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

人们可以这样计算“年龄”(即“西方人”的方式):

public static int AgeInYears(this System.DateTime source, System.DateTime target)
  => target.Year - source.Year is int age && age > 0 && source.AddYears(age) > target ? age - 1 : age < 0 && source.AddYears(age) < target ? age + 1 : age;

如果时间方向为“负”,则年龄也将为负。

可以添加一个分数,代表从目标到下一个生日的累计年龄:

public static double AgeInTotalYears(this System.DateTime source, System.DateTime target)
{
  var sign = (source <= target ? 1 : -1);

  var ageInYears = AgeInYears(source, target); // The method above.

  var last = source.AddYears(ageInYears);
  var next = source.AddYears(ageInYears + sign);

  var fractionalAge = (double)(target - last).Ticks / (double)(next - last).Ticks * sign;

  return ageInYears + fractionalAge;
}

分数是过去的时间(从上一个生日到下一个生日)与总时间的比率。

无论是向前还是向后,这两种方法都以相同的方式工作。

其他回答

看看这个:

TimeSpan ts = DateTime.Now.Subtract(Birthdate);
age = (byte)(ts.TotalDays / 365.25);

这里有一个DateTime扩展程序,它将年龄计算添加到DateTime对象。

public static class AgeExtender
{
    public static int GetAge(this DateTime dt)
    {
        int d = int.Parse(dt.ToString("yyyyMMdd"));
        int t = int.Parse(DateTime.Today.ToString("yyyyMMdd"));
        return (t-d)/10000;
    }
}

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;
}

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.
}