我认为TimeSpan包含了我们所需要的一切，而不必求助于365.25（或任何其他近似值）。扩展Aug的示例：

DateTime myBD = new DateTime(1980, 10, 10);
TimeSpan difference = DateTime.Now.Subtract(myBD);

textBox1.Text = difference.Years + " years " + difference.Months + " Months " + difference.Days + " days";

这里有一个单行线：

int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;

我们是否需要考虑小于1岁的人？作为中国文化，我们将小婴儿的年龄描述为2个月或4周。

下面是我的实现，它不像我想象的那么简单，尤其是处理2/28这样的日期。

public static string HowOld(DateTime birthday, DateTime now)
{
    if (now < birthday)
        throw new ArgumentOutOfRangeException("birthday must be less than now.");

    TimeSpan diff = now - birthday;
    int diffDays = (int)diff.TotalDays;

    if (diffDays > 7)//year, month and week
    {
        int age = now.Year - birthday.Year;

        if (birthday > now.AddYears(-age))
            age--;

        if (age > 0)
        {
            return age + (age > 1 ? " years" : " year");
        }
        else
        {// month and week
            DateTime d = birthday;
            int diffMonth = 1;

            while (d.AddMonths(diffMonth) <= now)
            {
                diffMonth++;
            }

            age = diffMonth-1;

            if (age == 1 && d.Day > now.Day)
                age--;

            if (age > 0)
            {
                return age + (age > 1 ? " months" : " month");
            }
            else
            {
                age = diffDays / 7;
                return age + (age > 1 ? " weeks" : " week");
            }
        }
    }
    else if (diffDays > 0)
    {
        int age = diffDays;
        return age + (age > 1 ? " days" : " day");
    }
    else
    {
        int age = diffDays;
        return "just born";
    }
}

此实现已通过以下测试用例。

[TestMethod]
public void TestAge()
{
    string age = HowOld(new DateTime(2011, 1, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("1 year", age);

    age = HowOld(new DateTime(2011, 11, 30), new DateTime(2012, 11, 30));
    Assert.AreEqual("1 year", age);

    age = HowOld(new DateTime(2001, 1, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("11 years", age);

    age = HowOld(new DateTime(2012, 1, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("10 months", age);

    age = HowOld(new DateTime(2011, 12, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("11 months", age);

    age = HowOld(new DateTime(2012, 10, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("1 month", age);

    age = HowOld(new DateTime(2008, 2, 28), new DateTime(2009, 2, 28));
    Assert.AreEqual("1 year", age);

    age = HowOld(new DateTime(2008, 3, 28), new DateTime(2009, 2, 28));
    Assert.AreEqual("11 months", age);

    age = HowOld(new DateTime(2008, 3, 28), new DateTime(2009, 3, 28));
    Assert.AreEqual("1 year", age);

    age = HowOld(new DateTime(2009, 1, 28), new DateTime(2009, 2, 28));
    Assert.AreEqual("1 month", age);

    age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 3, 1));
    Assert.AreEqual("1 month", age);

    // NOTE.
    // new DateTime(2008, 1, 31).AddMonths(1) == new DateTime(2009, 2, 28);
    // new DateTime(2008, 1, 28).AddMonths(1) == new DateTime(2009, 2, 28);
    age = HowOld(new DateTime(2009, 1, 31), new DateTime(2009, 2, 28));
    Assert.AreEqual("4 weeks", age);

    age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 2, 28));
    Assert.AreEqual("3 weeks", age);

    age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 3, 1));
    Assert.AreEqual("1 month", age);

    age = HowOld(new DateTime(2012, 11, 5), new DateTime(2012, 11, 30));
    Assert.AreEqual("3 weeks", age);

    age = HowOld(new DateTime(2012, 11, 1), new DateTime(2012, 11, 30));
    Assert.AreEqual("4 weeks", age);

    age = HowOld(new DateTime(2012, 11, 20), new DateTime(2012, 11, 30));
    Assert.AreEqual("1 week", age);

    age = HowOld(new DateTime(2012, 11, 25), new DateTime(2012, 11, 30));
    Assert.AreEqual("5 days", age);

    age = HowOld(new DateTime(2012, 11, 29), new DateTime(2012, 11, 30));
    Assert.AreEqual("1 day", age);

    age = HowOld(new DateTime(2012, 11, 30), new DateTime(2012, 11, 30));
    Assert.AreEqual("just born", age);

    age = HowOld(new DateTime(2000, 2, 29), new DateTime(2009, 2, 28));
    Assert.AreEqual("8 years", age);

    age = HowOld(new DateTime(2000, 2, 29), new DateTime(2009, 3, 1));
    Assert.AreEqual("9 years", age);

    Exception e = null;

    try
    {
        age = HowOld(new DateTime(2012, 12, 1), new DateTime(2012, 11, 30));
    }
    catch (ArgumentOutOfRangeException ex)
    {
        e = ex;
    }

    Assert.IsTrue(e != null);
}

希望这有帮助。

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。

这里有一个小的C#代码示例，我总结了一下，请注意边缘情况，特别是闰年，并不是所有上述解决方案都考虑到这些情况。将答案推出来作为DateTime可能会导致问题，因为你可能会在一个特定的月份中投入太多的时间，例如2月的30天。

public string LoopAge(DateTime myDOB, DateTime FutureDate)
{
    int years = 0;
    int months = 0;
    int days = 0;

    DateTime tmpMyDOB = new DateTime(myDOB.Year, myDOB.Month, 1);

    DateTime tmpFutureDate = new DateTime(FutureDate.Year, FutureDate.Month, 1);

    while (tmpMyDOB.AddYears(years).AddMonths(months) < tmpFutureDate)
    {
        months++;
        if (months > 12)
        {
            years++;
            months = months - 12;
        }
    }

    if (FutureDate.Day >= myDOB.Day)
    {
        days = days + FutureDate.Day - myDOB.Day;
    }
    else
    {
        months--;
        if (months < 0)
        {
            years--;
            months = months + 12;
        }
        days = days + (DateTime.DaysInMonth(FutureDate.AddMonths(-1).Year, FutureDate.AddMonths(-1).Month) + FutureDate.Day) - myDOB.Day;

    }

    //add an extra day if the dob is a leap day
    if (DateTime.IsLeapYear(myDOB.Year) && myDOB.Month == 2 && myDOB.Day == 29)
    {
        //but only if the future date is less than 1st March
        if(FutureDate >= new DateTime(FutureDate.Year, 3,1))
            days++;
    }

    return "Years: " + years + " Months: " + months + " Days: " + days;
}

还有一个答案：

public static int AgeInYears(DateTime birthday, DateTime today)
{
    return ((today.Year - birthday.Year) * 372 + (today.Month - birthday.Month) * 31 + (today.Day - birthday.Day)) / 372;
}

这已经过广泛的单元测试。它看起来确实有点“神奇”。数字372是如果每个月有31天，一年中会有多少天。

其工作原理的解释（此处省略）如下：

让我们设置Yn=DateTime.Now.Year，Yb=生日.Year，Mn=DateTime.Now.Month，Mb=生日.Month、Dn=DateTime.Now.Day，Db=生日.Day年龄=Yn-Yb+（31*（Mn-Mb）+（Dn-Db））/372我们知道，如果日期已经到达，我们需要的是Yn-Yb，如果日期尚未到达，则需要Yn-Yb-1。a） 如果Mn<Mb，我们有-341<=31*（Mn-Mb）<=-31和-30<=Dn-Db<=30-371<=31*（锰-Mb）+（Dn-Db）<=-1带整数除法（31*（Mn-Mb）+（Dn-Db））/372=-1b） 如果Mn=Mb和Dn<Db，则我们有31*（Mn-Mb）=0和-30<=Dn Db<=-1再次使用整数除法（31*（Mn-Mb）+（Dn-Db））/372=-1c） 如果Mn>Mb，我们有31<=31*（Mn-Mb）<=341和-30<=Dn-Db<=301<=31*（Mn-Mb）+（Dn-Db）<=371带整数除法（31*（Mn-Mb）+（Dn-Db））/372=0d） 如果Mn=Mb且Dn>Db，则我们有31*（Mn-Mb）=0且1<=Dn Db<=30再次使用整数除法（31*（Mn-Mb）+（Dn-Db））/372=0e） 如果Mn=Mb，Dn=Db，我们有31*（Mn-Mb）+Dn Db=0因此（31*（Mn-Mb）+（Dn-Db））/372=0