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

这是一个非常简单的方法:

int Age = DateTime.Today.Year - new DateTime(2000, 1, 1).Year;

其他回答

一句话的回答:

DateTime dateOfBirth = Convert.ToDateTime("01/16/1990");
var age = ((DateTime.Now - dateOfBirth).Days) / 365;

这是最准确的答案之一,它能够解决2月29日的生日,而不是2月28日的任何一年。

public int GetAge(DateTime birthDate)
{
    int age = DateTime.Now.Year - birthDate.Year;

    if (birthDate.DayOfYear > DateTime.Now.DayOfYear)
        age--;

    return age;
}




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.");

这里有一个单行线:

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

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);
}