blocks|key|3145650|text|要合并您链接的内容和您的链接，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145651|for+(i+=+1;+i+<+1050;+i%2B%2B)+{
++var+str+=+""+%2B+i
++var+pad+=+"0000"
++str+=+pad.substring(0,+pad.length+-+str.length)+%2B+str
++images.push('/images/'+%2B+str+%2B+'.jpg');
}|code-block|syntax|javascript|3145652|我希望这对你有帮助|3145653|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|$]]

To combine what you linked and yours:

<pre><code>for (i = 1; i &lt; 1050; i++) {
 var str = "" + i
 var pad = "0000"
 str = pad.substring(0, pad.length - str.length) + str
 images.push('/images/' + str + '.jpg');
}
</code></pre>

I hope that helps

blocks|key|1016578|text|您可以创建一个实用函数，用于将整数转换为4位字符串表示形式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1016579|function+intTo4digitsString(nb)+{
+++if(nb+>+0+&&+nb+<+10)
++++++++return+"000"%2Bnb;
+++else+if(nb+>=+10+&&+nb+<+100)
++++++++return+"00"%2Bnb;
+++else+if(nb+>=+100+&&+nb+<+1000)
++++++++return+"0"%2Bnb;
}|code-block|syntax|javascript|1016580|这个版本限制为4位数字(并且不像其他答案那样容易扩展)，但我认为它对您更具可读性:)然后您可以使用：|1016581|for+(i+=+1;+i+<+1050;+i%2B%2B)+{
++++images.push('/images/'+%2B+inTo4DigitsString(i)+%2B+'.jpg');
}|1016582|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can make a utility function which convert a integer to a 4 digits string representation:

<pre><code>function intTo4digitsString(nb) {
 if(nb &gt; 0 &amp;&amp; nb &lt; 10)
 return "000"+nb;
 else if(nb &gt;= 10 &amp;&amp; nb &lt; 100)
 return "00"+nb;
 else if(nb &gt;= 100 &amp;&amp; nb &lt; 1000)
 return "0"+nb;
}
</code></pre>

This version is limited to 4 digits (and not as easily extensible than other answers), but I think it's more readable fr you :) Then you can use:

<pre><code>for (i = 1; i &lt; 1050; i++) {
 images.push('/images/' + inTo4DigitsString(i) + '.jpg');
}
</code></pre>

blocks|key|4612995|text|我会做这样的事情|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4612996|for+(i+=+1;+i+<+1050;+i%2B%2B)
{
++++var+name+=+'0000'+%2B+i;
++++images.push('/images/'+%2B+name.substr(name.length+-+4)+%2B+'.jpg');
}|code-block|syntax|javascript|4612997|小提琴：http://jsfiddle.net/4z5Sd/|offset|length|4612998|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/4z5Sd/^0|0|0|4|Q|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

I would do something like this

<pre><code>for (i = 1; i &lt; 1050; i++)
{
 var name = '0000' + i;
 images.push('/images/' + name.substr(name.length - 4) + '.jpg');
}
</code></pre>

fiddle: <a href="http://jsfiddle.net/4z5Sd/" rel="nofollow">http://jsfiddle.net/4z5Sd/</a>

blocks|key|829722|text|谢谢，mattsjo。必须将'pad‘添加到数组中才能完全工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|829723|for+(i+=+1;+i+<+1050;+i%2B%2B)+{
++var+str+=+""+%2B+i
++var+pad+=+"0000"
++var+padd+=+pad.substring(0,+pad.length+-+str.length)+%2B+str
++images.push('/images/'+%2B+padd+%2B+'.jpg');
}|code-block|syntax|javascript|829724|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|$]]

Thanks mattsjo. Had to add 'pad' to array for it to work fully.

<pre><code>for (i = 1; i &lt; 1050; i++) {
 var str = "" + i
 var pad = "0000"
 var padd = pad.substring(0, pad.length - str.length) + str
 images.push('/images/' + padd + '.jpg');
}
</code></pre>

blocks|key|1010464|text|如果您需要一个通用的字符串格式化函数，那么sprintf库可能适合您。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1010465|var+i,+images+=+[];

for+(i+=+1;+i+<=+1050;+i%2B%2B)+{
++++images.push(sprintf('/images/%2504d.jpg',+i));
}|code-block|syntax|javascript|1010466|JSFiddle|1010467|entityMap|0|LINK|mutability|MUTABLE|url|http://www.diveintojavascript.com/projects/javascript-sprintf|1|http://jsfiddle.net/UvjzP/^0|L|7|L|7|0|0|0|0|8|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@$9|14|A|15|1|16]]|E|$]]|$1|M|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

If you want a general purpose string formatting function then the <a href="http://www.diveintojavascript.com/projects/javascript-sprintf" rel="nofollow"><code>sprintf</code></a> library might work for you.

<pre><code>var i, images = [];

for (i = 1; i &lt;= 1050; i++) {
 images.push(sprintf('/images/%04d.jpg', i));
}
</code></pre>

<a href="http://jsfiddle.net/UvjzP/" rel="nofollow">JSFiddle</a>

blocks|key|829664|text|zeropad+=+function(+n,l+){+
++++var+ret,+d,+i;
++++ret+=+n+===+null+%7C%7C+n+===+undefined+?+''+:+n.toString();
++++//+do+we+need+to+pad+or+truncate?
++++d+=+l+-+ret.length;
++++if+(d>0)+{
++++++++for+(i=0;i<d;+i%2B%2B)+{
++++++++++++ret+=+'0'%2Bret;
++++++++}
++++}
++++return(ret);
};
images.push('/images/'%2Bzeropad(i,4)+%2B+'.jpg');|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|829665|来自我前段时间写的一个库。https://github.com/deitch/jsorm-utilities|unstyled|offset|length|829666|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/deitch/jsorm-utilities^0|0|D|15|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>zeropad = function( n,l ){ 
 var ret, d, i;
 ret = n === null || n === undefined ? '' : n.toString();
 // do we need to pad or truncate?
 d = l - ret.length;
 if (d&gt;0) {
 for (i=0;i&lt;d; i++) {
 ret = '0'+ret;
 }
 }
 return(ret);
};
images.push('/images/'+zeropad(i,4) + '.jpg');
</code></pre>

From a library I wrote a while back. <a href="https://github.com/deitch/jsorm-utilities" rel="nofollow">https://github.com/deitch/jsorm-utilities</a>

blocks|key|3145710|text|function+padNum(num,length)
{
++++return+Array((length%2B1)-num.toString().length).join("0")%2Bnum;
}

padNum(1,4);//0001
padNum(250,4);//0250|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3145711|然后在你的代码中：|unstyled|3145712|images.push('/images/'+%2B+padNum(i,4)+%2B+'.jpg');|3145713|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>function padNum(num,length)
{
 return Array((length+1)-num.toString().length).join("0")+num;
}

padNum(1,4);//0001
padNum(250,4);//0250
</code></pre>

And then in your code:

<pre><code>images.push('/images/' + padNum(i,4) + '.jpg');
</code></pre>

blocks|key|3145717|text|不如..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145718|for+(i+=+1;+i+<+1050;+i%2B%2B)+{++++++++
++++images.push('/images/'+%2B+('0000'.substring(0,+i.toString().length)%2Bi)+%2B+'.jpg');
}|code-block|syntax|javascript|3145719|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...

<pre><code>for (i = 1; i &lt; 1050; i++) { 
 images.push('/images/' + ('0000'.substring(0, i.toString().length)+i) + '.jpg');
}
</code></pre>

blocks|key|3145721|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145722|for+(var+i+=+1;i<1050;i%2B%2B){
++++images.push("/images/"%2B(Array(5-(""%2Bi).length)).join("0")%2Bi%2B".jpg");
}+|code-block|syntax|javascript|3145723|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|$]]

What about this:

<pre><code>for (var i = 1;i&lt;1050;i++){
 images.push("/images/"+(Array(5-(""+i).length)).join("0")+i+".jpg");
} 
</code></pre>

blocks|key|3145736|text|我建议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145737|Number.prototype.leftPad+=+function+(len)+{
++++var+l+=+this.toString().length,
++++++++d+=+len+-+l;
++++return+new+Array(d+%2B+1).join('0')+%2B+this.toString();
};

for+(var+i+=+1,+len+=+1051;+i+<+len;+i%2B%2B)+{
++++console.log(i.leftPad(4)+%2B+'.jpg');
}|code-block|syntax|javascript|3145738|JS+Fiddle+demo。|offset|length|3145739|参考文献：|3145740|3145741|Array.|unordered-list-item|3145742|Array.join().|style|CODE|3145743|Number.toString().|3145744|3145745|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/davidThomas/PxC5B/|1|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array|2|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/join|3|https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toString^0|0|0|0|E|0|0|0|0|0|5|1|0|0|C|0|C|2|0|0|H|0|H|3|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1D|8|@]|9|@$I|1E|J|1F|1|1G]]|A|$]]|$1|K|3|L|5|6|7|1H|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|1J|8|@]|9|@$I|1K|J|1L|1|1M]]|A|$]]|$1|Q|3|R|5|P|7|1N|8|@$I|1O|J|1P|S|T]]|9|@$I|1Q|J|1R|1|1S]]|A|$]]|$1|U|3|V|5|P|7|1T|8|@$I|1U|J|1V|S|T]]|9|@$I|1W|J|1X|1|1Y]]|A|$]]|$1|W|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|20|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]|15|$5|10|11|12|A|$13|16]]|17|$5|10|11|12|A|$13|18]]|19|$5|10|11|12|A|$13|1A]]]]

I'd suggest:

<pre><code>Number.prototype.leftPad = function (len) {
 var l = this.toString().length,
 d = len - l;
 return new Array(d + 1).join('0') + this.toString();
};

for (var i = 1, len = 1051; i &lt; len; i++) {
 console.log(i.leftPad(4) + '.jpg');
}
</code></pre>

<a href="http://jsfiddle.net/davidThomas/PxC5B/" rel="nofollow">JS Fiddle demo</a>.

References:

<ul>
<li><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array" rel="nofollow">Array</a>.</li>
<li><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/join" rel="nofollow"><code>Array.join()</code></a>.</li>
<li><a href="https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toString" rel="nofollow"><code>Number.toString()</code></a>.</li>
</ul>

blocks|key|4613044|text|function+pad(val,+len)+{
++++return+("000000000000000000"+%2B+val).substr(len+*+-1);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4613045|用法：|unstyled|4613046|console.log(pad(1,+4)+%2B+".jpg");
>+0001.jpg|4613047|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>function pad(val, len) {
 return ("000000000000000000" + val).substr(len * -1);
}
</code></pre>

Usage:

<pre><code>console.log(pad(1, 4) + ".jpg");
&gt; 0001.jpg
</code></pre>

I have an array in javascript. It specifies URLs for images 1-1050. My images must be named 0001.jpg, 0002.jpg ect... However my script is calling up 1.jpg, 2.jpg... 
I am very new to JS and am struggling to incorporate an answer from <a href="https://stackoverflow.com/questions/5366849/convert-1-to-0001-in-javascript">here</a> with my script.

<pre><code>for (i = 0001; i &lt; 1050; i++) {
 images.push('/images/' + i + '.jpg');
}
</code></pre>

Converting 1 to 0001 in javascript array

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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我在javascript中有一个数组。它指定图像1-1050的URL。我的图片必须命名为0001.jpg，0002.jpg等...但是我的脚本调用的是1.jpg，2.jpg...我对JS非常陌生，正在努力将来自的答案与我的脚本结合起来。for (i = 0001; i < 1050; i++) { images.push('/images/' + i + '.jpg');}

问在javascript数组中将1转换为0001
EN

回答 11

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问在javascript数组中将1转换为0001EN

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问在javascript数组中将1转换为0001
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