blocks|key|5063859|text|将该月的天数乘以60+*+60+*+24。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5063860|entityMap^0|8|C|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Multiply the number of days in the month by <code>60 * 60 * 24</code>.

blocks|key|5064253|text|由于夏令时..。用您的语言创建一个很好的datetime库，计算每月的第一天0:00:00和下个月的第一天0:00:00之间的差值，并提取秒数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5064254|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Due to daylights savings... take a good datetime library in your language and calculate the difference between the first day of the month 0:00:00 and the first day of the next month 0:00:00 and extract the number of seconds.

blocks|key|5064286|text|你需要多精确？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5064287|一个月中的60秒*+60分钟*+24小时*Z天，为您提供了一个给定月份的准确数字。|5064288|如果你需要一个月的平均值，用一年中的秒数除以12。|5064289|在某些领域，例如账单或法律领域，一个“月”实际上可能恰好是30天。|5064290|如果您正在进行多年的工作或在分散的系统之间进行紧密集成，则需要参考参考资料来确定闰秒。对于历史数据，这可以是一个表，否则您会更适合于同步到可信的时间源。|5064291|http://en.wikipedia.org/wiki/Leap_second|offset|length|5064292|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|0|0|0|0|14|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|X|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|Z|8|@]|9|@$L|10|M|11|1|12]]|A|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|K]]]]

How accurate do you need to be?

60 seconds * 60 minutes * 24 hours * Z days in the month gives you an accurate number for a given month.

If you need an average month go for number of seconds in the year and divide by twelve.

In some domains, such as billing or legal domains a 'month' might actually be exactly 30 days.

If you are working across multiple years or doing tight integration between disperse systems, you'll need to consult resource to determine leap seconds. For historical data this could be a table, but otherwise you'd be better suited by synchronizing to a trusted time source.

<a href="http://en.wikipedia.org/wiki/Leap_second" rel="nofollow">http://en.wikipedia.org/wiki/Leap_second</a>

blocks|key|185103|text|60+(秒)*+60+(分钟)*+24+(小时)*+##+(月中天数)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|185104|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

60 (seconds) * 60 (minutes) * 24 (hours) * ## (days in the month)

blocks|key|5064100|text|假设一天有86,400秒，您可以将这个数字乘以DateTime.DaysInMonth函数的结果(在C#中)。下面的函数就是这样做的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5064101|public+double+SecondsInMonth(int+year,+int+month)
{
+++return+DateTime.DaysInMonth(year,+month)+*+86400;
}|code-block|syntax|javascript|5064102|例如，查找当月的秒数：|5064103|double+secondsInCurrentMonth+=+SecondsInMonth(DateTime.Now.Year,+DateTime.Now.Month);|5064104|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/system.datetime.daysinmonth.aspx^0|N|K|N|K|0|0|0|0|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]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Given that there are 86,400 seconds in a day, you can multiply this number by the result of the <a href="http://msdn.microsoft.com/en-us/library/system.datetime.daysinmonth.aspx" rel="nofollow"><code>DateTime.DaysInMonth</code></a> function (in C#). The following function does just that:

<pre><code>public double SecondsInMonth(int year, int month)
{
 return DateTime.DaysInMonth(year, month) * 86400;
}
</code></pre>

E.g., find the seconds in the current month:

<pre><code>double secondsInCurrentMonth = SecondsInMonth(DateTime.Now.Year, DateTime.Now.Month);
</code></pre>

blocks|key|2096604|text|给定月份的天数*小时/天*分钟/小时*秒/分钟|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2096605|是最好的方法。|2096606|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Number of days in the given month * hours/day * minutes/hour * seconds/minute

is the best way.

blocks|key|2950076|text|如果您是在纯数学中这样做的，那么它应该是60+*+60+*+24+*+<number+of+days+in+month>。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2950077|用例是什么？|2950078|entityMap^0|K|14|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

If you're doing this in pure math it would be <code>60 * 60 * 24 * &lt;number of days in month&gt;</code>. 

What's the use case?

blocks|key|5063999|text|否，使用特定语言可用的日期API并确定当前月份的天数。然后计算秒数。也要考虑到闰年。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5064000|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No, use the date API available for a particular lannguage and determine the number of days in the current month. Then calculate the number of seconds. Also take into account leap years.

blocks|key|2091785|text|取决于你是想要一个月平均值，还是想要一个特定的month....your方式得到一个平均值。对于特定的月份，计算天数并乘以86400+(每24小时一天的秒数)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2091786|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Depends on if you want an average month or a specific month....your way gets an average. For a specific month count days and multiply by 86400 (seconds per 24.0 hour day)

blocks|key|2950210|text|这不是一个真正的编程问题。月份有不同的长度，所以将一年中的秒数除以12不会得到任何有用的结果。很容易确定一个月中的日期--一个简单的查找表加上一个闰年的计算就可以做到这一点。然后乘以一天中的秒数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2950211|如果你真的很精确，你可能需要包括闰秒的计算，但由于它们是基于天文计算的不可预测的分配，并且不能提前预测，我可能会忽略它们。|2950212|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

This isn't really a programming question. Months have different lengths, so dividing the number of seconds in a year by 12 will give you nothing useful. It's easy to determine the days in a month - a simple lookup table plus a calcualation of leap years will do it. Then just multiply by the number of seconds in a day.

If you are being really precise you might need to include calculations of leap seconds, but since they are unpredictably assigned based on astronimical calculations, and not predictable in advance, I would probably ignore them.

blocks|key|185225|text|天数在每个month.Proper算法中都有所不同，因为这是获取天数并将其乘以86400+(一天中的秒数).You可能还需要计算平均计数或闰年...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|185226|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Number of days vary in each month.Proper algorithm for this is to get number of days in moth and multiply it with 86400 (number of seconds in a day).You might also need average count or leap years calculation ...

blocks|key|185275|text|简单的答案是找到一个月中的天数，然后乘以86400。如果您正在处理UTC格式的日期和时间，这将非常有效。但是，如果您使用的是本地时区，并且该时区遵守夏令时，则此方法会产生稍微不正确的结果。在一个月的时间内，误差有点小，但如果你需要在短时间内进行类似的计算，比如一天，误差就会放大。我绝对推荐使用UTC进行所有处理和存储，但根据应用程序的不同，您必须将UTC时间转换为最终用户期望的本地时区。您甚至可能需要使用本地时区来计算持续时间。同样，尽可能多地使用UTC，这样您就可以避免大多数问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|185276|我在C#中想出了这个解决方案。它与UTC和本地时区兼容。您只需告诉GetNumberOfSecondsInMonth您希望计算基于哪个时区。在我的例子中，我选择了2010年11月，因为在密苏里州我们这里是DST，这个月有一个额外的小时。夏令时规则发生了变化，因此我使用了一个从操作系统中提取DST信息的API，以便在2007年(即美国将DST扩展到大多数地区之前)之前的几年中计算都是正确的。|offset|length|style|CODE|185277|我应该指出的是，我的解决方案不能处理UTC中的闰秒。对我来说，这从来不是问题。但是，如果您确实需要超高精度的计时，那么使用查找表将很容易解决这一问题。|185278|public+class+Program
{
++++static+void+Main(string[]+args)
++++{
++++++++int+seconds+=+GetNumberOfSecondsInMonth(2010,+11,+DateTimeKind.Local);
++++}

++++public+static+int+GetNumberOfSecondsInMonth(int+year,+int+month,+DateTimeKind+kind)
++++{
++++++++DateTime+start+=+new+DateTime(year,+month,+1);
++++++++DateTime+end+=+start.AddMonths(1);
++++++++int+seconds+=+(int)(end+-+start).TotalSeconds;
++++++++if+(kind+==+DateTimeKind.Local)
++++++++{
++++++++++++DaylightTime+dt+=+TimeZone.CurrentTimeZone.GetDaylightChanges(year);
++++++++++++seconds+=+(dt.Start+>+start)+?+seconds+-+3600+:+seconds;
++++++++++++seconds+=+(dt.End+<+end)+?+seconds+%2B+3600+:+seconds;
++++++++}
++++++++return+seconds;
++++}
}|code-block|syntax|javascript|185279|entityMap^0|0|X|P|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|V|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The trivial answer is to find the number of days in the month and then multiply by 86400. That will work perfectly if you are dealing with dates and times in UTC. However, if you are using a local time zone then this approach yields a slightly incorrect result if that time zone observes daylight saving time. The error is somewhat small over a one month period, but will magnify if you need to make similiar calculation over short periods like a day. I definitely recommend doing all processing and storage in UTC, but depending on the application you will have to convert your UTC times to the local time zone that the end user is expecting. And it might even be plausible that you have to calculate durations using the local time zone. Again, use UTC as much as possible so that you avoid most of the problems.

I came up with this solution in C#. It is compatible with UTC and local time zones alike. You just have to tell the <code>GetNumberOfSecondsInMonth</code> which time zone you want the calculation to be based on. In my example I chose November of 2010 because here in Missouri we observe DST and there was one extra hour this month. Daylight saving time rules change so I used an API that pulls the DST information from the operating system so that the calculation will be correct for years prior to 2007 (that is when the United States expanded DST for most regions).

I should point out that my solution does not handle leap seconds in UTC. For me that is never an issue. But it would be easy to account for that by using a lookup table if you really needed ultra high precision timing.

<pre><code>public class Program
{
 static void Main(string[] args)
 {
 int seconds = GetNumberOfSecondsInMonth(2010, 11, DateTimeKind.Local);
 }

 public static int GetNumberOfSecondsInMonth(int year, int month, DateTimeKind kind)
 {
 DateTime start = new DateTime(year, month, 1);
 DateTime end = start.AddMonths(1);
 int seconds = (int)(end - start).TotalSeconds;
 if (kind == DateTimeKind.Local)
 {
 DaylightTime dt = TimeZone.CurrentTimeZone.GetDaylightChanges(year);
 seconds = (dt.Start &gt; start) ? seconds - 3600 : seconds;
 seconds = (dt.End &lt; end) ? seconds + 3600 : seconds;
 }
 return seconds;
 }
}
</code></pre>

blocks|key|2096805|text|这是年和月的问题，因为它们没有固定的天数。但经过深思熟虑后，我想出了怎么做。计算包含30天或31天的月份不是一个好主意，因为这看起来很糟糕，例如，如果我每个月有30天，则从1年转换为月将得到11个月25天的答案，或者如果我每个月有31天，则为12个月5天。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2096806|相反，我每个月循环执行一系列天:+30,30,31,30,31,30,31,30,31,30,31,30，一年总共365天。因此，如果我想要4个月的天数，我会加上30%2B30%2B31%2B30。如果我从23个月开始，它将经历几乎两次循环(23乘以30或31)。这是在while/until循环中完成的。对于每4年，我添加1天，使其为366天(列表中的前30天更改为31天)。这是相当复杂的，但它的工作和结果看起来更好。|2096807|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

It's a problem with years ang months as there is not a fixed number of days in them. But after a lot of thought I have figured out how to do it. It was not a good idea to calculate months with either 30 or 31 days in them, because it looks bad, for example converting from 1 year to months would give an answer of 11 months and 25 days if I had 30 days in each month, or 12 months and 5 days if I have 31 days in each month.

Instead I loop through a series of days per month: 30,30,31,30,31,30,31,30,31,30,31,30 which makes a total of 365 days in a year. So if I want the number of days in 4 months I add 30+30+31+30. And if I start with 23 months it would go through the loop almost twice (23 times 30 or 31). It's done in a while/until loop. For every 4 years I add 1 day, making it 366 days (the first 30 is changed to 31 in the list). It's rather complex but it works and the result looks better.

How do I accurately determine the number of seconds in a month using PHP? Is the best way to take the number of seconds in a year and divide by 12?

How do I calculate the number of seconds in a month in PHP?

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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如何使用PHP准确确定一个月的秒数？将一年中的秒数除以12是最好的方法吗？

问如何在PHP中计算一个月的秒数？
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