给定代表某人生日的DateTime,我如何计算他们的年龄(以年为单位)?


当前回答

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"; ;
}

其他回答

要使用最近的年龄计算年龄:

var ts = DateTime.Now - new DateTime(1988, 3, 19);
var age = Math.Round(ts.Days / 365.0);

我强烈建议使用名为AgeCalculator的NuGet软件包,因为在计算年龄(闰年、时间成分等)时需要考虑很多事情,而且只有两行代码不能削减它。该库给您的时间不只是一年。它甚至在计算时考虑了时间成分,这样你就可以得到一个包含年、月、日和时间成分的准确年龄。更为先进的是,可以选择将2月29日作为闰年,而将2月28日作为非闰年。

试试这个解决方案,它奏效了。

int age = (Int32.Parse(DateTime.Today.ToString("yyyyMMdd")) - 
           Int32.Parse(birthday.ToString("yyyyMMdd rawrrr"))) / 10000;
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
}
var startDate = new DateTime(2015, 04, 05);//your start date
var endDate = DateTime.Now;
var years = 0;
while(startDate < endDate) 
{
     startDate = startDate.AddYears(1);
     if(startDate < endDate) 
     {
         years++;
     }
}