==常见说法（从几个月到几岁）===

如果您只是为了通用，以下是代码作为您的信息：

DateTime today = DateTime.Today;
DateTime bday = DateTime.Parse("2016-2-14");
int age = today.Year - bday.Year;
var unit = "";

if (bday > today.AddYears(-age))
{
    age--;
}
if (age == 0)   // Under one year old
{
    age = today.Month - bday.Month;

    age = age <= 0 ? (12 + age) : age;  // The next year before birthday

    age = today.Day - bday.Day >= 0 ? age : --age;  // Before the birthday.day

    unit = "month";
}
else {
    unit = "year";
}

if (age > 1)
{
    unit = unit + "s";
}

测试结果如下：

The birthday: 2016-2-14

2016-2-15 =>  age=0, unit=month;
2016-5-13 =>  age=2, unit=months;
2016-5-14 =>  age=3, unit=months; 
2016-6-13 =>  age=3, unit=months; 
2016-6-15 =>  age=4, unit=months; 
2017-1-13 =>  age=10, unit=months; 
2017-1-14 =>  age=11, unit=months; 
2017-2-13 =>  age=11, unit=months; 
2017-2-14 =>  age=1, unit=year; 
2017-2-15 =>  age=1, unit=year; 
2017-3-13 =>  age=1, unit=year;
2018-1-13 =>  age=1, unit=year; 
2018-1-14 =>  age=1, unit=year; 
2018-2-13 =>  age=1, unit=year; 
2018-2-14 =>  age=2, unit=years; 

保持简单（可能是愚蠢的：）。

DateTime birth = new DateTime(1975, 09, 27, 01, 00, 00, 00);
TimeSpan ts = DateTime.Now - birth;
Console.WriteLine("You are approximately " + ts.TotalSeconds.ToString() + " seconds old.");

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
}

这是一种奇怪的方法，但如果您将日期设置为yyyymmdd，并从当前日期中减去出生日期，然后删除您获得的年龄的最后4位数字：）

我不知道C#，但我相信这在任何语言中都适用。

20080814 - 19800703 = 280111 

删除最后4位=28。

C#代码：

int now = int.Parse(DateTime.Now.ToString("yyyyMMdd"));
int dob = int.Parse(dateOfBirth.ToString("yyyyMMdd"));
int age = (now - dob) / 10000;

或者，也可以不进行扩展方法形式的所有类型转换。忽略错误检查：

public static Int32 GetAge(this DateTime dateOfBirth)
{
    var today = DateTime.Today;

    var a = (today.Year * 100 + today.Month) * 100 + today.Day;
    var b = (dateOfBirth.Year * 100 + dateOfBirth.Month) * 100 + dateOfBirth.Day;

    return (a - b) / 10000;
}

这个解决方案怎么样？

static string CalcAge(DateTime birthDay)
{
    DateTime currentDate = DateTime.Now;         
    int approximateAge = currentDate.Year - birthDay.Year;
    int daysToNextBirthDay = (birthDay.Month * 30 + birthDay.Day) - 
        (currentDate.Month * 30 + currentDate.Day) ;

    if (approximateAge == 0 || approximateAge == 1)
    {                
        int month =  Math.Abs(daysToNextBirthDay / 30);
        int days = Math.Abs(daysToNextBirthDay % 30);

        if (month == 0)
            return "Your age is: " + daysToNextBirthDay + " days";

        return "Your age is: " + month + " months and " + days + " days"; ;
    }

    if (daysToNextBirthDay > 0)
        return "Your age is: " + --approximateAge + " Years";

    return "Your age is: " + approximateAge + " Years"; ;
}

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。