blocks|key|2274070|text|Javascript中的%2B操作符根据操作数的不同，用于加法和级联。0000%2B1只是将0和1相加，这就给出了1。您可以将字符串"000"连接到数字1，它将为您提供字符串"0001"，这正是您要寻找的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2274071|这里有一个函数，它总是有4位数字，带有前导零。它的工作方式是将字符串"000"和您的数字连接起来，然后在需要时将其缩短为4个字符。|2274072|function+pad(n)+{
++++var+s+=+"000"+%2B+n;
++++return+s.substr(s.length-4);
}|code-block|syntax|javascript|2274073|entityMap^0|C|1|X|6|1Q|5|20|1|2B|6|0|Y|5|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

The <code>+</code> operator in Javascript is used for both addition and concatenating depending on the operands. <code>0000+1</code> is just adding zero and one, which gives one. You can concatenate the string <code>"000"</code> to the number <code>1</code> and it will give you the string <code>"0001"</code>, which is what you're looking for.

Here's a function that will always have 4 digits, with leading zeros. It works by concatenating the string <code>"000"</code> and your number, then shortening, if needed, to 4 characters.

<pre><code>function pad(n) {
 var s = "000" + n;
 return s.substr(s.length-4);
}
</code></pre>

blocks|key|2274013|text|你可以使用padStart|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|2274014|你可以像这样使用它：|2274015|var+count+=+1;
_.padStart(count,+4,+'0');
//this+will+give+you+0001
_.padStart(23,+4,+'0');
//this+will+give+you+0023|code-block|syntax|javascript|2274016|entityMap|0|LINK|mutability|MUTABLE|url|https://lodash.com/docs#padStart^0|5|8|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|C|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

You can use lodash's <a href="https://lodash.com/docs#padStart" rel="nofollow">padStart</a> 

You can use it like:

<pre><code>var count = 1;
_.padStart(count, 4, '0');
//this will give you 0001
_.padStart(23, 4, '0');
//this will give you 0023
</code></pre>

blocks|key|246731|text|使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|246732|var+count+=+1;
document.write(+'0000'.substr(+String(count).length+)+%2B+count++);|code-block|syntax|javascript|246733|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|$]]

Use:

<pre><code>var count = 1;
document.write( '0000'.substr( String(count).length ) + count );
</code></pre>

blocks|key|151543|text|使用整数是不可能的，但是您可以这样做，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|151544|"000"%2B1|151545|输出=>+"0001“|151546|也许如果你能分享你的用例，我可以帮你更好。|151547|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

This is not possible using integer, but you can do something like,

"000"+1

output => "0001"

P.S: Maybe if you share your use case, I can help you better.

blocks|key|2274090|text|添加零，然后删除剩余的0：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2274091|function+renameArtboards(){++
+++++var+docRef+=+app.activeDocument;++
+++++var+aB+=+docRef.artboards;++
+++++for(var+a=0;a<aB.length;a%2B%2B){++
++++++++++var+curAB+=+aB[a];++
++++++++++var+sNumber+=+"000"+%2B+[a%2B1];
++++++++++curAB.name+=+sNumber.substr(sNumber.length-4);
+++++}++
}++
renameArtboards();++|code-block|syntax|javascript|2274092|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|$]]

Add the zeros and then remove the ones that are left over:

<pre><code>function renameArtboards(){ 
 var docRef = app.activeDocument; 
 var aB = docRef.artboards; 
 for(var a=0;a&lt;aB.length;a++){ 
 var curAB = aB[a]; 
 var sNumber = "000" + [a+1];
 curAB.name = sNumber.substr(sNumber.length-4);
 } 
} 
renameArtboards(); 
</code></pre>

blocks|key|150429|text|我知道我迟到了一年半，但我现在恰好需要这种逻辑在我的应用中。在我看来，最“优雅”的解决方案是这样一个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|150430|const+yourNumber+=+32;
const+defaultFill+=+"0000"

const+formattedNumber+=+(defaultFill+%2B+yourNumber).substr(String(yourNumber).length)|code-block|syntax|javascript|150431|对于我的用例来说，这个很好。如果您必须处理超过4位数的数字，则可能会有一些限制--在这种情况下，您可能会在途中丢失一些信息。要注意这一点。|150432|希望这能帮到你，干杯！|150433|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I know I'm an year and a half late but I happen to need this kind of logic in my app right now. The most "elegant" solution in my opinion is this one:

<pre class="lang-js prettyprint-override"><code>const yourNumber = 32;
const defaultFill = "0000"

const formattedNumber = (defaultFill + yourNumber).substr(String(yourNumber).length)
</code></pre>

This one works just fine for my use-case. It may have some limits if you have to handle numbers with more than 4 digits – in that case you could lose some information on the way. Be aware of that.

I hope this helped you, cheers!

blocks|key|2274138|text|var+JobNumber+=+'JB-00000';
var+digistextraction+=+JobNumber.substr(3,5);

var+JobNumberTable=[];
for+(var+j+=+0;+j+<+5;+j%2B%2B)+{
++++k=1;
++++digistextraction+=+parseInt(digistextraction)+%2B+k;
++++JobNumberTable.push('JB-'+%2B+digistextraction.toLocaleString(undefined,+{useGrouping:+false,+minimumIntegerDigits:+5}));
++++k%2B%2B;
}
console.log(JobNumberTable);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2274139|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>var JobNumber = 'JB-00000';
var digistextraction = JobNumber.substr(3,5);

var JobNumberTable=[];
for (var j = 0; j &lt; 5; j++) {
 k=1;
 digistextraction = parseInt(digistextraction) + k;
 JobNumberTable.push('JB-' + digistextraction.toLocaleString(undefined, {useGrouping: false, minimumIntegerDigits: 5}));
 k++;
}
console.log(JobNumberTable);
</code></pre>

blocks|key|246735|text|最好的方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|246736|var+NumberWithLeadingZeroes+=+"000"+%2B+1;
alert(NumberWithLeadingZeroes);//Will+give+you+"0001"|code-block|syntax|javascript|246737|更新|offset|length|style|BOLD|246738|function+NumberWithLeadingZeroes+(n)
{
++++if+(+n<+10+)
++++{
++++++++return+(+'000'+%2B+n.toString+()+);
++++}
++++else+if+(+n<+100+)
++++{
++++++++return+(+'00'+%2B+n.toString+()+);
++++}
++++else+if+(+n<+1000+)
++++{
++++++++return+(+'0'+%2B+n.toString+()+);
++++}
++++else
++++{
++++++++return+(+n);
++++}
}|246739|entityMap^0|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Best way is

<pre><code>var NumberWithLeadingZeroes = "000" + 1;
alert(NumberWithLeadingZeroes);//Will give you "0001"
</code></pre>

UPDATE

<pre><code>function NumberWithLeadingZeroes (n)
{
 if ( n&lt; 10 )
 {
 return ( '000' + n.toString () );
 }
 else if ( n&lt; 100 )
 {
 return ( '00' + n.toString () );
 }
 else if ( n&lt; 1000 )
 {
 return ( '0' + n.toString () );
 }
 else
 {
 return ( n);
 }
}
</code></pre>

I want my output as 0000, 0001, 0002. If i use normal js math then the result come 0,1,2 

<pre><code>var count=0000+1;
document.write(count);
</code></pre>

Output is 1

But i want output as 0001

how to do this?

How to show number 4 digit (0000,0001) number using JavaScript

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我要我的输出值为0000,0001,0002。如果我使用普通的js数学，结果是0,1,2。var count=0000+1;document.write(count);输出为1但我希望输出为0001怎么做？

问如何使用JavaScript显示数字4位数(00000,0001)
EN

回答 8

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