blocks|key|1504731|text|我不认为JavaScript语言中有任何“内置”的东西可以做到这一点。下面是一个执行此操作的简单函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504732|function+FormatNumberLength(num,+length)+{
++++var+r+=+""+%2B+num;
++++while+(r.length+<+length)+{
++++++++r+=+"0"+%2B+r;
++++}
++++return+r;
}


FormatNumberLength(10000,+5)+outputs+'10000'
FormatNumberLength(1000,+5)++outputs+'01000'
FormatNumberLength(100,+5)+++outputs+'00100'
FormatNumberLength(10,+5)++++outputs+'00010'|code-block|syntax|javascript|1504733|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|$]]

I don't think there's anything "built" into the JavaScript language for doing this. Here's a simple function that does this:

<pre><code>function FormatNumberLength(num, length) {
 var r = "" + num;
 while (r.length &lt; length) {
 r = "0" + r;
 }
 return r;
}


FormatNumberLength(10000, 5) outputs '10000'
FormatNumberLength(1000, 5) outputs '01000'
FormatNumberLength(100, 5) outputs '00100'
FormatNumberLength(10, 5) outputs '00010'
</code></pre>

blocks|key|1290092|text|这可能会有所帮助：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1290093|String.prototype.padLeft+=+function+(length,+character)+{+
+++++return+new+Array(length+-+this.length+%2B+1).join(character+%7C%7C+'0')+%2B+this;+
}

var+num+=+'12';

alert(num.padLeft(4,+'0'));|code-block|syntax|javascript|1290094|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|$]]

This might help : 

<pre><code>String.prototype.padLeft = function (length, character) { 
 return new Array(length - this.length + 1).join(character || '0') + this; 
}

var num = '12';

alert(num.padLeft(4, '0'));
</code></pre>

blocks|key|1290151|text|你可以疯狂地使用这样的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1290152|function+PadDigits(input,+totalDigits)+
{+
++++var+result+=+input;
++++if+(totalDigits+>+input.length)+
++++{+
++++++++for+(i=0;+i+<+(totalDigits+-+input.length);+i%2B%2B)+
++++++++{+
++++++++++++result+=+'0'+%2B+result;+
++++++++}+
++++}+
++++return+result;
}+|code-block|syntax|javascript|1290153|但这并不会让生活变得更容易。C#在String类中有一个像PadLeft和PadRight这样的方法，不幸的是，Javascript没有内置这个功能|1290154|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|$]]

You could go crazy with methods like these:

<pre><code>function PadDigits(input, totalDigits) 
{ 
 var result = input;
 if (totalDigits &gt; input.length) 
 { 
 for (i=0; i &lt; (totalDigits - input.length); i++) 
 { 
 result = '0' + result; 
 } 
 } 
 return result;
} 
</code></pre>

But it won't make life easier. C# has a method like PadLeft and PadRight in the String class, unfortunately Javascript doesn't have this functionality build-in

blocks|key|1504810|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1504811|function+prefixZeros(number,+maxDigits)+
{++
++++var+length+=+maxDigits+-+number.toString().length;
++++if(length+<=+0)
++++++++return+number;

++++var+leadingZeros+=+new+Array(length+%2B+1);
++++return+leadingZeros.join('0')+%2B+number.toString();
}
//Call+it+like+prefixZeros(100,+5);+//Alerts+00100|code-block|syntax|javascript|1504812|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|$]]

How about something like this:

<pre><code>function prefixZeros(number, maxDigits) 
{ 
 var length = maxDigits - number.toString().length;
 if(length &lt;= 0)
 return number;

 var leadingZeros = new Array(length + 1);
 return leadingZeros.join('0') + number.toString();
}
//Call it like prefixZeros(100, 5); //Alerts 00100
</code></pre>

blocks|key|1505013|text|一种有趣(但很有趣)的数字前缀为零的方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505014|function+FormatInteger(num,+length)+{

++++return+(num+/+Math.pow(10,+length)).toFixed(length).substr(2);
}|code-block|syntax|javascript|1505015|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|$]]

A funny (but interesting) way to prefix numbers with zeros:

<pre><code>function FormatInteger(num, length) {

 return (num / Math.pow(10, length)).toFixed(length).substr(2);
}
</code></pre>

blocks|key|1290448|text|一个"while“循环应该会让这件事变得足够简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1290449|function+formatLength(a)+{
++++var+num+=+document.getElementById(a)
++++while+(num.value.length+<+4)+{
++++++++num.value+=+'0'+%2B+num.value
++++}
}|code-block|syntax|javascript|1290450|这将循环执行，直到num值的长度达到4位(假设您已将number元素的id作为参数传入)|1290451|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|$]]

A "while" loop should make this easy enough. 

<pre><code>function formatLength(a) {
 var num = document.getElementById(a)
 while (num.value.length &lt; 4) {
 num.value = '0' + num.value
 }
}
</code></pre>

That would loop through until the length of the num value reached 4 digits (assuming you have passed in the id of the number element as an argument)

blocks|key|1290532|text|我遇到了几乎相同的问题，我找到了一种简洁的方法来解决它。如果我必须在我的代码中多次使用它，或者如果我使用它的位数超过四位，我会使用其他建议的解决方案之一，但这种方式让我可以将其全部放在一个表达式中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1290533|((x<10)?"000":+(x<100)?"00":+(x<1000)?"0":+"")+%2B+x|code-block|syntax|javascript|1290534|它实际上与您的代码相同，但使用三元运算符而不是"if-else“语句(并将"%2B+x”移到条件代码之外，它将始终是表达式的一部分)。|1290535|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 ran into much the same problem and I found a compact way to solve it. If I had to use it multiple times in my code or if I was doing it for any more than four digits, I'd use one of the other suggested solutions, but this way lets me put it all in an expression:

<pre><code>((x&lt;10)?"000": (x&lt;100)?"00": (x&lt;1000)?"0": "") + x
</code></pre>

It's actually the same as your code, but using the ternary operator instead of "if-else" statements (and moving the "+ x", which will always be part of the expression, outside of the conditional code).

blocks|key|1505234|text|我能想到的最简单的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505235|("000"+%2B+num).slice(-4)|code-block|syntax|javascript|1505236|填充的数字是一个字符串。|1505237|将数字添加到字符串时，该数字将转换为字符串。|1505238|string具有方法slice，该方法返回字符串的一段固定长度。|1505239|如果length为负，则从字符串的末尾开始对返回的字符串进行切片。|1505240|要测试：|1505241|var+num=12;
console.log(("000"+%2B+num).slice(-4));+//+Will+show+"0012"|1505242|当然，这只适用于最多4位的正整数。一个稍微复杂一点的解决方案，将处理正整数：|1505243|'0'.repeat(+Math.max(4+-+num.toString().length,+0))+%2B+num|1505244|通过重复添加零来创建字符串，只要数字的数量(字符串的长度)小于4就添加数字，然后将其转换为字符串。|1505245|entityMap^0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

The simplest way I can think of is this:

<pre><code>("000" + num).slice(-4)
</code></pre>

A padded number is a string. 
When you add a number to a string, it is converted to a string. 
Strings has the method slice, that retuns a fixed length piece of the string. 
If length is negative the returned string is sliced from the end of the string.

to test:

<pre><code>var num=12;
console.log(("000" + num).slice(-4)); // Will show "0012"
</code></pre>

Of cause this only works for positive integers of up to 4 digits. A slightly more complex solution, will handle positive integers:

<pre><code>'0'.repeat( Math.max(4 - num.toString().length, 0)) + num
</code></pre>

Create a string by repeat adding zeros, as long as the number of digits (length of string) is less than 4
Add the number, that is then converted to a string also.

blocks|key|1505253|text|另一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505254|function+format_number_length(num,+length)+{
++++var+r+=+num.toString();
++++if+(r.length<length)+r+=+Array(length+-+r.length+%2B+1).join('0')+%2B+r;
++++return+r;
}|code-block|syntax|javascript|1505255|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|$]]

Another one:

<pre><code>function format_number_length(num, length) {
 var r = num.toString();
 if (r.length&lt;length) r = Array(length - r.length + 1).join('0') + r;
 return r;
}
</code></pre>

blocks|key|1505323|text|我认为最简洁而又直观的方式是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505324|function+toFixedLength(input,+length,+padding)+{
++++padding+=+String(padding+%7C%7C+"0");
++++return+(padding.repeat(length)+%2B+input).slice(-length);
}|code-block|syntax|javascript|1505325|这里的slice方法可以替换为substr，如果这对程序员更直观的话。|offset|length|style|CODE|1505326|entityMap^0|0|0|3|5|F|6|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

I think the most compact yet intuitive way is:

<pre><code>function toFixedLength(input, length, padding) {
 padding = String(padding || "0");
 return (padding.repeat(length) + input).slice(-length);
}
</code></pre>

The <code>slice</code> method here could be replaced with <code>substr</code> if that is more intuitive to the coder.

blocks|key|1505527|text|我来寻找答案，但我认为这是一种更好的函数方法(ES6)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1505528|const+formatNumberToString+=+(num,+minChars)+=>+{
++return+num.toString().length+<+minChars
+++?+formatNumberToString(`0${num}`,+minChars)
+++:+num.toString()
}
//+formatNumberToString(12,+4)+//+"0012"
//+formatNumberToString(12,+5)+//+"00012"
//+formatNumberToString(1,+4)+//+"0001"
//+formatNumberToString(1,+2)+//+"01"
//+formatNumberToString(12,+2)+//+"12"
//+formatNumberToString(12,+1)+//+"12"|code-block|syntax|javascript|1505529|同样，这也可以在一行代码中实现|1505530|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 came looking for the answer, but I think this is a better functional approach (ES6):

<pre><code>const formatNumberToString = (num, minChars) =&gt; {
 return num.toString().length &lt; minChars
 ? formatNumberToString(`0${num}`, minChars)
 : num.toString()
}
// formatNumberToString(12, 4) // "0012"
// formatNumberToString(12, 5) // "00012"
// formatNumberToString(1, 4) // "0001"
// formatNumberToString(1, 2) // "01"
// formatNumberToString(12, 2) // "12"
// formatNumberToString(12, 1) // "12"
</code></pre>

also, this can be implemented just in one line

blocks|key|1290981|text|因为可以使用String.prototype.padStart和String.prototype.padEnd简单地将ES2017填充到最小长度|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1290982|let+num+=+3
let+str+=+num.toString().padStart(3,+"0")
console.log(str)+//+"003"|code-block|syntax|javascript|1290983|或者，如果浮点数的整个部分应该是固定的长度：|1290984|let+num+=+3.141
let+arr+=+num.toString().split(".")
arr[0]+=+arr[0].padStart(3,+"0")
let+str+=+arr.join(".")
console.log(str)+//+"003.141"|1290985|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd^0|6|P|0|W|N|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|14|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|15|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

Since ES2017 padding to a minimum length can be done simply with
<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart" rel="noreferrer">String.prototype.padStart</a>
and
<a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padEnd" rel="noreferrer">String.prototype.padEnd</a>:

<pre class="lang-js prettyprint-override"><code>let num = 3
let str = num.toString().padStart(3, "0")
console.log(str) // "003"
</code></pre>

Or if only the whole part of a float should be a fixed length:

<pre class="lang-js prettyprint-override"><code>let num = 3.141
let arr = num.toString().split(".")
arr[0] = arr[0].padStart(3, "0")
let str = arr.join(".")
console.log(str) // "003.141"
</code></pre>

blocks|key|1505587|text|ES6+repeat()方法的最新版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1505588|++++const+length_required+=+5;
++++let+num+=+10;
++++num+=+"0".repeat(length_required+-+String(num).length)+%2B+num;
++++console.log(num)
++++//+output:+00010

++++let+num+=+1000;
++++num+=+"0".repeat(length_required+-+String(num).length)+%2B+num;
++++console.log(num)
++++//+output:+01000|code-block|syntax|javascript|1505589|entityMap|0|LINK|mutability|MUTABLE|url|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat^0|4|8|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Latest with ES6 <a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat" rel="nofollow noreferrer">repeat()</a> method:

<pre><code> const length_required = 5;
 let num = 10;
 num = "0".repeat(length_required - String(num).length) + num;
 console.log(num)
 // output: 00010

 let num = 1000;
 num = "0".repeat(length_required - String(num).length) + num;
 console.log(num)
 // output: 01000
</code></pre>

blocks|key|1291022|text|我知道这个问题有点老生常谈了，但对于任何想在Java或Python上寻找类似字符串格式化的东西的人，我有以下帮助方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1291023|String.format+=+(...args)+=>+{
++++if(+args.length+==+0+){
++++++++throw+new+Error("String+format+error:+You+must+provide+at+least+one+argument");
++++}
++++const+delimiter+=+"@LIMIT*";
++++const+format+=+String(args.shift(1,0)).replace(/(%25[0-9]{0,}[sd])/g,+delimiter%2B"$1"%2Bdelimiter).split(delimiter);+//+First+element+is+the+format
++++if(+[...format].filter(el=>el.indexOf("%25")>-1).length+!=+args.length+){
++++++++throw+new+Error("String+format+error:+Arguments+must+match+pattern");
++++}
++++if(+format.length+==+1+&&+args.length+==+0+){
++++++++return+String(format);
++++}
++++let+formattedString+=+"";
++++//+patterns
++++const+decimalPattern+=+/%25[0-9]{0,}d/;
++++const+stringPattern++=+/%25s/;
++++if(+format.length+==+0+){
++++++++throw+new+Error("String+format+error:+Invalid+format");
++++}
++++let+value++++++++=+null;
++++let+indexOfParam+=+0;
++++let+currPattern++=+null;
++++while(+args.length+>+0+)+{
++++++++currPattern+=+format[indexOfParam];
++++++++indexOfParam%2B%2B;
++++++++if(+currPattern.indexOf("%25")<0+){
++++++++++++formattedString%2B=currPattern;
++++++++++++continue;
++++++++}
++++++++value+=+args.shift(0,1);
++++++++if(+decimalPattern.test(currPattern)+){
++++++++++++let+numberLength+=+parseInt(currPattern.replace(/[%5E0-9]/g,''));
++++++++++++if(+isNaN(numberLength)+){
++++++++++++++++numberLength+=+0;
++++++++++++}
++++++++++++formattedString%2B=numberToLength(value,+numberLength);
++++++++}+else+if(+stringPattern.test(currPattern)+)+{
++++++++++++if(+typeof+value+===+'object'+&&+value.toSource+){
++++++++++++++++formattedString%2B=String(value.toSource());
++++++++++++}+else+{
++++++++++++++++formattedString%2B=String(value);
++++++++++++}
++++++++}+else+{
++++++++++++throw+new+Error("String+format+error:+Unrecognized+pattern:"%2BcurrPattern);
++++++++}
++++}
++++return+formattedString;
}

const+numberToLength+=+(number,+length)+=>+{
++++length+=+parseInt(length);
++++number+=+String(number);
++++if(+isNaN(length)+%7C%7C+isNaN(parseInt(number))+){
++++++++throw+new+Error("Invalid+number+passed");
++++}
++++while(+number.length+<+length+)+{
++++++++number+=+"0"+%2B+number;
++++}
++++return+number;
}|code-block|syntax|javascript|1291024|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|$]]

I know this question is kind of old, but for anyone looking for something similar to String formatting on Java or Python, I have these helper methods:

<pre><code>String.format = (...args) =&gt; {
 if( args.length == 0 ){
 throw new Error("String format error: You must provide at least one argument");
 }
 const delimiter = "@LIMIT*";
 const format = String(args.shift(1,0)).replace(/(%[0-9]{0,}[sd])/g, delimiter+"$1"+delimiter).split(delimiter); // First element is the format
 if( [...format].filter(el=&gt;el.indexOf("%")&gt;-1).length != args.length ){
 throw new Error("String format error: Arguments must match pattern");
 }
 if( format.length == 1 &amp;&amp; args.length == 0 ){
 return String(format);
 }
 let formattedString = "";
 // patterns
 const decimalPattern = /%[0-9]{0,}d/;
 const stringPattern = /%s/;
 if( format.length == 0 ){
 throw new Error("String format error: Invalid format");
 }
 let value = null;
 let indexOfParam = 0;
 let currPattern = null;
 while( args.length &gt; 0 ) {
 currPattern = format[indexOfParam];
 indexOfParam++;
 if( currPattern.indexOf("%")&lt;0 ){
 formattedString+=currPattern;
 continue;
 }
 value = args.shift(0,1);
 if( decimalPattern.test(currPattern) ){
 let numberLength = parseInt(currPattern.replace(/[^0-9]/g,''));
 if( isNaN(numberLength) ){
 numberLength = 0;
 }
 formattedString+=numberToLength(value, numberLength);
 } else if( stringPattern.test(currPattern) ) {
 if( typeof value === 'object' &amp;&amp; value.toSource ){
 formattedString+=String(value.toSource());
 } else {
 formattedString+=String(value);
 }
 } else {
 throw new Error("String format error: Unrecognized pattern:"+currPattern);
 }
 }
 return formattedString;
}

const numberToLength = (number, length) =&gt; {
 length = parseInt(length);
 number = String(number);
 if( isNaN(length) || isNaN(parseInt(number)) ){
 throw new Error("Invalid number passed");
 }
 while( number.length &lt; length ) {
 number = "0" + number;
 }
 return number;
}
</code></pre>

I have a number in Javascript, that I know is less than 10000 and also non-negative. I want to display it as a four-digit number, with leading zeroes. Is there anything more elegant than the following?

<pre><code>if(num&lt;10) num="000"+num;
else if(num&lt;100) num="00"+num;
else if(num&lt;1000) num="0"+num;
</code></pre>

I want something that is built into Javascript, but I can't seem to find anything.

How can I format an integer to a specific length in javascript?

我在Javascript中有一个数字，我知道它小于10000，而且也是非负的。我想将它显示为一个四位数的数字，并带有前导零。有什么比下面的更优雅的吗？if(num<10) num="000"+num;else if(num<100) num="00"+num;else if(num<1000) num="0"+num;...

问如何在javascript中将整数格式化为特定长度？
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问如何在javascript中将整数格式化为特定长度？
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