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
一句话的回答:
DateTime dateOfBirth = Convert.ToDateTime("01/16/1990");
var age = ((DateTime.Now - dateOfBirth).Days) / 365;
其他回答
因为闰年和所有事情,我知道的最好的方法是:
DateTime birthDate = new DateTime(2000,3,1);
int age = (int)Math.Floor((DateTime.Now - birthDate).TotalDays / 365.25D);
2需要解决的主要问题有:
1.计算准确年龄-以年、月、日等为单位。
2.计算人们普遍认为的年龄——人们通常不关心自己到底多大,他们只关心自己当年的生日是什么时候。
1的解决方案显而易见:
DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today; //we usually don't care about birth time
TimeSpan age = today - birth; //.NET FCL should guarantee this as precise
double ageInDays = age.TotalDays; //total number of days ... also precise
double daysInYear = 365.2425; //statistical value for 400 years
double ageInYears = ageInDays / daysInYear; //can be shifted ... not so precise
2的解决方案在确定总年龄时并不那么精确,但人们认为它是精确的。当人们“手动”计算年龄时,通常也会使用它:
DateTime birth = DateTime.Parse("1.1.2000");
DateTime today = DateTime.Today;
int age = today.Year - birth.Year; //people perceive their age in years
if (today.Month < birth.Month ||
((today.Month == birth.Month) && (today.Day < birth.Day)))
{
age--; //birthday in current year not yet reached, we are 1 year younger ;)
//+ no birthday for 29.2. guys ... sorry, just wrong date for birth
}
注释2.:
这是我的首选解决方案我们不能使用DateTime.DayOfYear或TimeSpans,因为它们会在闰年中改变天数为了可读性,我只增加了几行
还有一个提示。。。我将为它创建两个静态重载方法,一个用于通用,另一个用于使用友好:
public static int GetAge(DateTime bithDay, DateTime today)
{
//chosen solution method body
}
public static int GetAge(DateTime birthDay)
{
return GetAge(birthDay, DateTime.Now);
}
这个经典问题值得野田时间来解决。
static int GetAge(LocalDate dateOfBirth)
{
Instant now = SystemClock.Instance.Now;
// The target time zone is important.
// It should align with the *current physical location* of the person
// you are talking about. When the whereabouts of that person are unknown,
// then you use the time zone of the person who is *asking* for the age.
// The time zone of birth is irrelevant!
DateTimeZone zone = DateTimeZoneProviders.Tzdb["America/New_York"];
LocalDate today = now.InZone(zone).Date;
Period period = Period.Between(dateOfBirth, today, PeriodUnits.Years);
return (int) period.Years;
}
用法:
LocalDate dateOfBirth = new LocalDate(1976, 8, 27);
int age = GetAge(dateOfBirth);
您可能还对以下改进感兴趣:
将时钟作为IClock传递,而不是使用SystemClock.Instance,将提高可测试性。目标时区可能会更改,因此您也需要DateTimeZone参数。
另请参阅我关于这个主题的博客文章:处理生日和其他周年纪念日
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;
}
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