++++++++++++++}
console.log("After+:+\n+string+:"%2Bstr%2B"\n+Length+:"%2Bstr.length);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|263872|它对我起作用了！要增加零，请更新'max‘变量|unstyled|offset|length|style|BOLD|263873|工作地址：Adding+extra+zeros+in+front+of+a+number+using+jQuery?|263874|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/karthikeyanbss/1wtn4gn3/^0|0|H|3|0|5|1H|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@$G|Z|H|10|1|11]]|A|$]]|$1|M|3|-4|5|F|7|12|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<pre><code>var str = "43215"; 
console.log("Before : \n string :"+str+"\n Length :"+str.length);
var max = 9;
while(str.length &lt; max ){
 str = "0" + str;

 }
console.log("After : \n string :"+str+"\n Length :"+str.length);
</code></pre>

It worked for me ! 
To increase the zeroes, update the 'max' variable

Working Fiddle URL : <a href="http://jsfiddle.net/karthikeyanbss/1wtn4gn3/" rel="nofollow">Adding extra zeros in front of a number using jQuery?</a>:

blocks|key|2956866|text|注意:如果您使用的是最新的ECMAScript，请参阅更新2+...|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2956867|2956868|这里有一个我喜欢的来自an+answer+to+a+similar+question的简单的解决方案|2956869|var+n+=+123

String('00000'+%2B+n).slice(-5);+//+returns+00123
('00000'+%2B+n).slice(-5);+++++++//+returns+00123|code-block|syntax|javascript|2956870|2956871|更新|2956872|正如@RWC建议的那样，您可以将其很好地封装在一个泛型函数中，如下所示：|2956873|function+leftPad(value,+length)+{+
++++return+('0'.repeat(length)+%2B+value).slice(-length);+
}

leftPad(123,+5);+//+returns+00123|2956874|对于那些不喜欢这个切片的人：|2956875|function+leftPad(value,+length)+{
++++value+=+String(value);
++++length+=+length+-+value.length;
++++return+('0'.repeat(length)+%2B+value)
}|2956876|但如果性能很重要，我建议在选择建议的解决方案之一之前，先通读链接答案。|2956877|更新2|2956878|在ES6中，string类现在提供了一个内置的padStart方法，可以将前导字符添加到字符串中。在String.prototype.padStart()上供参考的Check+MDN+here。还有一个用于结束字符的padEnd方法。|CODE|2956879|所以有了ES6，事情就变得很简单了：|2956880|var+n+=+123;
n.padStart(5,+'0');+//+returns+00123|2956881|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/14760377/1697459|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart^0|0|Y|0|0|0|1E|B|V|0|0|0|0|0|2|0|0|10|0|0|0|E|0|0|0|Z|0|0|3|0|0|39|N|8|1E|R|30|6|2A|E|1|0|0|I|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]|$1|G|3|H|5|6|7|1N|8|@$9|1O|A|1P|B|C]]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|I|3|J|5|K|7|1T|8|@]|D|@]|E|$L|M]]|$1|N|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1V|8|@$9|1W|A|1X|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|S|3|T|5|K|7|21|8|@]|D|@]|E|$L|M]]|$1|U|3|V|5|6|7|22|8|@$9|23|A|24|B|C]]|D|@]|E|$]]|$1|W|3|X|5|K|7|25|8|@]|D|@]|E|$L|M]]|$1|Y|3|Z|5|6|7|26|8|@$9|27|A|28|B|C]]|D|@]|E|$]]|$1|10|3|11|5|6|7|29|8|@$9|2A|A|2B|B|C]]|D|@]|E|$]]|$1|12|3|13|5|6|7|2C|8|@$9|2D|A|2E|B|C]|$9|2F|A|2G|B|14]|$9|2H|A|2I|B|14]|$9|2J|A|2K|B|14]]|D|@$9|2L|A|2M|1|2N]]|E|$]]|$1|15|3|16|5|6|7|2O|8|@$9|2P|A|2Q|B|C]]|D|@]|E|$]]|$1|17|3|18|5|K|7|2R|8|@]|D|@]|E|$L|M]]|$1|19|3|-4|5|6|7|2S|8|@]|D|@]|E|$]]]|1A|$1B|$5|1C|1D|1E|E|$1F|1G]]|1H|$5|1C|1D|1E|E|$1F|1I]]]]

Note: see Update 2 if you are using latest ECMAScript...

<hr>

Here a solution I liked for its simplicity from <a href="https://stackoverflow.com/a/14760377/1697459">an answer to a similar question</a>:

<pre><code>var n = 123

String('00000' + n).slice(-5); // returns 00123
('00000' + n).slice(-5); // returns 00123
</code></pre>

<hr>

<h2>UPDATE</h2>

As @RWC suggested you can wrap this of course nicely in a generic function like this:

<pre><code>function leftPad(value, length) { 
 return ('0'.repeat(length) + value).slice(-length); 
}

leftPad(123, 5); // returns 00123
</code></pre>

And for those who don't like the slice:

<pre><code>function leftPad(value, length) {
 value = String(value);
 length = length - value.length;
 return ('0'.repeat(length) + value)
}
</code></pre>

But if performance matters I recommend reading through the linked answer before choosing one of the solutions suggested.

<h2>UPDATE 2</h2>

In ES6 the String class now comes with a inbuilt <code>padStart</code> method which adds leading characters to a string. <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart" rel="noreferrer">Check MDN here</a> for reference on <code>String.prototype.padStart()</code>. And there is also a <code>padEnd</code> method for ending characters.

So with ES6 it became as simple as: 

<pre><code>var n = 123;
n.padStart(5, '0'); // returns 00123
</code></pre>

blocks|key|265524|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|265525|字符串可以是数字或字符串。|blockquote|265526|265527|265528|formatting("hi",3);
function+formatting(str,len)
{
+++return+("000000"%2Bstr).slice(-len);
}|code-block|syntax|javascript|265529|如果需要较大的数字，则添加更多零|offset|length|style|BOLD|265530|entityMap^0|0|0|0|0|0|F|1|0^^$0|@$1|2|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|T|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|U|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|V|7|@]|8|@]|9|$]]|$1|F|3|G|4|H|6|W|7|@]|8|@]|9|$I|J]]|$1|K|3|L|4|5|6|X|7|@$M|Y|N|Z|O|P]]|8|@]|9|$]]|$1|Q|3|-4|4|5|6|10|7|@]|8|@]|9|$]]]|R|$]]

<blockquote>
 str could be a number or a string.
</blockquote>

<pre><code>formatting("hi",3);
function formatting(str,len)
{
 return ("000000"+str).slice(-len);
}
</code></pre>

Add more zeros if needs large digits

blocks|key|2956924|text|试试下面的方法，它会将单位数和双位数的数字转换为3位数字，方法是在前面加上零。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2956925|var+base_number+=+2;
var+zero_prefixed_string+=+("000"+%2B+base_number).slice(-3);|code-block|syntax|javascript|2956926|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try following, which will convert convert single and double digit numbers to 3 digit numbers by prefixing zeros.

<pre><code>var base_number = 2;
var zero_prefixed_string = ("000" + base_number).slice(-3);
</code></pre>

blocks|key|267124|text|在数字上加上100，然后运行一个从索引1到右边最后一个位置的子串函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|267125|var+dt+=+new+Date();
var+month+=+(100+%2B+dt.getMonth()%2B1).toString().substr(1,+2);
var+day+=+(100+%2B+dt.getDate()).toString().substr(1,+2);

console.log(month,day);|code-block|syntax|javascript|267126|您将从2020年11月3日起获得以下结果|267127|11,03|267128|我希望答案是有用的。|267129|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

By adding 100 to the number, then run a substring function from index 1 to the last position in right.
<pre><code>var dt = new Date();
var month = (100 + dt.getMonth()+1).toString().substr(1, 2);
var day = (100 + dt.getDate()).toString().substr(1, 2);

console.log(month,day);
</code></pre>
you will got this result from the date of 2020-11-3
<pre><code>11,03
</code></pre>
I hope the answer is useful

I have file that are uploaded which are formatted like so

<blockquote>
 MR 1
 
 MR 2
 
 MR 100
 
 MR 200
 
 MR 300 
 
 ETC.
</blockquote>

What i need to do is add extra two 00s before anything before MR 10 and add one extra 0 before MR10-99

So files are formatted

<blockquote>
 MR 001
 
 MR 010
 
 MR 076
 
 ETC.
</blockquote>

Any help would be great!

Adding extra zeros in front of a number using jQuery?

我上传的文件格式是这样的 MR 1MR 2MR 100MR 200MR 300等。我需要做的是在MR10之前添加额外的两个00，并且在MR10-99之前添加额外的0所以文件是格式化的 MR 001MR 010076先生等。任何帮助都是最好的！

问使用jQuery在数字前面添加额外的零？
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