为什么不减去两个本地DateTime值似乎占夏令时？

Question

我正在玩一些C#代码,试图了解如何在夏令时中减去C#中的DateTime对象.

根据谷歌和其他消息来源,2017年东部标准时区的夏令时"春晚"活动是在3月12日凌晨2点.因此,当天的前几个小时是:

   12:00am - 1:00am
    1:00am - 2:00am
   (There was no 2:00am - 3:00am hour due to the "spring ahead")
    3:00am - 4:00am

所以,如果我要计算该日期在该时区的凌晨1点到凌晨4点之间的时间差,我希望结果是2小时.

但是,我试图模拟这个问题的代码是返回3小时TimeSpan.

码:

TimeZoneInfo easternStandardTime = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");

DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);
DateTime fourAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 04, 00, 00), easternStandardTime);

TimeSpan difference = (fourAm - oneAm);

Console.WriteLine(oneAm);
Console.WriteLine(fourAm);
Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(oneAm));
Console.WriteLine(TimeZoneInfo.Local.IsDaylightSavingTime(fourAm));
Console.WriteLine(difference);

在我的电脑上,这会产生:

2017-03-12 01:00:00.000 -5
2017-03-12 04:00:00.000 -4
False
True
03:00:00

所有这些输出都是预期的 - 除了3小时的最终值,如上所述,我预计将是2小时.

显然,我的代码没有正确模拟我想到的情况.缺陷是什么？

Answer 1

注意:

// These are just plain unspecified DateTimes
DateTime dtOneAm = new DateTime(2017, 03, 12, 01, 00, 00);
DateTime dtFourAm = new DateTime(2017, 03, 12, 04, 00, 00);

// The difference is not going to do anything other than 4-1=3
TimeSpan difference1 = dtFourAm - dtOneAm;

// ... but we have a time zone to consider!
TimeZoneInfo eastern = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");

// Use that time zone to get DateTimeOffset values.
// The GetUtcOffset method has what we need.
DateTimeOffset dtoOneAmEastern = new DateTimeOffset(dtOneAm, eastern.GetUtcOffset(dtOneAm));
DateTimeOffset dtoFourAmEastern = new DateTimeOffset(dtFourAm, eastern.GetUtcOffset(dtFourAm));

// Subtracting these will take the offset into account!
// It essentially does this: [4-(-4)]-[1-(-5)] = 8-6 = 2
TimeSpan difference2 = dtoFourAmEastern - dtoOneAmEastern;

// Let's see the results
Console.WriteLine("dtOneAm: {0:o} (Kind: {1})", dtOneAm, dtOneAm.Kind);
Console.WriteLine("dtFourAm: {0:o} (Kind: {1})", dtFourAm, dtOneAm.Kind);
Console.WriteLine("difference1: {0}", difference1);

Console.WriteLine("dtoOneAmEastern: {0:o})", dtoOneAmEastern);
Console.WriteLine("dtoFourAmEastern: {0:o})", dtoFourAmEastern);
Console.WriteLine("difference2: {0}", difference2);

结果:

dtOneAm: 2017-03-12T01:00:00.0000000 (Kind: Unspecified)
dtFourAm: 2017-03-12T04:00:00.0000000 (Kind: Unspecified)
difference1: 03:00:00

dtoOneAmEastern: 2017-03-12T01:00:00.0000000-05:00)
dtoFourAmEastern: 2017-03-12T04:00:00.0000000-04:00)
difference2: 02:00:00

请注意,默认情况下在其属性中DateTime包含a .它不属于任何特定时区. 没有一种,它有一个,告诉你本地时间偏离UTC的距离. 这些都没有给你时区.这就是对象正在做的事情.请参阅timezone标记wiki中的 "time zone!= offset" .DateTimeKindKindUnspecifiedDateTimeOffset OffsetTimeZoneInfo

我想你也许是沮丧的,该部分是几个历史原因,DateTime对象不会永远不懂时区做数学的时候,即使你可能有DateTimeKind.Local.可以实现它来观察本地时区的转换,但是没有这样做.

您也可能对Noda Time感兴趣,它以更加明智和有目的的方式为.NET中的日期和时间提供了一个非常不同的API.

using NodaTime;

...

// Start with just the local values.
// They are local to *somewhere*, who knows where?  We didn't say.
LocalDateTime ldtOneAm = new LocalDateTime(2017, 3, 12, 1, 0, 0);
LocalDateTime ldtFourAm = new LocalDateTime(2017, 3, 12, 4, 0, 0);

// The following won't compile, because LocalDateTime does not reference
// a linear time scale!
// Duration difference = ldtFourAm - ldtOneAm;

// We can get the 3 hour period, but what does that really tell us?
Period period = Period.Between(ldtOneAm, ldtFourAm, PeriodUnits.Hours);

// But now lets introduce a time zone
DateTimeZone eastern = DateTimeZoneProviders.Tzdb["America/New_York"];

// And apply the zone to our local values.
// We'll choose to be lenient about DST gaps & overlaps.
ZonedDateTime zdtOneAmEastern = ldtOneAm.InZoneLeniently(eastern);
ZonedDateTime zdtFourAmEastern = ldtFourAm.InZoneLeniently(eastern);

// Now we can get the difference as an exact elapsed amount of time
Duration difference = zdtFourAmEastern - zdtOneAmEastern;


// Dump the output
Console.WriteLine("ldtOneAm: {0}", ldtOneAm);
Console.WriteLine("ldtFourAm: {0}", ldtFourAm);
Console.WriteLine("period: {0}", period);

Console.WriteLine("zdtOneAmEastern: {0}", zdtOneAmEastern);
Console.WriteLine("zdtFourAmEastern: {0}", zdtFourAmEastern);
Console.WriteLine("difference: {0}", difference);

ldtOneAm: 3/12/2017 1:00:00 AM
ldtFourAm: 3/12/2017 4:00:00 AM
period: PT3H

zdtOneAmEastern: 2017-03-12T01:00:00 America/New_York (-05)
zdtFourAmEastern: 2017-03-12T04:00:00 America/New_York (-04)
difference: 0:02:00:00

我们可以看到三个小时的时间段,但它并不意味着与经过的时间相同.它只是意味着两个本地值在时钟上相隔三个小时.NodaTime理解这些概念之间的区别,而.Net的内置类型则不然.

一些后续阅读:

• 无论如何,DateTime有什么问题？
• 使用DateTime更有趣
• 针对DateTime.Now的案例
• .NET开发人员的五种常见夏令时反模式

哦,还有一件事.你的代码有这个......

DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);

由于DateTime您创建的类型未指定,因此您要求从计算机的本地时区转换为东部时间.如果您恰好不在东部时间,您的oneAm变量可能根本不是凌晨1点!