使用c[副本]的年份日期差异

我已经编写了一个实现,它可以正确地处理相隔一年的日期。

但是,与其他算法不同,它不能优雅地处理负时间跨度。它也不使用自己的日期算法,而是依赖于标准库。

因此,不用再多费吹灰之力,下面是代码:

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);

用途:

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);
}

我们必须对支票进行编码,以确定两个日期之间的差异是否大于2年,即开始日期和结束日期。

由于上面的提示,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 .....

如果你需要它来了解某人的年龄是因为一些微不足道的原因,那么时间跨度是可以的,但是如果你需要计算退休金、长期存款或其他财务、科学或法律上的目的,那么时间跨度恐怕不够准确,因为时间跨度假定每年有相同的天数、相同的小时和相同的秒数)。

事实上,有些年份的长度会有所不同(因为不同的原因,这些原因不在这个答案的范围之内)。为了克服时间跨度的限制,您可以模拟Excel的功能,即:

    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;
    }

var totalYears = 
    (DateTime.Today - new DateTime(2007, 03, 11)).TotalDays
    / 365.2425;

开始的平均天数 Wikipedia/Leap_year .

目前还不清楚您希望如何处理分数年份,但可能是这样:

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);

我实现了一个扩展方法来获取两个日期之间的年数,用整个月份四舍五入。

    /// <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;
    }

这是一个让系统自动处理闰年的巧妙技巧。它给出了所有日期组合的准确答案。

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);

如果你想让别人的年龄,看看这个

How do I calculate someone's age in C#?

    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);
    }

int Age = new DateTime((DateTime.Now - BirthDateTime).Ticks).Year;

我发现了这个 TimeSpan for years, months and days :

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;

如果你处理的是月份和年份,你需要知道每个月有多少天,哪些年份是闰年。

进入 Gregorian Calendar (以及其他特定文化 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,但它确实有一些方法,例如

DateTime AddWeeks(DateTime time, int weeks)
DateTime AddMonths(DateTime time, int months)
DateTime AddYears(DateTime time, int years)

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

作品完美:

    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;
    }

简单解决方案:

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);
}

也许这将有助于回答以下问题: 给定年份的天数 ,

new DateTime(anyDate.Year, 12, 31).DayOfYear //will include leap years too

关于 DateTime.DayOfYear Property .

这是计算年和月差异的最佳代码:

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;
    }
}

下面是基于Dana的简单代码,它在大多数情况下都能产生正确的答案。但它没有考虑到两个日期之间不到一年的时间。下面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;
}

希望下面的链接有帮助

MSDN - DateTime.Subtract.Method (DateTime)

甚至还有一些关于C的例子。只需单击“C语言”选项卡。

祝你好运