blocks|key|1493725|text|简单地说：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1493726|yourDate.setDate(yourDate.getDate()+-+daysToSubtract);|code-block|syntax|javascript|1493727|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|$]]

Simply:

<pre><code>yourDate.setDate(yourDate.getDate() - daysToSubtract);
</code></pre>

blocks|key|1279058|text|这里有一个例子，但是这不做任何检查(例如，如果你在2009/7/1使用它，它将使用负数天或抛出错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279059|function+subDate(o,+days)+{
//+keep+in+mind,+months+in+javascript+are+0-11
return+new+Date(o.getFullYear(),+o.getMonth(),+o.getDate()+-+days);;
}|code-block|syntax|javascript|1279060|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's an example, however this does no kind of checking (for example if you use it on 2009/7/1 it'll use a negative day or throw an error.

<pre><code>function subDate(o, days) {
// keep in mind, months in javascript are 0-11
return new Date(o.getFullYear(), o.getMonth(), o.getDate() - days);;
}
</code></pre>

blocks|key|1493934|text|这就是我会做的。注意，你可以简化这个表达式，我写出来是为了清楚地表明你是用一天中的毫秒数乘以天数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1493935|+var+newDate+=+new+Date(+yourDate.getTime()+-+(days+*+24+*+60+*+60+*+1000)+);|code-block|syntax|javascript|1493936|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|$]]

This is what I would do. Note you can simplify the expression, I've just written it out to make it clear you are multiplying the number of days by the number of milliseconds in a day.

<pre><code> var newDate = new Date( yourDate.getTime() - (days * 24 * 60 * 60 * 1000) );
</code></pre>

blocks|key|1493972|text|这只是我写的另一种选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1493973|DP_DateExtensions+Library|offset|length|1493974|如果你只想做一次计算，这可能有点夸张，但如果你要做更多的日期操作，你可能会发现它很有用。|1493975|支持日期/时间格式，日期数学(加/减日期部分)，日期比较，日期解析等。|1493976|entityMap|0|LINK|mutability|MUTABLE|url|http://depressedpress.com/javascript-extensions/dp_dateextensions/^0|0|0|P|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

Just another option, which I wrote:

<a href="http://depressedpress.com/javascript-extensions/dp_dateextensions/" rel="nofollow noreferrer">DP_DateExtensions Library</a>

It's probably overkill if ALL you want to do is one calculation, but if you're going to do more date manipulation you might find it useful.

Supports date/time formatting, date math (add/subtract date parts), date compare, date parsing, etc.

blocks|key|1279190|text|function+date_by_subtracting_days(date,+days)+{
++++return+new+Date(
++++++++date.getFullYear(),+
++++++++date.getMonth(),+
++++++++date.getDate()+-+days,
++++++++date.getHours(),
++++++++date.getMinutes(),
++++++++date.getSeconds(),
++++++++date.getMilliseconds()
++++);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1279191|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>function date_by_subtracting_days(date, days) {
 return new Date(
 date.getFullYear(), 
 date.getMonth(), 
 date.getDate() - days,
 date.getHours(),
 date.getMinutes(),
 date.getSeconds(),
 date.getMilliseconds()
 );
}
</code></pre>

blocks|key|1279247|text|永远不要使用这个解决方案yourDate.setDate(yourDate.getDate()+-+daysToSubtract)；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279248|如果你的日期是任何一个月的1号，并且你想删除一些日子，比如1号，它就不起作用了。|1279249|取而代之的是下面的解决方案，它总是有效的|1279250|var天数=新日期(+yourDate.getTime()+-(+newDate+*+24+*+60+*+60+*+1000)+)；|1279251|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|$]]

Never go for this solution 
yourDate.setDate(yourDate.getDate() - daysToSubtract);

it wont work in case your date is 1st of any month and you want to delete some days say 1.

Instead go for below solution which will work always

var newDate = new Date( yourDate.getTime() - (days * 24 * 60 * 60 * 1000) );

blocks|key|1279290|text|我已经创建了一个用于日期操作的函数。您可以添加或减去任意天数、小时数、分钟数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1279291|function+dateManipulation(date,+days,+hrs,+mins,+operator)+{
++++date+=+new+Date(date);
++++if+(operator+==+"-")+{
+++++++var+durationInMs+=+(((24+*+days)+*+60)+%2B+(hrs+*+60)+%2B+mins)+*+60000;
+++++++var+newDate+=+new+Date(date.getTime()+-+durationInMs);
++++}+else+{
+++++++var+durationInMs+=+(((24+*+days)+*+60)+%2B+(hrs+*+60)+%2B+mins)+*+60000;
+++++++var+newDate+=+new+Date(date.getTime()+%2B+durationInMs);
++++}
+++return+newDate;
}|code-block|syntax|javascript|1279292|现在，通过传递参数来调用此函数。例如，这里有一个函数调用，用于获取从今天起3天前的日期。|1279293|var+today+=+new+Date();
var+newDate+=+dateManipulation(today,+3,+0,+0,+"-");|1279294|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|$]]

I have created a function for date manipulation. you can add or subtract any number of days, hours, minutes. 

<pre><code>function dateManipulation(date, days, hrs, mins, operator) {
 date = new Date(date);
 if (operator == "-") {
 var durationInMs = (((24 * days) * 60) + (hrs * 60) + mins) * 60000;
 var newDate = new Date(date.getTime() - durationInMs);
 } else {
 var durationInMs = (((24 * days) * 60) + (hrs * 60) + mins) * 60000;
 var newDate = new Date(date.getTime() + durationInMs);
 }
 return newDate;
}
</code></pre>

Now, call this function by passing parameters. For example, here is a function call for getting date before 3 days from today.

<pre><code>var today = new Date();
var newDate = dateManipulation(today, 3, 0, 0, "-");
</code></pre>

I want to find date by subtracting X number of days from a particular date in JavaScript. My JavaScript function accepts 2 parameters. One is the date value and the other is the number of days that needs to be subtracted.

For example, I pass my argument date as 27 July 2009 and i pass my other argument as 3. So i want to calculate the date 3 days before 27 July 2009. So the resultant date that we should get is 24 July 2009. How is this possible in JavaScript. Thanks for any help.

Finding date by subtracting X number of days from a particular date in Javascript

我想通过从JavaScript中的特定日期中减去X天数来计算日期。我的JavaScript函数接受2个参数。一个是日期值，另一个是需要减去的天数。例如，我将我的参数日期传递为2009年7月27日，我将另一个参数传递为3，因此我希望在2009年7月27日之前3天计算日期。因此，我们应该得到的结果日期是2009年7月24日...

问通过从Javascript中的特定日期中减去X天数来查找日期
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