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|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|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|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|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|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>


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>


<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。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

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
EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用JavaScript遍历日期范围EN

回答 8

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用JavaScript遍历日期范围
EN