如何计算两年中两个日期之间的日期差异?
例如:(Datetime.Now.Today() - 11/03/2007)
年。
答案 0 :(得分:101)
我编写了一个适用于相隔一年的日期的实现。
然而,与其他算法不同,它没有优雅地处理负时间跨度。它也不使用自己的日期算术,而是依赖于标准库。
所以不用多说,这里是代码:
DateTime zeroTime = new DateTime(1, 1, 1);
DateTime a = new DateTime(2007, 1, 1);
DateTime b = new DateTime(2008, 1, 1);
TimeSpan span = b - a;
// Because we start at year 1 for the Gregorian
// calendar, we must subtract a year here.
int years = (zeroTime + span).Year - 1;
// 1, where my other algorithm resulted in 0.
Console.WriteLine("Yrs elapsed: " + years);
答案 1 :(得分:39)
使用:
int Years(DateTime start, DateTime end)
{
return (end.Year - start.Year - 1) +
(((end.Month > start.Month) ||
((end.Month == start.Month) && (end.Day >= start.Day))) ? 1 : 0);
}
答案 2 :(得分:21)
我们必须对支票进行编码,以确定两个日期之间的差异,即开始日期和结束日期是否大于2年。
感谢上面的提示,它完成如下:
DateTime StartDate = Convert.ToDateTime("01/01/2012");
DateTime EndDate = Convert.ToDateTime("01/01/2014");
DateTime TwoYears = StartDate.AddYears(2);
if EndDate > TwoYears .....
答案 3 :(得分:13)
如果您因为琐碎的原因需要了解某人的年龄,那么Timespan可以,但如果您需要计算退休金,长期存款或任何其他用于财务,科学或法律目的的话,那么我担心Timespan会不准确因为Timespan假设每年都有相同的天数,相同的小时数和相同的秒数。
实际上,某些年份的长度会有所不同(出于不同的原因,超出了本答案的范围)。为了解决Timespan的限制,你可以模仿Excel的作用:
public int GetDifferenceInYears(DateTime startDate, DateTime endDate)
{
//Excel documentation says "COMPLETE calendar years in between dates"
int years = endDate.Year - startDate.Year;
if (startDate.Month == endDate.Month &&// if the start month and the end month are the same
endDate.Day < startDate.Day// AND the end day is less than the start day
|| endDate.Month < startDate.Month)// OR if the end month is less than the start month
{
years--;
}
return years;
}
答案 4 :(得分:5)
var totalYears =
(DateTime.Today - new DateTime(2007, 03, 11)).TotalDays
/ 365.2425;
Wikipedia/Leap_year的平均天数。
答案 5 :(得分:4)
目前还不清楚你想如何处理分数年,但也许是这样:
DateTime now = DateTime.Now;
DateTime origin = new DateTime(2007, 11, 3);
int calendar_years = now.Year - origin.Year;
int whole_years = calendar_years - ((now.AddYears(-calendar_years) >= origin)? 0: 1);
int another_method = calendar_years - ((now.Month - origin.Month) * 32 >= origin.Day - now.Day)? 0: 1);
答案 6 :(得分:3)
这是一个巧妙的技巧,让系统自动处理闰年。它为所有日期组合提供了准确的答案。
DateTime dt1 = new DateTime(1987, 9, 23, 13, 12, 12, 0);
DateTime dt2 = new DateTime(2007, 6, 15, 16, 25, 46, 0);
DateTime tmp = dt1;
int years = -1;
while (tmp < dt2)
{
years++;
tmp = tmp.AddYears(1);
}
Console.WriteLine("{0}", years);
答案 7 :(得分:3)
我实施了一种扩展方法,以获得两个日期之间的年数,四舍五入。
/// <summary>
/// Gets the total number of years between two dates, rounded to whole months.
/// Examples:
/// 2011-12-14, 2012-12-15 returns 1.
/// 2011-12-14, 2012-12-14 returns 1.
/// 2011-12-14, 2012-12-13 returns 0,9167.
/// </summary>
/// <param name="start">
/// Stardate of time period
/// </param>
/// <param name="end">
/// Enddate of time period
/// </param>
/// <returns>
/// Total Years between the two days
/// </returns>
public static double DifferenceTotalYears(this DateTime start, DateTime end)
{
// Get difference in total months.
int months = ((end.Year - start.Year) * 12) + (end.Month - start.Month);
// substract 1 month if end month is not completed
if (end.Day < start.Day)
{
months--;
}
double totalyears = months / 12d;
return totalyears;
}
答案 8 :(得分:2)
如果你想让某个人的年龄看到这个
答案 9 :(得分:1)
public string GetAgeText(DateTime birthDate)
{
const double ApproxDaysPerMonth = 30.4375;
const double ApproxDaysPerYear = 365.25;
int iDays = (DateTime.Now - birthDate).Days;
int iYear = (int)(iDays / ApproxDaysPerYear);
iDays -= (int)(iYear * ApproxDaysPerYear);
int iMonths = (int)(iDays / ApproxDaysPerMonth);
iDays -= (int)(iMonths * ApproxDaysPerMonth);
return string.Format("{0} år, {1} måneder, {2} dage", iYear, iMonths, iDays);
}
答案 10 :(得分:1)
int Age = new DateTime((DateTime.Now - BirthDateTime).Ticks).Year;
答案 11 :(得分:1)
我在 TimeSpan for years, months and days 找到了这个:
DateTime target_dob = THE_DOB;
DateTime true_age = DateTime.MinValue + ((TimeSpan)(DateTime.Now - target_dob )); // Minimum value as 1/1/1
int yr = true_age.Year - 1;
答案 12 :(得分:0)
如果您要处理数月和数年,您需要知道每个月有多少天以及哪些年是闰年。
输入Gregorian Calendar(以及其他特定于文化的Calendar实施)。
虽然日历不提供直接计算两个时间点之间差异的方法,但它确实有
等方法DateTime AddWeeks(DateTime time, int weeks)
DateTime AddMonths(DateTime time, int months)
DateTime AddYears(DateTime time, int years)
答案 13 :(得分:0)
DateTime musteriDogum = new DateTime(dogumYil, dogumAy, dogumGun);
int additionalDays = ((DateTime.Now.Year - dogumYil) / 4); //Count of the years with 366 days
int extraDays = additionalDays + ((DateTime.Now.Year % 4 == 0 || musteriDogum.Year % 4 == 0) ? 1 : 0); //We add 1 if this year or year inserted has 366 days
int yearsOld = ((DateTime.Now - musteriDogum).Days - extraDays ) / 365; // Now we extract these extra days from total days and we can divide to 365
答案 14 :(得分:0)
简单的解决方案:
public int getYearDiff(DateTime startDate, DateTime endDate){
int y = Year(endDate) - Year(startDate);
int startMonth = Month(startDate);
int endMonth = Month(endDate);
if (endMonth < startMonth)
return y - 1;
if (endMonth > startMonth)
return y;
return (Day(endDate) < Day(startDate) ? y - 1 : y);
}
答案 15 :(得分:0)
也许这有助于回答这个问题:给定年份的天数,
new DateTime(anyDate.Year, 12, 31).DayOfYear //will include leap years too
答案 16 :(得分:0)
这是计算年月差异的最佳代码:
DateTime firstDate = DateTime.Parse("1/31/2019");
DateTime secondDate = DateTime.Parse("2/1/2016");
int totalYears = firstDate.Year - secondDate.Year;
int totalMonths = 0;
if (firstDate.Month > secondDate.Month)
totalMonths = firstDate.Month - secondDate.Month;
else if (firstDate.Month < secondDate.Month)
{
totalYears -= 1;
int monthDifference = secondDate.Month - firstDate.Month;
totalMonths = 12 - monthDifference;
}
if ((firstDate.Day - secondDate.Day) == 30)
{
totalMonths += 1;
if (totalMonths % 12 == 0)
{
totalYears += 1;
totalMonths = 0;
}
}
答案 17 :(得分:0)
完美的作品:
internal static int GetDifferenceInYears(DateTime startDate)
{
int finalResult = 0;
const int DaysInYear = 365;
DateTime endDate = DateTime.Now;
TimeSpan timeSpan = endDate - startDate;
if (timeSpan.TotalDays > 365)
{
finalResult = (int)Math.Round((timeSpan.TotalDays / DaysInYear), MidpointRounding.ToEven);
}
return finalResult;
}
答案 18 :(得分:0)
以下是基于Dana的简单代码,它在大多数情况下会产生正确的答案。但它并没有考虑到日期之间不到一年的时间。所以这是我用来产生一致结果的代码:
public static int DateDiffYears(DateTime startDate, DateTime endDate)
{
var yr = endDate.Year - startDate.Year - 1 +
(endDate.Month >= startDate.Month && endDate.Day >= startDate.Day ? 1 : 0);
return yr < 0 ? 0 : yr;
}
答案 19 :(得分:-2)