blocks|key|859094|text|如果startDate和endDate确实是date对象，则可以将它们转换为自1970年1月1日午夜以来的毫秒数，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|859095|var+startTime+=+startDate.getTime(),+endTime+=+endDate.getTime();|code-block|syntax|javascript|859096|然后，您可以从一个循环到另一个循环，将loopTime递增86400000+(1000*60*60*24)+-一天中的毫秒数：|859097|for(loopTime+=+startTime;+loopTime+<+endTime;+loopTime+%2B=+86400000)
{
++++var+loopDay=new+Date(loopTime)
++++//use+loopDay+as+you+wish
}|859098|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|$]]

<p>If startDate and endDate are indeed date objects you could convert them to number of milliseconds since midnight Jan 1, 1970, like this:</p>

<pre><code>var startTime = startDate.getTime(), endTime = endDate.getTime();
</code></pre>

<p>Then you could loop from one to another incrementing loopTime by 86400000 (1000*60*60*24) - number of milliseconds in one day:</p>

<pre><code>for(loopTime = startTime; loopTime &lt; endTime; loopTime += 86400000)
{
    var loopDay=new Date(loopTime)
    //use loopDay as you wish
}
</code></pre>


blocks|key|859136|text|基于Jayarjo的回答：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|859137|var+loopDate+=+new+Date();
loopDate.setTime(datFrom.valueOf());

while+(loopDate.valueOf()+<+datTo.valueOf()+%2B+86400000)+{

++++alert(loopDay);

++++loopDate.setTime(loopDate.valueOf()+%2B+86400000);
}|code-block|syntax|javascript|859138|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|$]]

<p>Based on Jayarjo's answer:</p>

<pre><code>var loopDate = new Date();
loopDate.setTime(datFrom.valueOf());

while (loopDate.valueOf() &lt; datTo.valueOf() + 86400000) {

    alert(loopDay);

    loopDate.setTime(loopDate.valueOf() + 86400000);
}
</code></pre>


blocks|key|1073716|text|根据Tom+Gullen的回答。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1073717|var+start+=+new+Date("02/05/2013");
var+end+=+new+Date("02/10/2013");


var+loop+=+new+Date(start);
while(loop+<=+end){
+++alert(loop);+++++++++++

+++var+newDate+=+loop.setDate(loop.getDate()+%2B+1);
+++loop+=+new+Date(newDate);
}|code-block|syntax|javascript|1073718|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|$]]

<p>Based on Tom Gullen´s answer.</p>

<pre><code>var start = new Date("02/05/2013");
var end = new Date("02/10/2013");


var loop = new Date(start);
while(loop &lt;= end){
   alert(loop);           

   var newDate = loop.setDate(loop.getDate() + 1);
   loop = new Date(newDate);
}
</code></pre>


blocks|key|1073731|text|var+start+=+new+Date("2014-05-01");+//yyyy-mm-dd
var+end+=+new+Date("2014-05-05");+//yyyy-mm-dd

while(start+<=+end){

++++var+mm+=+((start.getMonth()%2B1)>=10)?(start.getMonth()%2B1):'0'%2B(start.getMonth()%2B1);
++++var+dd+=+((start.getDate())>=10)?+(start.getDate())+:+'0'+%2B+(start.getDate());
++++var+yyyy+=+start.getFullYear();
++++var+date+=+dd%2B"/"%2Bmm%2B"/"%2Byyyy;+//yyyy-mm-dd

++++alert(date);+

++++start+=+new+Date(start.setDate(start.getDate()+%2B+1));+//date+increase+by+1
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1073732|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 start = new Date("2014-05-01"); //yyyy-mm-dd
var end = new Date("2014-05-05"); //yyyy-mm-dd

while(start &lt;= end){

    var mm = ((start.getMonth()+1)&gt;=10)?(start.getMonth()+1):'0'+(start.getMonth()+1);
    var dd = ((start.getDate())&gt;=10)? (start.getDate()) : '0' + (start.getDate());
    var yyyy = start.getFullYear();
    var date = dd+"/"+mm+"/"+yyyy; //yyyy-mm-dd

    alert(date); 

    start = new Date(start.setDate(start.getDate() + 1)); //date increase by 1
}
</code></pre>


blocks|key|1073780|text|根据Tabare的回答，我不得不在最后再加一天，因为之前的周期被削减了|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1073781|var+start+=+new+Date("02/05/2013");
var+end+=+new+Date("02/10/2013");
var+newend+=+end.setDate(end.getDate()%2B1);
var+end+=+new+Date(newend);
while(start+<+end){
+++alert(start);+++++++++++

+++var+newDate+=+start.setDate(start.getDate()+%2B+1);
+++start+=+new+Date(newDate);
}|code-block|syntax|javascript|1073782|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|$]]

<p>Based on Tabare's Answer, 
I had to add one more day at the end, since the cycle is cut before</p>

<pre><code>var start = new Date("02/05/2013");
var end = new Date("02/10/2013");
var newend = end.setDate(end.getDate()+1);
var end = new Date(newend);
while(start &lt; end){
   alert(start);           

   var newDate = start.setDate(start.getDate() + 1);
   start = new Date(newDate);
}
</code></pre>


blocks|key|1073836|text|这里有简单的工作代码，对我很有效|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1073837|​|1073838|var+from+=+new+Date(2012,0,1);
var+to+=+new+Date(2012,1,20);
++++
//+loop+for+every+day
for+(var+day+=+from;+day+<=+to;+day.setDate(day.getDate()+%2B+1))+{
++++++
+++//+your+day+is+here

}|code-block|syntax|javascript|1073839|1073840|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

<p>Here simple working code, worked for me</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>var from = new Date(2012,0,1);
var to = new Date(2012,1,20);
    
// loop for every day
for (var day = from; day &lt;= to; day.setDate(day.getDate() + 1)) {
      
   // your day is here

}</code></pre>
</div>
</div>
</p>


blocks|key|1073899|text|我想我找到了一个更简单的答案，如果你允许自己使用Moment.js|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1073900|​|1073901|//+cycle+through+last+five+days,+today+included
//+you+could+also+cycle+through+any+dates+you+want,+mostly+for
//+making+this+snippet+not+time+aware
const+currentMoment+=+moment().subtract(4,+'days');
const+endMoment+=+moment().add(1,+'days');
while+(currentMoment.isBefore(endMoment,+'day'))+{
++console.log(`Loop+at+${currentMoment.format('YYYY-MM-DD')}`);
++currentMoment.add(1,+'days');
}|code-block|syntax|javascript|1073902|<script+src="https://cdn.jsdelivr.net/npm/moment@2/moment.min.js"></script>|1073903|1073904|entityMap|0|LINK|mutability|MUTABLE|url|https://momentjs.com/^0|O|9|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|6|7|Z|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|10|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|H|7|11|8|@]|9|@]|C|$I|J]]|$1|M|3|E|5|6|7|12|8|@]|9|@]|C|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|O|$P|$5|Q|R|S|C|$T|U]]]]

<p>I think I found an even simpler answer, if you allow yourself to use <a href="https://momentjs.com/" rel="noreferrer">Moment.js</a>:</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>// cycle through last five days, today included
// you could also cycle through any dates you want, mostly for
// making this snippet not time aware
const currentMoment = moment().subtract(4, 'days');
const endMoment = moment().add(1, 'days');
while (currentMoment.isBefore(endMoment, 'day')) {
  console.log(`Loop at ${currentMoment.format('YYYY-MM-DD')}`);
  currentMoment.add(1, 'days');
}</code></pre>
<pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;script src="https://cdn.jsdelivr.net/npm/moment@2/moment.min.js"&gt;&lt;/script&gt;</code></pre>
</div>
</div>
</p>


blocks|key|1073966|text|如果您想要一种高效的毫秒级方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1073967|var+daysOfYear+=+[];
for+(var+d+=+begin;+d+<=+end;+d+=+d+%2B+86400000)+{
++++daysOfYear.push(new+Date(d));
}|code-block|syntax|javascript|1073968|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|$]]

<p>If you want an efficient way with milliseconds:</p>

<pre><code>var daysOfYear = [];
for (var d = begin; d &lt;= end; d = d + 86400000) {
    daysOfYear.push(new Date(d));
}
</code></pre>


<p>Given two <code>Date()</code> objects, where one is less than the other, how do I loop every day between the dates?</p>

<pre><code>for(loopDate = startDate; loopDate &lt; endDate; loopDate += 1)
{

}
</code></pre>

<p>Would this sort of loop work?  But how can I add one day to the loop counter?</p>

<p>Thanks!</p>


Loop through a date range with 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。

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给定两个Date()对象，其中一个小于另一个，我如何在这两个日期之间每天循环？for(loopDate = startDate; loopDate < endDate; loopDate += 1){}这样的循环会起作用吗？但是我怎么才能给循环计数器增加一天呢？谢谢!

问使用JavaScript遍历日期范围
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