计算年、月、周和日

Question

在我的应用程序中，用户输入两个日期。 预定的开始日期和预定的结束日期。 我们必须获取这些日期，并根据差异填充 4 个字段。

所以，假设他选择 2010 年 1 月 1 日作为开始，2011 年 3 月 2 日作为结束，我们需要以：

年：1 月：2 周：0 天 1

这意味着总持续时间为 1 年 2 个月零 1 天。

有这样做的标准方法吗？ 或者我是否需要编写一个具有很多非常棘手的逻辑的方法来解决这个问题？ 我希望我很幸运，并且会有一个日期差异类型.Net class 可用。

Answer 1

这是一个完整的方法，周数不包括在内，但可以相对简单地添加。 这是一个有点复杂的问题（在 stackoverflow 上以多种方式提出并以多种方式回答得很差），但仍然可以回答。 TimeSpan object 为我们提供了部分所需内容，但只能持续几天。 我已经针对此方法编写了大量测试，如果您发现漏洞，请发表评论。

这将做的是比较 2 个日期，获取年、月、日、小时和分钟。 （例如，某些事件发生在 1 年、6 个月、3 天、4 小时和 7 分钟前）

因为这个问题已经被问过很多次并试图回答很多次，所以我不确定它是否会被注意到，但如果是这样的话，它应该会提供价值。

    public static void TimeSpanToDateParts(DateTime d1, DateTime d2, out int years, out int months, out int days, out int hours, out int minutes)
    {
        if (d1 < d2)
        {
            var d3 = d2;
            d2 = d1;
            d1 = d3;
        }

        var span = d1 - d2;

        months = 12 * (d1.Year - d2.Year) + (d1.Month - d2.Month);

        //month may need to be decremented because the above calculates the ceiling of the months, not the floor.
        //to do so we increase d2 by the same number of months and compare.
        //(500ms fudge factor because datetimes are not precise enough to compare exactly)
        if (d1.CompareTo(d2.AddMonths(months).AddMilliseconds(-500)) <= 0)
        {
            --months;
        }

        years = months / 12;
        months -= years * 12;

        if (months == 0 && years == 0)
        {
            days = span.Days;
        }
        else
        {
            var md1 = new DateTime(d1.Year, d1.Month, d1.Day);
            // Fixed to use d2.Day instead of d1.Day
            var md2 = new DateTime(d2.Year, d2.Month, d2.Day);
            var mDays = (int) (md1 - md2).TotalDays;

            if (mDays > span.Days)
            {
                mDays = (int)(md1.AddMonths(-1) - md2).TotalDays;
            }

            days = span.Days - mDays;


        }
        hours = span.Hours;
        minutes = span.Minutes;
    }

Answer 2

您可以使用此免费库的DateDiff class：

// ----------------------------------------------------------------------
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
} // DateDiffSample

Answer 3

我也需要这个，但就我而言，没有周部分（所以只有年、月和日）。 鉴于此，这就是我所做的：

DateTime startDate = DateTime.ParseExact (start, "dd/MM/yyyy",CultureInfo.InvariantCulture);
DateTime endDate = DateTime.ParseExact (end, "dd/MM/yyyy",CultureInfo.InvariantCulture);
int days=0;
int months = 0;
int years = 0;
//calculate days
if (endDate.Day >= startDate.Day) {
    days = endDate.Day - startDate.Day;
} else {
    var tempDate = endDate.AddMonths (-1);
    int daysInMonth = DateTime.DaysInMonth (tempDate.Year, tempDate.Month);
    days = daysInMonth - (startDate.Day - endDate.Day);
    months--;
}
//calculate months
if (endDate.Month >= startDate.Month) {
    months+=endDate.Month - startDate.Month;
} else {
    months+= 12 - (startDate.Month - endDate.Month);
    years--;
}
//calculate years
years+=endDate.Year - startDate.Year;
Debug.WriteLine (string.Format("{0} years, {1} months, {2} days",years,months,days));

如果你想更动态地显示这个，你也可以使用这个代码：

//build the string
var result = "";
if (years!=0){
    result = years == 1 ? years + " year" : years + " years";
}
if (months != 0) {
    if (result != "") {
        result += ", ";
    }
    result += months == 1 ? months + " month" : months + " months";
}
if (days != 0) {
    if (result != "") {
        result += ", ";
    }
    result += days == 1 ? days + " day" : days + " days";
}
Debug.WriteLine (result);

Answer 4

应该这样做。 关键是如果给定的 2 个日期之间有奇数个闰日，则将天数减 1。

    /// <summary>
    /// //Assume DateTime dt1 < DateTime dt2, print out difference between dt1 to dt2 in years, months, weeks and days
    /// </summary>
    /// <param name="dt1"></param>
    /// <param name="dt2"></param>
    static void DateDiff(DateTime dt1, DateTime dt2)
    {
        DateTime zeroTime = new DateTime(1, 1, 1);

        int leapDaysInBetween = CountLeapDays(dt1, dt2);

        TimeSpan span = dt2 - dt1;

        int years = (zeroTime + span).Year - 1;
        int months = (zeroTime + span).Month - 1;
        int days = (zeroTime + span).Day - (leapDaysInBetween % 2 == 1 ? 1 : 0);
        int weeks = days / 7;
        int remainingdays = days % 7;

        Console.WriteLine(String.Format("\nThe difference between date {0} and date {1} is: \n\t {2} year(s), {3} month(s), and {4} day(s).", dt1, dt2, years, months, days));
        Console.WriteLine(String.Format("\nThe difference between date {0} and date {1} is: \n\t {2} year(s), {3} month(s), {4} week(s) and {5} day(s).", dt1, dt2, years, months, weeks, remainingdays));
    }

    private static int CountLeapDays(DateTime dt1, DateTime dt2)
    {
        int leapDaysInBetween = 0;
        int year1 = dt1.Year, year2 = dt2.Year;
        DateTime dateValue;

        for (int i = year1; i <= year2; i++)
        {
            if (DateTime.TryParse("02/29/" + i.ToString(), out dateValue))
            {
                if (dateValue >= dt1 && dateValue <= dt2)
                    leapDaysInBetween++;
            }
        }

        return leapDaysInBetween;
    }

运行了这些测试：

static void Main(string[] args)
{
    DateDiff(new DateTime(2010, 1, 1), new DateTime(2012, 2, 9));
    DateDiff(new DateTime(2010, 1, 1), new DateTime(2012, 4, 9));
    DateDiff(new DateTime(2010, 1, 1), new DateTime(2020, 2, 9));
    DateDiff(new DateTime(2010, 1, 1), new DateTime(2020, 4, 9));
    DateDiff(new DateTime(2020, 2, 29), new DateTime(2021, 2, 28));
    DateDiff(new DateTime(2019, 2, 28), new DateTime(2021, 2, 28));
}

这些是打印输出：

日期 1/1/2010 12:00:00 AM 和日期 4/9/2012 12:00:00 AM 之间的差异是：2 年、3 个月和 9 天。

日期 1/1/2010 12:00:00 AM 和日期 4/9/2012 12:00:00 AM 之间的差异是：2 年、3 个月、1 周和 2 天(s).

日期 1/1/2010 12:00:00 AM 和日期 2/9/2020 12:00:00 AM 之间的差异是：10 年、1 个月和 9 天。

日期 1/1/2010 12:00:00 AM 和日期 2/9/2020 12:00:00 AM 之间的差异是：10 年、1 个月、1 周和 2 天(s).

日期 1/1/2010 12:00:00 AM 和日期 4/9/2020 12:00:00 AM 之间的差异是：10 年、3 个月和 9 天。

日期 1/1/2010 12:00:00 AM 和日期 4/9/2020 12:00:00 AM 之间的差异是：10 年、3 个月、1 周和 2 天(s).

日期 2/29/2020 12:00:00 AM 和日期 2/28/2021 12:00:00 AM 之间的差异是：1 年、0 个月和 0 天。

日期 2/29/2020 12:00:00 AM 和日期 2/28/2021 12:00:00 AM 之间的差异是：1 年、0 个月、0 周和 0 天(s).

日期 2/28/2019 12:00:00 AM 和日期 2/28/2021 12:00:00 AM 之间的差异是：2 年、0 个月和 1 天。

日期 2/28/2019 12:00:00 AM 和日期 2/28/2021 12:00:00 AM 之间的差异是：2 年、0 个月、0 周和 1 天(s).

Answer 5

 public partial class Age1 : System.Web.UI.Page
    {
        private int Years;
        private int Months;
        private int Days;
        DateTime Cday;
        DateTime Bday;


        protected void Page_Load(object sender, EventArgs e)
        {
            txtCurrentDate.Enabled = false;
            txtCurrentDate.Text = DateTime.Now.ToString("g");
            Cday = Convert.ToDateTime(txtCurrentDate.Text);
        }

        protected void Button1_Click(object sender, EventArgs e)
        {
            Bday = Convert.ToDateTime(txtBirthdate.Text);
            AgeCaluclation(Bday, Cday);
            txtBirthdate.Text = "";
            txtCurrentDate.Text = "";
            lblAge.Text = this.Years+"  Years "+this.Months+"  Months " +this.Days+ "Days";
        }

        private Age1 AgeCaluclation(DateTime Bday, DateTime Cday)
        {

            if ((Cday.Year - Bday.Year) > 0 || 
               (((Cday.Year - Bday.Year) == 0) && 
               ((Bday.Month < Cday.Month) ||
               ((Bday.Month == Cday.Month) && 
               (Bday.Day <= Cday.Day)))))
            {

                int DaysInBdayMonth = DateTime.DaysInMonth(Bday.Year, Bday.Month);
                int DaysRemain = Cday.Day + (DaysInBdayMonth - Bday.Day);

                    if(Cday.Month > Bday.Month)
                    {
                        this.Years = Cday.Year - Bday.Year;
                        this.Months = Cday.Month - (Bday.Month + 1) + Math.Abs(DaysRemain / DaysInBdayMonth);
                        this.Days = (DaysRemain % DaysInBdayMonth + DaysInBdayMonth) % DaysInBdayMonth;
                    }
                    else if (Cday.Month == Bday.Month)
                    {
                        if (Cday.Day >= Bday.Day)
                        {
                            this.Years = Cday.Year - Bday.Year;
                            this.Months = 0;
                            this.Days = Cday.Day - Bday.Day;
                        }
                        else
                        {
                            this.Years = (Cday.Year - 1) - Bday.Year;
                            this.Months = 11;
                            this.Days = DateTime.DaysInMonth(Bday.Year, Bday.Month) - (Bday.Day - Cday.Day);

                        }
                    }
                    else
                    {
                        this.Years = (Cday.Year - 1) - Bday.Year;
                        this.Months = Cday.Month + (11 - Bday.Month) + Math.Abs(DaysRemain / DaysInBdayMonth);
                        this.Days = (DaysRemain % DaysInBdayMonth + DaysInBdayMonth) % DaysInBdayMonth;
                    }
            }
            else
            {
                throw new ArgumentException("Birthday date must be earlier than current date");
            }
            return this;
        }

    }

Answer 6

我创建了这个用于返回 2 个日期之间的年、月和日差异。

public static Dictionary<string, int> TimeSpanToDateParts(DateTime fromDate,DateTime toDate)
        {
            int years;
            int months;
            int days;
            Dictionary<string, int> dateParts = new Dictionary<string, int>();
            if (toDate < fromDate)
            {
               return TimeSpanToDateParts(toDate,fromDate);
            }

            var span = toDate - fromDate;

            months = 12 * (toDate.Year - fromDate.Year) + (toDate.Month - fromDate.Month);

            if (toDate.CompareTo(fromDate.AddMonths(months).AddMilliseconds(-500)) <= 0)
            {
                --months;
            }

            years = months / 12;
            months -= years * 12;

            if (months == 0 && years == 0)
            {
                days = span.Days;
            }
            else
            {
                days = toDate.Day;
                if (fromDate.Day > toDate.Day)
                    days = days + (DateTime.DaysInMonth(toDate.Year, toDate.Month - 1) - fromDate.Day);
                else
                    days = days - fromDate.Day;
            }
            dateParts.Add("Years", years);
            dateParts.Add("Months", months);
            dateParts.Add("Days", days);

            return dateParts;
        }

Answer 7

我认为TimeSpan是你要找的，但它不会做几年或几个月，因为它们的长度不同。

下面的例子来自上面的链接；

// Define two dates.
DateTime date1 = new DateTime(2010, 1, 1, 8, 0, 15);
DateTime date2 = new DateTime(2010, 8, 18, 13, 30, 30);
// Calculate the interval between the two dates.
TimeSpan interval = date2 - date1;
Console.WriteLine("{0} - {1} = {2}", date2, date1, interval.ToString());
// Display individual properties of the resulting TimeSpan object.
Console.WriteLine("   {0,-35} {1,20}", "Value of Days Component:", interval.Days);
Console.WriteLine("   {0,-35} {1,20}", "Total Number of Days:", interval.TotalDays);
Console.WriteLine("   {0,-35} {1,20}", "Value of Hours Component:", interval.Hours);
Console.WriteLine("   {0,-35} {1,20}", "Total Number of Hours:", interval.TotalHours);
Console.WriteLine("   {0,-35} {1,20}", "Value of Minutes Component:", interval.Minutes);
Console.WriteLine("   {0,-35} {1,20}", "Total Number of Minutes:", interval.TotalMinutes);
Console.WriteLine("   {0,-35} {1,20:N0}", "Value of Seconds Component:", interval.Seconds);
Console.WriteLine("   {0,-35} {1,20:N0}", "Total Number of Seconds:", interval.TotalSeconds);
Console.WriteLine("   {0,-35} {1,20:N0}", "Value of Milliseconds Component:", interval.Milliseconds);
Console.WriteLine("   {0,-35} {1,20:N0}", "Total Number of Milliseconds:", interval.TotalMilliseconds);
Console.WriteLine("   {0,-35} {1,20:N0}", "Ticks:", interval.Ticks);
// the example displays the following output:
//       8/18/2010 1:30:30 PM - 1/1/2010 8:00:15 AM = 229.05:30:15
//          Value of Days Component:                             229
//          Total Number of Days:                   229.229340277778
//          Value of Hours Component:                              5
//          Total Number of Hours:                  5501.50416666667
//          Value of Minutes Component:                           30
//          Total Number of Minutes:                       330090.25
//          Value of Seconds Component:                           15
//          Total Number of Seconds:                      19,805,415
//          Value of Milliseconds Component:                       0
//          Total Number of Milliseconds:             19,805,415,000
//          Ticks:                               198,054,150,000,000

Answer 8

调查 C# TimeSpan结构。 它不会持续数月或数年，但会持续数天。 这让事情变得更容易。

Answer 9

如果您首先计算总月数和总天数，则逻辑并不过分复杂。 然后很容易用除法和模法做一些 integer 的数学运算。

DateTime start_date = new DateTime(2010, 1, 1);
DateTime end_date = new DateTime(2011, 3, 2);

int diff_years = end_date.Year - start_date.Year;
int diff_months = end_date.Month - start_date.Month;

int total_months = 12 * diff_years + diff_months;
DateTime near_end_date = start_date.AddMonths(total_months);
int total_days = (int)end_date.Subtract(near_end_date).TotalDays;

int years = total_months / 12;
int months = total_months % 12;
int weeks = total_days / 7;
int days = total_days % 7;

Console.WriteLine($"Years: {years} Months: {months} Weeks: {weeks} Days: {days}");

这给出了Years: 1 Months: 2 Weeks: 0 Days: 1正如预期的那样。

Answer 10

这是我刚刚构建的转换器。 我没有尝试上面的所有解决方案，但我尝试过的所有解决方案在某种程度上仍然略有偏差。 这考虑了年、闰年、月、周、日等，然后获取最后两个最有用的值 YM 或 MW 或 WD 或 DH。 这很简单，但我的项目需要它，也许对以后的人有帮助。 当然，如果出于某种原因这不好，我相信你们都会告诉我。

        DateTime beforeDate = topic.lastActive.ToLocalTime();
        DateTime futureDate = DateTime.Now.ToLocalTime();

        int minutes = 0;
        int hours = 0;
        int days = 0;
        int weeks = 0;
        int months = 0;
        int years = 0;

        Dictionary<int, int> dictMonths = new Dictionary<int, int> { };
        dictMonths.Add(1, 31);
        if (DateTime.IsLeapYear(futureDate.Year))
            dictMonths.Add(2, 29);
        else
            dictMonths.Add(2, 28);
        dictMonths.Add(3, 31);
        dictMonths.Add(4, 30);
        dictMonths.Add(5, 31);
        dictMonths.Add(6, 30);
        dictMonths.Add(7, 31);
        dictMonths.Add(8, 31);
        dictMonths.Add(9, 30);
        dictMonths.Add(10, 31);
        dictMonths.Add(11, 30);
        dictMonths.Add(12, 31);

        //Time difference between dates
        TimeSpan span = futureDate - beforeDate;
        hours = span.Hours;
        minutes = span.Minutes;

        //Days total
        days = span.Days;
        //Find how many years
        DateTime zeroTime = new DateTime(1, 1, 1);

        // Because we start at year 1 for the Gregorian
        // calendar, we must subtract a year here.
        years = (zeroTime + span).Year - 1;
        //find difference of days of years already found
        int startYear = futureDate.Year - years;
        for (int i = 0; i < years; i++)
        {
            if (DateTime.IsLeapYear(startYear))
                days -= 366;
            else
                days -= 365;

            startYear++;
        }
        //Find months by multiplying months in year by difference of datetime years then add difference of current year months
        months = 12 * (futureDate.Year - beforeDate.Year) + (futureDate.Month - beforeDate.Month);
        //month may need to be decremented because the above calculates the ceiling of the months, not the floor.
        //to do so we increase before by the same number of months and compare.
        //(500ms fudge factor because datetimes are not precise enough to compare exactly)
        if (futureDate.CompareTo(beforeDate.AddMonths(months).AddMilliseconds(-500)) <= 0)
        {
            --months;
        }
        //subtract months from how many years we have already accumulated
        months -= (12 * years);
        //find how many days by compared to our month dictionary
        int startMonth = beforeDate.Month;
        for (int i = 0; i < months; i++)
        {
            //check if faulty leap year
            if (startMonth == 2 && (months - 1 > 10))
                days -= 28;
            else
                days -= dictMonths[startMonth];
            startMonth++;
            if (startMonth > 12)
            {
                startMonth = 1;
            }
        }
        //Find if any weeks are within our now total days
        weeks = days / 7;
        if (weeks > 0)
        {
            //remainder is days left
            days = days % 7;
        }

        Console.WriteLine(years + " " + months + " " + weeks + " " + days + " " + span.Hours + " " + span.Minutes);

        if (years > 0)
        {
            if (years > 1 && months > 1)
                return $"Latest Reply: {years} years, {months} months ago.";
            else if (years > 1 && months == 0)
                return $"Latest Reply: {years} years ago.";
            else if (years == 1 && months == 0)
                return $"Latest Reply: {years} year ago.";
            else if (years > 1)
                return $"Latest Reply: {years} years, {months} month ago.";
            else if (months > 1)
                return $"Latest Reply: {years} year, {months} months ago.";
            else
                return $"Latest Reply: {years} year, {months} month ago.";
        }
        else if (months > 0)
        {
            if (months > 1 && weeks > 1)
                return $"Latest Reply: {months} months, {weeks} weeks ago.";
            else if (months > 1 && weeks == 0)
                return $"Latest Reply: {months} months ago.";
            else if (months == 1 && weeks == 0)
                return $"Latest Reply: {months} month ago.";
            else if (months > 1)
                return $"Latest Reply: {months} months, {weeks} week ago.";
            else if (weeks > 1)
                return $"Latest Reply: {months} month, {weeks} weeks ago.";
            else
                return $"Latest Reply: {months} month, {weeks} week ago.";
        }
        else if (weeks > 0)
        {
            if (weeks > 1 && days > 1)
                return $"Latest Reply: {weeks} weeks, {days} days ago.";
            else if (weeks > 1 && days == 0)
                return $"Latest Reply: {weeks} weeks ago.";
            else if (weeks == 1 && days == 0)
                return $"Latest Reply: {weeks} week ago.";
            else if (weeks > 1)
                return $"Latest Reply: {weeks} weeks, {days} day ago.";
            else if (days > 1)
                return $"Latest Reply: {weeks} week, {days} days ago.";
            else
                return $"Latest Reply: {weeks} week, {days} day ago.";
        }
        else if (days > 0)
        {
            if (days > 1 && hours > 1)
                return $"Latest Reply: {days} days, {hours} hours ago.";
            else if (days > 1 && hours == 0)
                return $"Latest Reply: {days} days ago.";
            else if (days == 1 && hours == 0)
                return $"Latest Reply: {days} day ago.";
            else if (days > 1)
                return $"Latest Reply: {days} days, {hours} hour ago.";
            else if (hours > 1)
                return $"Latest Reply: {days} day, {hours} hours ago.";
            else
                return $"Latest Reply: {days} day, {hours} hour ago.";
        }
        else if (hours > 0)
        {
            if (hours > 1 && minutes > 1)
                return $"Latest Reply: {hours} hours, {minutes} minutes ago.";
            else if (hours > 1 && minutes == 0)
                return $"Latest Reply: {hours} hours ago.";
            else if (hours == 1 && minutes == 0)
                return $"Latest Reply: {hours} hour ago.";
            else if (hours > 1)
                return $"Latest Reply: {hours} hours, {minutes} minute ago.";
            else if (minutes > 1)
                return $"Latest Reply: {hours} hour, {minutes} minutes ago.";
            else
                return $"Latest Reply: {hours} hour, {minutes} minute ago.";
        }
        else if (minutes > 0)
        {
            if (minutes > 1)
                return $"Latest Reply: {minutes} minutes ago.";
            else
                return $"Latest Reply: {minutes} minute ago.";
        }
        else
        {
            return $"Latest Reply: Just now.";
        }

Answer 11

这是一个很老的问题，但还是时候问问 go。

重要的是要考虑月份有不同的天数。

    public static (int, int, int) GetYearsMonthsDaysDifference(DateTime date1, DateTime date2)
    {
        DateTime startDate = date1;
        DateTime endDate = date2;

        if (date1 > date2)
        {
            startDate = date2;
            endDate = date1;
        }

        var months = (endDate.Month + (12 - startDate.Month)) - (endDate.Day < startDate.Day ? 1 : 0);
        var years = (endDate.Year - startDate.Year - (months < 12 ? 1 : 0));
        months -= (months >= 12 ? 12 : 0);
        var days = endDate.DayOfYear - startDate.AddYears(years).AddMonths(months).DayOfYear;

        return (years, months, days);
    }

Answer 12

在我的应用程序中，用户输入了两个日期。 预定的开始日期和预定的结束日期。 我们必须采用这些日期，并根据差异填充 4 个字段。

所以，假设他选择 2010 年 1 月 1 日作为开始，2011 年 3 月 2 日作为结束，我们需要结束：

年：1 月：2 周：0 天 1

意味着总持续时间为1年2个月1天。

有这样做的标准方法吗？ 或者我是否需要编写一个具有很多非常棘手的逻辑的方法来解决它？ 我希望我会很幸运，并且会有一个 date-diff type.Net class 可用。