C# 如何获得年/月/周/日中两个日期之间的差异？
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原文地址: http://stackoverflow.com/questions/1083955/
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How to get difference between two dates in Year/Month/Week/Day?
提问by Ahmed Atia
How to get difference between two dates in Year/Month/Week/Day in an efficient way?
如何以有效的方式获得年/月/周/日中两个日期之间的差异？
eg. difference between two dates is 1 Year, 2 Months, 3 Weeks, 4 Days.
例如。两个日期之间的差异是 1 年、2 个月、3 周、4 天。
Difference represents count of year(s), month(s), week(s) and day(s) between two dates.
差异表示两个日期之间的年、月、周和日数。
采纳答案by Jon Skeet
This is actually quite tricky. A different total number of days can result in the same result. For example:
这实际上非常棘手。不同的总天数可能会导致相同的结果。例如：
19th June 2008 to 19th June 2010 = 2 years, but also 365 * 2 days
19th June 2006 to 19th June 2008 = 2 years, but also 365 + 366 days due to leap years
2008 年 6 月 19 日至 2010 年 6 月 19 日 = 2 年，但也是 365 * 2 天
2006 年 6 月 19 日至 2008 年 6 月 19 日 = 2 年，但由于闰年也是 365 + 366 天
You may well want to subtract years until you get to the point where you've got two dates which are less than a year apart. Then subtract months until you get to the point where you've got two dates which are less than a month apart.
您可能想减去年数，直到您得到两个相隔不到一年的日期。然后减去几个月，直到你得到两个相隔不到一个月的日期。
Further confusion: subtracting (or adding) months is tricky when you might start with a date of "30th March" - what's a month earlier than that?
进一步的困惑：当您从“3 月 30 日”开始时，减去（或增加）月份很棘手 - 比这早一个月是什么？
Even further confusion (maynot be relevant): even a day isn't always 24 hours. Daylight saving anyone?
更进一步的混淆（可能不相关）：即使一天也不总是 24 小时。夏令时有人吗？
Even further confusion (almost certainly notrelevant): even a minute isn't always 60 seconds. Leap seconds are highly confusing...
更进一步的混乱（几乎肯定不相关）：即使一分钟也不总是 60 秒。闰秒非常令人困惑......
I don't have the time to work out the exact right way of doing this right now - this answer is mostly to raise the fact that it's not nearly as simple as it might sound.
我现在没有时间找出正确的方法来做这件事——这个答案主要是为了提出这样一个事实，即它并不像听起来那么简单。
EDIT: Unfortunately I'm not going to have enough time to answer this fully. I would suggest you start off by defining a struct representing a Period:
编辑：不幸的是，我没有足够的时间来完全回答这个问题。我建议您首先定义一个表示 a 的结构Period：
public struct Period
{
private readonly int days;
public int Days { get { return days; } }
private readonly int months;
public int Months { get { return months; } }
private readonly int years;
public int Years { get { return years; } }
public Period(int years, int months, int days)
{
this.years = years;
this.months = months;
this.days = days;
}
public Period WithDays(int newDays)
{
return new Period(years, months, newDays);
}
public Period WithMonths(int newMonths)
{
return new Period(years, newMonths, days);
}
public Period WithYears(int newYears)
{
return new Period(newYears, months, days);
}
public static DateTime operator +(DateTime date, Period period)
{
// TODO: Implement this!
}
public static Period Difference(DateTime first, DateTime second)
{
// TODO: Implement this!
}
}
I suggest you implement the + operator first, which should inform the Differencemethod - you should make sure that first + (Period.Difference(first, second)) == secondfor all first/secondvalues.
我建议你首先实现 + 运算符，它应该通知Difference方法 - 你应该确保first + (Period.Difference(first, second)) == second所有first/second值。
Start with writing a whole slew of unit tests - initially "easy" cases, then move on to tricky ones involving leap years. I know the normal approach is to write one test at a time, but I'd personally brainstorm a bunch of them before you start any implementation work.
从编写大量单元测试开始——最初是“简单”的案例，然后转向涉及闰年的棘手案例。我知道通常的方法是一次编写一个测试，但在您开始任何实施工作之前，我会亲自对其中的一堆进行集思广益。
Allow yourself a day to implement this properly. It's tricky stuff.
给自己一天的时间来正确地实现这一点。这是棘手的事情。
Note that I've omitted weeks here - that value at least is easy, because it's always 7 days. So given a (positive) period, you'd have:
请注意，我在这里省略了几周 - 该值至少很容易，因为它总是 7 天。所以给定一个（积极的）时期，你会有：
int years = period.Years;
int months = period.Months;
int weeks = period.Days / 7;
int daysWithinWeek = period.Days % 7;
(I suggest you avoid even thinking about negative periods - make sure everything is positive, all the time.)
（我建议你甚至避免考虑消极时期 - 确保一切都是积极的，一直。）
回答by Andy
回答by Andy
Use the Subtractmethod of the DateTimeobject which returns a TimeSpan...
使用返回一个...Subtract的DateTime对象的方法TimeSpan
DateTime dt1 = new DateTime(2009, 3, 14);
DateTime dt2 = new DateTime(2008, 3, 15);
TimeSpan ts = dt1.Subtract(dt2);
Double days = ts.TotalDays;
Double hours = ts.TotalHours;
Double years = ts.TotalDays / 365;
回答by Richard
Subtract two DateTimeinstances to give you a TimeSpanwhich has a Daysproperty. (E.g. in PowerShell):
减去两个DateTime实例，得到一个TimeSpan有Days属性的。（例如在 PowerShell 中）：
PS > ([datetime]::today - [datetime]"2009-04-07") Days : 89 Hours : 0 Minutes : 0 Seconds : 0 Milliseconds : 0 Ticks : 76896000000000 TotalDays : 89 TotalHours : 2136 TotalMinutes : 128160 TotalSeconds : 7689600 TotalMilliseconds : 7689600000
Converting days into years or weeks is relatively easy (days in a year could be 365, 365.25, ... depending on context). Months is much harder, because without a base date you don't know which month lengths apply.
将天数转换为年数或周数相对容易（一年中的天数可能是 365、365.25...，具体取决于上下文）。月份要困难得多，因为没有基准日期，您不知道适用的月份长度。
Assuming you want to start with your base date, you can incrementally substract while counting first years (checking for leap years), then month lengths (indexing from startDate.Month), then weeks (remaining days divided by 7) and then days (remainder).
假设您想从基准日期开始，您可以在计算第一年（检查闰年），然后是月份长度（从 startDate.Month 开始索引），然后是周（剩余天数除以 7），然后是天数（余数）的同时逐渐减去）。
There are a lot of edge cases to consider, e.g. 2005-03-01 is one year from 2004-03-01, and from 2004-02-29depending on what you mean by "Year".
有很多边缘情况需要考虑，例如 2005-03-01 是 2004-03-01和 2004-02-29 的一年，具体取决于“年”的含义。
回答by lc.
Leap years and uneven months actually make this a non-trivial problem. I'm sure someone can come up with a more efficient way, but here's one option - approximate on the small side first and adjust up (untested):
闰年和不均匀的月份实际上使这成为一个不平凡的问题。我敢肯定有人可以想出更有效的方法，但这里有一个选择 - 首先在小的方面近似并调整（未经测试）：
public static void GetDifference(DateTime date1, DateTime date2, out int Years,
out int Months, out int Weeks, out int Days)
{
//assumes date2 is the bigger date for simplicity
//years
TimeSpan diff = date2 - date1;
Years = diff.Days / 366;
DateTime workingDate = date1.AddYears(Years);
while(workingDate.AddYears(1) <= date2)
{
workingDate = workingDate.AddYears(1);
Years++;
}
//months
diff = date2 - workingDate;
Months = diff.Days / 31;
workingDate = workingDate.AddMonths(Months);
while(workingDate.AddMonths(1) <= date2)
{
workingDate = workingDate.AddMonths(1);
Months++;
}
//weeks and days
diff = date2 - workingDate;
Weeks = diff.Days / 7; //weeks always have 7 days
Days = diff.Days % 7;
}
回答by Jon Skeet
DateTime dt1 = new DateTime(2009, 3, 14);
DateTime dt2 = new DateTime(2008, 3, 15);
int diffMonth = Math.Abs((dt2.Year - dt1.Year)*12 + dt1.Month - dt2.Month)
回答by sheir
What about using the System.Data.Linqnamespace and its SqlMethods.DateDiffMonthmethod?
使用System.Data.Linq命名空间及其SqlMethods.DateDiffMonth方法怎么样？
For example, say:
例如，说：
DateTime starDT = {01-Jul-2009 12:00:00 AM}
DateTime endDT = {01-Nov-2009 12:00:00 AM}
Then:
然后：
int monthDiff = System.Data.Linq.SqlClient.SqlMethods.DateDiffMonth(startDT, endDT);
==> 4
==> 4
There are other DateDiffstatic methods in the SqlMethodsclass.
类中还有其他DateDiff静态方法SqlMethods。
回答by sheir
For the correct difference calculation of Years/Months/Weeks, the Calendarof the CultureInfomust be considered:
多年来/月/周的正确差计算，日历中的CultureInfo的必须考虑：
- leap vs. non-leap years
- months with different count of days
- years with different count of weeks (varying by the first day of week and the calendar week rule)
- 闰年与非闰年
- 天数不同的月份
- 具有不同周数的年份（因一周的第一天和日历周规则而异）
The DateDiffclass of the Time Period Library for .NETrespects all these factors:
.NET 时间段库的DateDiff类考虑所有这些因素：
// ----------------------------------------------------------------------
public void DateDiffSample()
{
DateTime date1 = new DateTime( 2009, 11, 8, 7, 13, 59 );
Console.WriteLine( "Date1: {0}", date1 );
// > Date1: 08.11.2009 07:13:59
DateTime date2 = new DateTime( 2011, 3, 20, 19, 55, 28 );
Console.WriteLine( "Date2: {0}", date2 );
// > Date2: 20.03.2011 19:55:28
DateDiff dateDiff = new DateDiff( date1, date2 );
// differences
Console.WriteLine( "DateDiff.Years: {0}", dateDiff.Years );
// > DateDiff.Years: 1
Console.WriteLine( "DateDiff.Quarters: {0}", dateDiff.Quarters );
// > DateDiff.Quarters: 5
Console.WriteLine( "DateDiff.Months: {0}", dateDiff.Months );
// > DateDiff.Months: 16
Console.WriteLine( "DateDiff.Weeks: {0}", dateDiff.Weeks );
// > DateDiff.Weeks: 70
Console.WriteLine( "DateDiff.Days: {0}", dateDiff.Days );
// > DateDiff.Days: 497
Console.WriteLine( "DateDiff.Weekdays: {0}", dateDiff.Weekdays );
// > DateDiff.Weekdays: 71
Console.WriteLine( "DateDiff.Hours: {0}", dateDiff.Hours );
// > DateDiff.Hours: 11940
Console.WriteLine( "DateDiff.Minutes: {0}", dateDiff.Minutes );
// > DateDiff.Minutes: 716441
Console.WriteLine( "DateDiff.Seconds: {0}", dateDiff.Seconds );
// > DateDiff.Seconds: 42986489
// elapsed
Console.WriteLine( "DateDiff.ElapsedYears: {0}", dateDiff.ElapsedYears );
// > DateDiff.ElapsedYears: 1
Console.WriteLine( "DateDiff.ElapsedMonths: {0}", dateDiff.ElapsedMonths );
// > DateDiff.ElapsedMonths: 4
Console.WriteLine( "DateDiff.ElapsedDays: {0}", dateDiff.ElapsedDays );
// > DateDiff.ElapsedDays: 12
Console.WriteLine( "DateDiff.ElapsedHours: {0}", dateDiff.ElapsedHours );
// > DateDiff.ElapsedHours: 12
Console.WriteLine( "DateDiff.ElapsedMinutes: {0}", dateDiff.ElapsedMinutes );
// > DateDiff.ElapsedMinutes: 41
Console.WriteLine( "DateDiff.ElapsedSeconds: {0}", dateDiff.ElapsedSeconds );
// > DateDiff.ElapsedSeconds: 29
// description
Console.WriteLine( "DateDiff.GetDescription(1): {0}", dateDiff.GetDescription( 1 ) );
// > DateDiff.GetDescription(1): 1 Year
Console.WriteLine( "DateDiff.GetDescription(2): {0}", dateDiff.GetDescription( 2 ) );
// > DateDiff.GetDescription(2): 1 Year 4 Months
Console.WriteLine( "DateDiff.GetDescription(3): {0}", dateDiff.GetDescription( 3 ) );
// > DateDiff.GetDescription(3): 1 Year 4 Months 12 Days
Console.WriteLine( "DateDiff.GetDescription(4): {0}", dateDiff.GetDescription( 4 ) );
// > DateDiff.GetDescription(4): 1 Year 4 Months 12 Days 12 Hours
Console.WriteLine( "DateDiff.GetDescription(5): {0}", dateDiff.GetDescription( 5 ) );
// > DateDiff.GetDescription(5): 1 Year 4 Months 12 Days 12 Hours 41 Mins
Console.WriteLine( "DateDiff.GetDescription(6): {0}", dateDiff.GetDescription( 6 ) );
// > DateDiff.GetDescription(6): 1 Year 4 Months 12 Days 12 Hours 41 Mins 29 Secs
} // DateDiffSample
DateDiffalso calculates the difference of Quarters.
DateDiff还计算 Quarters 的差值。
回答by Colin
Based on Joaquim's answer, but fixing the calculation when end date month is less than start date month, and adding code to handle end date before start date:
基于 Joaquim 的回答，但在结束日期月份小于开始日期月份时修复计算，并添加代码以在开始日期之前处理结束日期：
public static class GeneralHelper
{
public static int GetYears(DateTime startDate, DateTime endDate)
{
if (endDate < startDate)
return -GetYears(endDate, startDate);
int years = (endDate.Year - startDate.Year);
if (endDate.Year == startDate.Year)
return years;
if (endDate.Month < startDate.Month)
return years - 1;
if (endDate.Month == startDate.Month && endDate.Day < startDate.Day)
return years - 1;
return years;
}
public static int GetMonths(DateTime startDate, DateTime endDate)
{
if (startDate > endDate)
return -GetMonths(endDate, startDate);
int months = 12 * GetYears(startDate, endDate);
if (endDate.Month > startDate.Month)
months = months + endDate.Month - startDate.Month;
else
months = 12 - startDate.Month + endDate.Month;
if (endDate.Day < startDate.Day)
months = months - 1;
return months;
}
}
[TestClass()]
public class GeneralHelperTest
{
[TestMethod]
public void GetYearsTest()
{
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2000, 5, 5), new DateTime(2000, 12, 31)));
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2000, 5, 5), new DateTime(2001, 4, 4)));
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2000, 5, 5), new DateTime(2001, 5, 4)));
Assert.AreEqual(1, GeneralHelper.GetYears(new DateTime(2000, 5, 5), new DateTime(2001, 5, 5)));
Assert.AreEqual(1, GeneralHelper.GetYears(new DateTime(2000, 5, 5), new DateTime(2001, 12, 31)));
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2000, 12, 31), new DateTime(2000, 5, 5)));
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2001, 4, 4), new DateTime(2000, 5, 5)));
Assert.AreEqual(0, GeneralHelper.GetYears(new DateTime(2001, 5, 4), new DateTime(2000, 5, 5)));
Assert.AreEqual(-1, GeneralHelper.GetYears(new DateTime(2001, 5, 5), new DateTime(2000, 5, 5)));
Assert.AreEqual(-1, GeneralHelper.GetYears(new DateTime(2001, 12, 31), new DateTime(2000, 5, 5)));
}
[TestMethod]
public void GetMonthsTest()
{
Assert.AreEqual(0, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2000, 6, 4)));
Assert.AreEqual(1, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2000, 6, 5)));
Assert.AreEqual(1, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2000, 6, 6)));
Assert.AreEqual(11, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2001, 5, 4)));
Assert.AreEqual(12, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2001, 5, 5)));
Assert.AreEqual(13, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2001, 6, 6)));
Assert.AreEqual(0, GeneralHelper.GetMonths(new DateTime(2000, 6, 4), new DateTime(2000, 5, 5)));
Assert.AreEqual(-1, GeneralHelper.GetMonths(new DateTime(2000, 6, 5), new DateTime(2000, 5, 5)));
Assert.AreEqual(-1, GeneralHelper.GetMonths(new DateTime(2000, 6, 6), new DateTime(2000, 5, 5)));
Assert.AreEqual(-11, GeneralHelper.GetMonths(new DateTime(2001, 5, 4), new DateTime(2000, 5, 5)));
Assert.AreEqual(-12, GeneralHelper.GetMonths(new DateTime(2001, 5, 5), new DateTime(2000, 5, 5)));
Assert.AreEqual(-13, GeneralHelper.GetMonths(new DateTime(2001, 6, 6), new DateTime(2000, 5, 5)));
}
}
EDITNo that still doesn't work. It fails this test:
编辑不，这仍然不起作用。它没有通过这个测试：
Assert.AreEqual(24, GeneralHelper.GetMonths(new DateTime(2000, 5, 5), new DateTime(2003, 5, 5)));
Expected:<24>. Actual:<12>
预期：<24>。实际：<12>
回答by pk.
I came across this post while looking to solve a similar problem. I was trying to find the age of an animal in units of Years, Months, Weeks, and Days. Those values are then displayed in SpinEdits where the user can manually change the values to find/estimate a birth date. When my form was passed a birth date from a month with less than 31 days, the value calculated was 1 day off. I based my solution off of Ic's answer above.
我在寻找解决类似问题时遇到了这篇文章。我试图以年、月、周和天为单位找出动物的年龄。这些值然后显示在 SpinEdits 中，用户可以在其中手动更改值以查找/估计出生日期。当我的表单通过一个少于 31 天的月份的出生日期时，计算的值是 1 天。我的解决方案基于 Ic 上面的回答。
Main calculation method that is called after my form loads.
在我的表单加载后调用的主要计算方法。
birthDateDisplay.Text = birthDate.ToString("MM/dd/yyyy");
DateTime currentDate = DateTime.Now;
Int32 numOfDays = 0;
Int32 numOfWeeks = 0;
Int32 numOfMonths = 0;
Int32 numOfYears = 0;
// changed code to follow this model http://stackoverflow.com/posts/1083990/revisions
//years
TimeSpan diff = currentDate - birthDate;
numOfYears = diff.Days / 366;
DateTime workingDate = birthDate.AddYears(numOfYears);
while (workingDate.AddYears(1) <= currentDate)
{
workingDate = workingDate.AddYears(1);
numOfYears++;
}
//months
diff = currentDate - workingDate;
numOfMonths = diff.Days / 31;
workingDate = workingDate.AddMonths(numOfMonths);
while (workingDate.AddMonths(1) <= currentDate)
{
workingDate = workingDate.AddMonths(1);
numOfMonths++;
}
//weeks and days
diff = currentDate - workingDate;
numOfWeeks = diff.Days / 7; //weeks always have 7 days
// if bday month is same as current month and bday day is after current day, the date is off by 1 day
if(DateTime.Now.Month == birthDate.Month && DateTime.Now.Day < birthDate.Day)
numOfDays = diff.Days % 7 + 1;
else
numOfDays = diff.Days % 7;
// If the there are fewer than 31 days in the birth month, the date calculated is 1 off
// Dont need to add a day for the first day of the month
int daysInMonth = 0;
if ((daysInMonth = DateTime.DaysInMonth(birthDate.Year, birthDate.Month)) != 31 && birthDate.Day != 1)
{
startDateforCalc = DateTime.Now.Date.AddDays(31 - daysInMonth);
// Need to add 1 more day if it is a leap year and Feb 29th is the date
if (DateTime.IsLeapYear(birthDate.Year) && birthDate.Day == 29)
startDateforCalc = startDateforCalc.AddDays(1);
}
yearsSpinEdit.Value = numOfYears;
monthsSpinEdit.Value = numOfMonths;
weeksSpinEdit.Value = numOfWeeks;
daysSpinEdit.Value = numOfDays;
And then, in my spinEdit_EditValueChanged event handler, I calculate the new birth date starting from my startDateforCalc based on the values in the spin edits. (SpinEdits are constrained to only allow >=0)
然后，在我的 spinEdit_EditValueChanged 事件处理程序中，我根据旋转编辑中的值从 startDateforCalc 开始计算新的出生日期。（SpinEdits 被限制为只允许 >=0）
birthDate = startDateforCalc.Date.AddYears(-((Int32)yearsSpinEdit.Value)).AddMonths(-((Int32)monthsSpinEdit.Value)).AddDays(-(7 * ((Int32)weeksSpinEdit.Value) + ((Int32)daysSpinEdit.Value)));
birthDateDisplay.Text = birthDate.ToString("MM/dd/yyyy");
I know its not the prettiest solution, but it seems to be working for me for all month lengths and years.
我知道它不是最漂亮的解决方案，但它似乎对我所有月份和年份都有效。
