blocks|key|2538008|text|const+oneDay+=+24+*+60+*+60+*+1000;+//+hours*minutes*seconds*milliseconds
const+firstDate+=+new+Date(2008,+1,+12);
const+secondDate+=+new+Date(2008,+1,+22);

const+diffDays+=+Math.round(Math.abs((firstDate+-+secondDate)+/+oneDay));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2538009|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>const oneDay = 24 * 60 * 60 * 1000; // hours*minutes*seconds*milliseconds
const firstDate = new Date(2008, 1, 12);
const secondDate = new Date(2008, 1, 22);

const diffDays = Math.round(Math.abs((firstDate - secondDate) / oneDay));
</code></pre>

blocks|key|2538002|text|下面是一个执行此操作的函数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2538003|function+days_between(date1,+date2)+{

++++//+The+number+of+milliseconds+in+one+day
++++const+ONE_DAY+=+1000+*+60+*+60+*+24;

++++//+Calculate+the+difference+in+milliseconds
++++const+differenceMs+=+Math.abs(date1+-+date2);

++++//+Convert+back+to+days+and+return
++++return+Math.round(differenceMs+/+ONE_DAY);

}|code-block|syntax|javascript|2538004|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|$]]

Here is a function that does this:

<pre><code>function days_between(date1, date2) {

 // The number of milliseconds in one day
 const ONE_DAY = 1000 * 60 * 60 * 24;

 // Calculate the difference in milliseconds
 const differenceMs = Math.abs(date1 - date2);

 // Convert back to days and return
 return Math.round(differenceMs / ONE_DAY);

}
</code></pre>

blocks|key|2748024|text|下面是我使用的代码。如果只减去日期，它将无法跨越夏令时界限(例如，4月1日至4月30日或10月1日至10月31日)。这减少了所有小时，以确保您有一天，并通过使用UTC消除了任何DST问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2748025|var+nDays+=+(++++Date.UTC(EndDate.getFullYear(),+EndDate.getMonth(),+EndDate.getDate())+-
+++++++++++++++++Date.UTC(StartDate.getFullYear(),+StartDate.getMonth(),+StartDate.getDate()))+/+86400000;|code-block|syntax|javascript|2748026|作为函数：|2748027|function+DaysBetween(StartDate,+EndDate)+{
++//+The+number+of+milliseconds+in+all+UTC+days+(no+DST)
++const+oneDay+=+1000+*+60+*+60+*+24;

++//+A+day+in+UTC+always+lasts+24+hours+(unlike+in+other+time+formats)
++const+start+=+Date.UTC(EndDate.getFullYear(),+EndDate.getMonth(),+EndDate.getDate());
++const+end+=+Date.UTC(StartDate.getFullYear(),+StartDate.getMonth(),+StartDate.getDate());

++//+so+it's+safe+to+divide+by+24+hours
++return+(start+-+end)+/+oneDay;
}|2748028|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|$]]

Here's what I use. If you just subtract the dates, it won't work across the Daylight Savings Time Boundary (eg April 1 to April 30 or Oct 1 to Oct 31). This drops all the hours to make sure you get a day and eliminates any DST problem by using UTC.

<pre><code>var nDays = ( Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) -
 Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;
</code></pre>

as a function:

<pre><code>function DaysBetween(StartDate, EndDate) {
 // The number of milliseconds in all UTC days (no DST)
 const oneDay = 1000 * 60 * 60 * 24;

 // A day in UTC always lasts 24 hours (unlike in other time formats)
 const start = Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate());
 const end = Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate());

 // so it's safe to divide by 24 hours
 return (start - end) / oneDay;
}
</code></pre>

blocks|key|2748066|text|下面是我的实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2748067|function+daysBetween(one,+another)+{
++return+Math.round(Math.abs((%2Bone)+-+(%2Banother))/8.64e7);
}|code-block|syntax|javascript|2748068|%2B<date>对整数表示进行类型强制，并具有与<date>.getTime()相同的效果，而8.64e7是一天中的毫秒数。|offset|length|style|CODE|2748069|entityMap^0|0|0|0|7|N|G|1A|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]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

Here is my implementation:

<pre><code>function daysBetween(one, another) {
 return Math.round(Math.abs((+one) - (+another))/8.64e7);
}
</code></pre>

<code>+&lt;date&gt;</code> does the type coercion to the integer representation and has the same effect as <code>&lt;date&gt;.getTime()</code> and <code>8.64e7</code> is the number of milliseconds in a day.

blocks|key|2538017|text|调整以允许夏令时差异。试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2538018|++function+daysBetween(date1,+date2)+{

+//+adjust+diff+for+for+daylight+savings
+var+hoursToAdjust+=+Math.abs(date1.getTimezoneOffset()+/60)+-+Math.abs(date2.getTimezoneOffset()+/60);
+//+apply+the+tz+offset
+date2.addHours(hoursToAdjust);+

++++//+The+number+of+milliseconds+in+one+day
++++var+ONE_DAY+=+1000+*+60+*+60+*+24

++++//+Convert+both+dates+to+milliseconds
++++var+date1_ms+=+date1.getTime()
++++var+date2_ms+=+date2.getTime()

++++//+Calculate+the+difference+in+milliseconds
++++var+difference_ms+=+Math.abs(date1_ms+-+date2_ms)

++++//+Convert+back+to+days+and+return
++++return+Math.round(difference_ms/ONE_DAY)

}

//+you'll+want+this+addHours+function+too+

Date.prototype.addHours=+function(h){
++++this.setHours(this.getHours()%2Bh);
++++return+this;
}|code-block|syntax|javascript|2538019|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|$]]

Adjusted to allow for daylight saving differences. try this:

<pre><code> function daysBetween(date1, date2) {

 // adjust diff for for daylight savings
 var hoursToAdjust = Math.abs(date1.getTimezoneOffset() /60) - Math.abs(date2.getTimezoneOffset() /60);
 // apply the tz offset
 date2.addHours(hoursToAdjust); 

 // The number of milliseconds in one day
 var ONE_DAY = 1000 * 60 * 60 * 24

 // Convert both dates to milliseconds
 var date1_ms = date1.getTime()
 var date2_ms = date2.getTime()

 // Calculate the difference in milliseconds
 var difference_ms = Math.abs(date1_ms - date2_ms)

 // Convert back to days and return
 return Math.round(difference_ms/ONE_DAY)

}

// you'll want this addHours function too 

Date.prototype.addHours= function(h){
 this.setHours(this.getHours()+h);
 return this;
}
</code></pre>

blocks|key|2538029|text|我为另一个帖子写了这个解决方案，他问道，如何计算两个日期之间的差异，所以我分享了我准备的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2538030|//+Here+are+the+two+dates+to+compare
var+date1+=+'2011-12-24';
var+date2+=+'2012-01-01';

//+First+we+split+the+values+to+arrays+date1[0]+is+the+year,+[1]+the+month+and+[2]+the+day
date1+=+date1.split('-');
date2+=+date2.split('-');

//+Now+we+convert+the+array+to+a+Date+object,+which+has+several+helpful+methods
date1+=+new+Date(date1[0],+date1[1],+date1[2]);
date2+=+new+Date(date2[0],+date2[1],+date2[2]);

//+We+use+the+getTime()+method+and+get+the+unixtime+(in+milliseconds,+but+we+want+seconds,+therefore+we+divide+it+through+1000)
date1_unixtime+=+parseInt(date1.getTime()+/+1000);
date2_unixtime+=+parseInt(date2.getTime()+/+1000);

//+This+is+the+calculated+difference+in+seconds
var+timeDifference+=+date2_unixtime+-+date1_unixtime;

//+in+Hours
var+timeDifferenceInHours+=+timeDifference+/+60+/+60;

//+and+finaly,+in+days+:)
var+timeDifferenceInDays+=+timeDifferenceInHours++/+24;

alert(timeDifferenceInDays);|code-block|syntax|javascript|2538031|你可以跳过代码中的一些步骤，我这样写是为了让它更容易理解。|2538032|你可以在这里找到一个运行的例子：http://jsfiddle.net/matKX/|offset|length|2538033|entityMap|0|LINK|mutability|MUTABLE|url|http://jsfiddle.net/matKX/^0|0|0|0|G|Q|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

I have written this solution for another post who asked, how to calculate the difference between two dates, so I share what I have prepared:

<pre><code>// Here are the two dates to compare
var date1 = '2011-12-24';
var date2 = '2012-01-01';

// First we split the values to arrays date1[0] is the year, [1] the month and [2] the day
date1 = date1.split('-');
date2 = date2.split('-');

// Now we convert the array to a Date object, which has several helpful methods
date1 = new Date(date1[0], date1[1], date1[2]);
date2 = new Date(date2[0], date2[1], date2[2]);

// We use the getTime() method and get the unixtime (in milliseconds, but we want seconds, therefore we divide it through 1000)
date1_unixtime = parseInt(date1.getTime() / 1000);
date2_unixtime = parseInt(date2.getTime() / 1000);

// This is the calculated difference in seconds
var timeDifference = date2_unixtime - date1_unixtime;

// in Hours
var timeDifferenceInHours = timeDifference / 60 / 60;

// and finaly, in days :)
var timeDifferenceInDays = timeDifferenceInHours / 24;

alert(timeDifferenceInDays);
</code></pre>

You can skip some steps in the code, I have written it so to make it easy to understand.

You'll find a running example here: <a href="http://jsfiddle.net/matKX/" rel="noreferrer">http://jsfiddle.net/matKX/</a>

blocks|key|53196|text|来自我的小日期差异计算器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|53197|var+startDate+=+new+Date(2000,+1-1,+1);++//+2000-01-01
var+endDate+=+++new+Date();++++++++++++++//+Today

//+Calculate+the+difference+of+two+dates+in+total+days
function+diffDays(d1,+d2)
{
++var+ndays;
++var+tv1+=+d1.valueOf();++//+msec+since+1970
++var+tv2+=+d2.valueOf();

++ndays+=+(tv2+-+tv1)+/+1000+/+86400;
++ndays+=+Math.round(ndays+-+0.5);
++return+ndays;
}|code-block|syntax|javascript|53198|因此，您可以调用：|53199|var+nDays+=+diffDays(startDate,+endDate);|53200|(完整源代码请访问http://david.tribble.com/src/javascript/jstimespan.html。)|offset|length|53201|附录|style|BOLD|53202|可以通过更改以下行来改进代码：|53203|++var+tv1+=+d1.getTime();++//+msec+since+1970
++var+tv2+=+d2.getTime();|53204|entityMap|0|LINK|mutability|MUTABLE|url|http://david.tribble.com/src/javascript/jstimespan.html^0|0|0|0|0|9|1J|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|18|8|@]|9|@$M|19|N|1A|1|1B]]|A|$]]|$1|O|3|P|5|6|7|1C|8|@$M|1D|N|1E|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

From my little date difference calculator:

<pre><code>var startDate = new Date(2000, 1-1, 1); // 2000-01-01
var endDate = new Date(); // Today

// Calculate the difference of two dates in total days
function diffDays(d1, d2)
{
 var ndays;
 var tv1 = d1.valueOf(); // msec since 1970
 var tv2 = d2.valueOf();

 ndays = (tv2 - tv1) / 1000 / 86400;
 ndays = Math.round(ndays - 0.5);
 return ndays;
}
</code></pre>

So you would call:

<pre><code>var nDays = diffDays(startDate, endDate);
</code></pre>

(Full source at <a href="http://david.tribble.com/src/javascript/jstimespan.html" rel="nofollow">http://david.tribble.com/src/javascript/jstimespan.html</a>.)

Addendum

The code can be improved by changing these lines:

<pre><code> var tv1 = d1.getTime(); // msec since 1970
 var tv2 = d2.getTime();
</code></pre>

<ol>
<li>I am calculating the number of days between the 'from' and 'to' date. For example, if the from date is 13/04/2010 and the to date is 15/04/2010 the result should be </li>
<li>How do I get the result using JavaScript? </li>
</ol>

How to calculate the number of days between two dates?

我正在计算“from”和“to”日期之间的天数。例如，如果开始日期是13/04/2010，结束日期是15/04/2010，则结果应为How do I get the result using JavaScript?

问如何计算两个日期之间的天数？
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问如何计算两个日期之间的天数？EN

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问如何计算两个日期之间的天数？
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