下面是一个使用VB的更简洁的解决方案。只适用于年、月、日的净日期。你也可以在c#中加载DateDiff库。

Date1必须<= date2

VB。网

Dim date1 = Now.AddDays(-2000)
Dim date2 = Now
Dim diffYears = DateDiff(DateInterval.Year, date1, date2) - If(date1.DayOfYear > date2.DayOfYear, 1, 0)
Dim diffMonths = DateDiff(DateInterval.Month, date1, date2) - diffYears * 12 - If(date1.Day > date2.Day, 1, 0)
Dim diffDays = If(date2.Day >= date1.Day, date2.Day - date1.Day, date2.Day + (Date.DaysInMonth(date1.Year, date1.Month) - date1.Day))

C#

DateTime date1 = Now.AddDays(-2000);
DateTime date2 = Now;
int diffYears = DateDiff(DateInterval.Year, date1, date2) - date1.DayOfYear > date2.DayOfYear ? 1 : 0;
int diffMonths = DateDiff(DateInterval.Month, date1, date2) - diffYears * 12 - date1.Day > date2.Day ? 1 : 0;
int diffDays = date2.Day >= date1.Day ? date2.Day - date1.Day : date2.Day + (System.DateTime.DaysInMonth(date1.Year, date1.Month) - date1.Day);

要获得月份的差异(包括开始和结束)，而不考虑日期:

DateTime start = new DateTime(2013, 1, 1);
DateTime end = new DateTime(2014, 2, 1);
var diffMonths = (end.Month + end.Year * 12) - (start.Month + start.Year * 12);

这个简单的静态函数计算两个Datetimes之间的月份分数。

1.1. 到31.1。= 1.0 1.4. 到15.4。= 0.5 16.4. 到30.4。= 0.5 1.3. 到1.4。= 1 + 1/30

该函数假设第一个日期比第二个日期小。要处理负时间间隔，可以通过在开始时引入符号和变量交换来轻松地修改函数。

public static double GetDeltaMonths(DateTime t0, DateTime t1)
{
     DateTime t = t0;
     double months = 0;
     while(t<=t1)
     {
         int daysInMonth = DateTime.DaysInMonth(t.Year, t.Month);
         DateTime endOfMonth = new DateTime(t.Year, t.Month, daysInMonth);
         int cutDay = endOfMonth <= t1 ? daysInMonth : t1.Day;
         months += (cutDay - t.Day + 1) / (double) daysInMonth;
         t = new DateTime(t.Year, t.Month, 1).AddMonths(1);
     }
     return Math.Round(months,2);
 }

您可以使用以下扩展: 代码

public static class Ext
{
    #region Public Methods

    public static int GetAge(this DateTime @this)
    {
        var today = DateTime.Today;
        return ((((today.Year - @this.Year) * 100) + (today.Month - @this.Month)) * 100 + today.Day - @this.Day) / 10000;
    }

    public static int DiffMonths(this DateTime @from, DateTime @to)
    {
        return (((((@to.Year - @from.Year) * 12) + (@to.Month - @from.Month)) * 100 + @to.Day - @from.Day) / 100);
    }

    public static int DiffYears(this DateTime @from, DateTime @to)
    {
        return ((((@to.Year - @from.Year) * 100) + (@to.Month - @from.Month)) * 100 + @to.Day - @from.Day) / 10000;
    }

    #endregion Public Methods
}

实现!

int Age;
int years;
int Months;
//Replace your own date
var d1 = new DateTime(2000, 10, 22);
var d2 = new DateTime(2003, 10, 20);
//Age
Age = d1.GetAge();
Age = d2.GetAge();
//positive
years = d1.DiffYears(d2);
Months = d1.DiffMonths(d2);
//negative
years = d2.DiffYears(d1);
Months = d2.DiffMonths(d1);
//Or
Months = Ext.DiffMonths(d1, d2);
years = Ext.DiffYears(d1, d2); 

这是我所需要的。对我来说，一个月的哪一天并不重要，因为它总是碰巧是一个月的最后一天。

public static int MonthDiff(DateTime d1, DateTime d2){
    int retVal = 0;

    if (d1.Month<d2.Month)
    {
        retVal = (d1.Month + 12) - d2.Month;
        retVal += ((d1.Year - 1) - d2.Year)*12;
    }
    else
    {
        retVal = d1.Month - d2.Month;
        retVal += (d1.Year - d2.Year)*12;
    }
    //// Calculate the number of years represented and multiply by 12
    //// Substract the month number from the total
    //// Substract the difference of the second month and 12 from the total
    //retVal = (d1.Year - d2.Year) * 12;
    //retVal = retVal - d1.Month;
    //retVal = retVal - (12 - d2.Month);

    return retVal;
}

我只是需要一些简单的东西来满足例如，只输入月份/年的就业日期，所以希望工作的年份和月份不同。这就是我所使用的，只是为了实用

public static YearsMonths YearMonthDiff(DateTime startDate, DateTime endDate) {
    int monthDiff = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month) + 1;
    int years = (int)Math.Floor((decimal) (monthDiff / 12));
    int months = monthDiff % 12;
    return new YearsMonths {
        TotalMonths = monthDiff,
            Years = years,
            Months = months
    };
}

net小提琴