private int GetYearDiff(DateTime start, DateTime end)
{
    int diff = end.Year - start.Year;
    if (end.DayOfYear < start.DayOfYear) { diff -= 1; }
    return diff;
}
[Fact]
public void GetYearDiff_WhenCalls_ShouldReturnCorrectYearDiff()
{
    //arrange
    var now = DateTime.Now;
    //act
    //assert
    Assert.Equal(24, GetYearDiff(new DateTime(1992, 7, 9), now)); // passed
    Assert.Equal(24, GetYearDiff(new DateTime(1992, now.Month, now.Day), now)); // passed
    Assert.Equal(23, GetYearDiff(new DateTime(1992, 12, 9), now)); // passed
}

这是我们在这里使用的版本。它有效，而且相当简单。这与Jeff的想法相同，但我认为它更清晰一点，因为它分离了减法的逻辑，所以更容易理解。

public static int GetAge(this DateTime dateOfBirth, DateTime dateAsAt)
{
    return dateAsAt.Year - dateOfBirth.Year - (dateOfBirth.DayOfYear < dateAsAt.DayOfYear ? 0 : 1);
}

如果你认为这类事情不清楚，你可以扩展三元运算符使其更清晰。

显然，这是作为DateTime上的一个扩展方法完成的，但很明显，您可以抓取一行代码来完成工作并将其放在任何位置。这里我们有另一个传入DateTime的Extension方法重载。

我使用ScArcher2的解决方案来精确计算人的年龄，但我需要进一步计算他们的月和日以及年。

    public static Dictionary<string,int> CurrentAgeInYearsMonthsDays(DateTime? ndtBirthDate, DateTime? ndtReferralDate)
    {
        //----------------------------------------------------------------------
        // Can't determine age if we don't have a dates.
        //----------------------------------------------------------------------
        if (ndtBirthDate == null) return null;
        if (ndtReferralDate == null) return null;

        DateTime dtBirthDate = Convert.ToDateTime(ndtBirthDate);
        DateTime dtReferralDate = Convert.ToDateTime(ndtReferralDate);

        //----------------------------------------------------------------------
        // Create our Variables
        //----------------------------------------------------------------------
        Dictionary<string, int> dYMD = new Dictionary<string,int>();
        int iNowDate, iBirthDate, iYears, iMonths, iDays;
        string sDif = "";

        //----------------------------------------------------------------------
        // Store off current date/time and DOB into local variables
        //---------------------------------------------------------------------- 
        iNowDate = int.Parse(dtReferralDate.ToString("yyyyMMdd"));
        iBirthDate = int.Parse(dtBirthDate.ToString("yyyyMMdd"));

        //----------------------------------------------------------------------
        // Calculate Years
        //----------------------------------------------------------------------
        sDif = (iNowDate - iBirthDate).ToString();
        iYears = int.Parse(sDif.Substring(0, sDif.Length - 4));

        //----------------------------------------------------------------------
        // Store Years in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Years", iYears);

        //----------------------------------------------------------------------
        // Calculate Months
        //----------------------------------------------------------------------
        if (dtBirthDate.Month > dtReferralDate.Month)
            iMonths = 12 - dtBirthDate.Month + dtReferralDate.Month - 1;
        else
            iMonths = dtBirthDate.Month - dtReferralDate.Month;

        //----------------------------------------------------------------------
        // Store Months in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Months", iMonths);

        //----------------------------------------------------------------------
        // Calculate Remaining Days
        //----------------------------------------------------------------------
        if (dtBirthDate.Day > dtReferralDate.Day)
            //Logic: Figure out the days in month previous to the current month, or the admitted month.
            //       Subtract the birthday from the total days which will give us how many days the person has lived since their birthdate day the previous month.
            //       then take the referral date and simply add the number of days the person has lived this month.

            //If referral date is january, we need to go back to the following year's December to get the days in that month.
            if (dtReferralDate.Month == 1)
                iDays = DateTime.DaysInMonth(dtReferralDate.Year - 1, 12) - dtBirthDate.Day + dtReferralDate.Day;       
            else
                iDays = DateTime.DaysInMonth(dtReferralDate.Year, dtReferralDate.Month - 1) - dtBirthDate.Day + dtReferralDate.Day;       
        else
            iDays = dtReferralDate.Day - dtBirthDate.Day;             

        //----------------------------------------------------------------------
        // Store Days in Return Value
        //----------------------------------------------------------------------
        dYMD.Add("Days", iDays);

        return dYMD;
}

这是最准确的答案之一，它能够解决2月29日的生日，而不是2月28日的任何一年。

public int GetAge(DateTime birthDate)
{
    int age = DateTime.Now.Year - birthDate.Year;

    if (birthDate.DayOfYear > DateTime.Now.DayOfYear)
        age--;

    return age;
}

简单易懂的解决方案。

// Save today's date.
var today = DateTime.Today;

// Calculate the age.
var age = today.Year - birthdate.Year;

// Go back to the year in which the person was born in case of a leap year
if (birthdate.Date > today.AddYears(-age)) age--;

然而，这假设你在寻找西方的时代观念，而不是使用东亚的推算法。

我创建了一个Age结构，如下所示：

public struct Age : IEquatable<Age>, IComparable<Age>
{
    private readonly int _years;
    private readonly int _months;
    private readonly int _days;

    public int Years  { get { return _years; } }
    public int Months { get { return _months; } }
    public int Days { get { return _days; } }

    public Age( int years, int months, int days ) : this()
    {
        _years = years;
        _months = months;
        _days = days;
    }

    public static Age CalculateAge( DateTime dateOfBirth, DateTime date )
    {
        // Here is some logic that ressembles Mike's solution, although it
        // also takes into account months & days.
        // Ommitted for brevity.
        return new Age (years, months, days);
    }

    // Ommited Equality, Comparable, GetHashCode, functionality for brevity.
}