如何获得两个日期之间的差异在年/月/周/日？ [英] How to get difference between two dates in Year/Month/Week/Day?
问题描述
如何获得两个日期之间的差异在年/月/周/日以有效的方式？
How to get difference between two dates in Year/Month/Week/Day in an efficient way?
如。两个日期之间的差为1年，2月，3周，4天。
eg. difference between two dates is 1 Year, 2 Months, 3 Weeks, 4 Days.
今年（S）的差异再presents算，两个日期之间一个月（县），周（S）和天（S）。
Difference represents count of year(s), month(s), week(s) and day(s) between two dates.
推荐答案
这其实是相当棘手。天的不同总数会导致同样的结果。例如：
This is actually quite tricky. A different total number of days can result in the same result. For example:
-
19日2008年6月至2010年6月19日= 2年，而且365 * 2天
19th June 2008 to 19th June 2010 = 2 years, but also 365 * 2 days
19日2006年6月至2008年6月19日= 2年，但是由于闰年也是365 +366天
19th June 2006 to 19th June 2008 = 2 years, but also 365 + 366 days due to leap years
您可能想减去年，直到你得到的地方，你有两个日期这是不到一年的时间间隔点。然后减去个月，直到你得到的地方，你有两个日期这是不到一个月的时间间隔点。
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.
进一步的混乱：减去（或增加）个月是棘手的，当你不妨先从3月30日的日期 - 什么是一个月前比
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?
甚至进一步的混乱（可能的是不相关的）：甚至一天不总是24小时。夏令人？
Even further confusion (may not be relevant): even a day isn't always 24 hours. Daylight saving anyone?
甚至进一步的混乱（几乎可以肯定的没有的相关）：即使一分钟并不总是60秒。闰秒是非常混乱...
Even further confusion (almost certainly not relevant): even a minute isn't always 60 seconds. Leap seconds are highly confusing...
我没有工作了，现在这样做的权利的确切正确的方式的时候 - 这个答案多半是提高的事实，它并不像它听起来几乎一样简单
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.
编辑：不幸的是，我不会有足够的时间来充分回答这个问题。我建议你开始定义一个结构重presenting一个周期：
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:
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!
}
}
我建议你先实施+运营商，这应告知区别方法 - 你应该确保第一+（Period.Difference（第一，第二））==第二个所有第一 / 第二值。
I suggest you implement the + operator first, which should inform the Difference method - you should make sure that first + (Period.Difference(first, second)) == second for all first/second values.
开始用编写单元测试整体转换 - 开始轻松的情况下，然后转移到棘手的那些涉及闰年。我知道正常的方法是在同一时间编写一个测试，但我个人头脑风暴一群人在开始执行任何工作之前。
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.
让自己每天要正确地实现这一点。这是棘手的东西。
Allow yourself a day to implement this properly. It's tricky stuff.
请注意，我在这里省略了周 - 该值至少是很容易的，因为它总是7天。因此，给予（阳性）期间，你必须：
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:
int years = period.Years;
int months = period.Months;
int weeks = period.Days / 7;
int daysWithinWeek = period.Days % 7;
（我建议你避免甚至想着消极的时期 - 确保一切都是积极的，所有的时间。）
(I suggest you avoid even thinking about negative periods - make sure everything is positive, all the time.)
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