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3月12日凌晨2点。因此，该日当天的前几个小时是：< / p>

   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上午1:00到凌晨4:00之间的时间差，我希望结果是 2 小时。

但是，我试图模拟此问题的代码是返回 3 小时TimeSpan。

代码：

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

所有输出都符合预期 - 除了最终值为3小时，如上所述，我预计会有2个小时。

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

观察：

// 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DateTime在DateTimeKind属性中带有Kind，默认为Unspecified。它不属于任何特定时区。 DateTimeOffset 没有有一种，它有一个Offset，告诉你当地时间偏离UTC的距离。这些都没有为您提供时区。这就是TimeZoneInfo对象正在做的事情。请参阅＆＃34;时区！=偏移＆＃34;在the timezone tag wiki。

我认为你可能感到沮丧的部分是，由于几个历史原因，DateTime对象在进行数学运算时不会曾理解时区，即使你可能{ {1}}。 可以实现来观察本地时区的转换，但是没有这样做。

您可能也对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。

DateTimeKind.Local

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);

我们可以看到三个小时的时间段，但它与经过的时间并不一样。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。

为您提供一些后续阅读材料：

What's wrong with DateTime anyway?

More Fun with DateTime

The case against DateTime.Now

Five Common Daylight Saving Time Antipatterns of .NET Developers

哦，还有一件事。你的代码有这个......

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

由于您创建的DateTime oneAm = TimeZoneInfo.ConvertTime(new DateTime(2017, 03, 12, 01, 00, 00), easternStandardTime);具有未指定类型，因此您要求将从计算机的本地时区转换为东部时间。如果您在东部时间碰巧不，那么您的DateTime变量可能根本不是凌晨1点！

Answer 2

所以这在MSDN文档中已得到解决。

基本上，当从另一个日期中减去一个日期时，你应该使用DateTimeOffset.Subtract()，而不是像在这里那样使用算术减法。

TimeSpan difference = fourAm.Subtract(oneAm);

产生预期的2小时时差。

Answer 3

好的，所以我对你的代码做了一些小改动。不确定这是否是你想要实现的目标，但这会给你你想要的东西......

static void Main() {
        TimeZoneInfo easternStandardTime = TimeZoneInfo.FindSystemTimeZoneById("Eastern Standard Time");
        TimeZone timeZone = TimeZone.CurrentTimeZone;

        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);

        DaylightTime time = timeZone.GetDaylightChanges(fourAm.Year);

        TimeSpan difference = ((fourAm - time.Delta) - oneAm);

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

为什么不减去两个本地DateTime值似乎占夏令时？

3 个答案: