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
对此的简单答案是应用AddYears,如下所示,因为这是唯一一种将年份添加到闰年2月29日的本地方法,并获得普通年份2月28日的正确结果。
有些人认为3月1日是勒普林斯的生日,但.Net和任何官方规则都不支持这一点,也没有常见的逻辑解释为什么一些出生在2月的人应该在另一个月拥有75%的生日。
此外,Age方法可以作为DateTime的扩展添加。由此,您可以以最简单的方式获得年龄:
列表项目
int age=出生日期.age();
public static class DateTimeExtensions
{
/// <summary>
/// Calculates the age in years of the current System.DateTime object today.
/// </summary>
/// <param name="birthDate">The date of birth</param>
/// <returns>Age in years today. 0 is returned for a future date of birth.</returns>
public static int Age(this DateTime birthDate)
{
return Age(birthDate, DateTime.Today);
}
/// <summary>
/// Calculates the age in years of the current System.DateTime object on a later date.
/// </summary>
/// <param name="birthDate">The date of birth</param>
/// <param name="laterDate">The date on which to calculate the age.</param>
/// <returns>Age in years on a later day. 0 is returned as minimum.</returns>
public static int Age(this DateTime birthDate, DateTime laterDate)
{
int age;
age = laterDate.Year - birthDate.Year;
if (age > 0)
{
age -= Convert.ToInt32(laterDate.Date < birthDate.Date.AddYears(age));
}
else
{
age = 0;
}
return age;
}
}
现在,运行此测试:
class Program
{
static void Main(string[] args)
{
RunTest();
}
private static void RunTest()
{
DateTime birthDate = new DateTime(2000, 2, 28);
DateTime laterDate = new DateTime(2011, 2, 27);
string iso = "yyyy-MM-dd";
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 3; j++)
{
Console.WriteLine("Birth date: " + birthDate.AddDays(i).ToString(iso) + " Later date: " + laterDate.AddDays(j).ToString(iso) + " Age: " + birthDate.AddDays(i).Age(laterDate.AddDays(j)).ToString());
}
}
Console.ReadKey();
}
}
关键日期示例如下:
出生日期:2000-02-29出生日期:2011-02-28年龄:11
输出:
{
Birth date: 2000-02-28 Later date: 2011-02-27 Age: 10
Birth date: 2000-02-28 Later date: 2011-02-28 Age: 11
Birth date: 2000-02-28 Later date: 2011-03-01 Age: 11
Birth date: 2000-02-29 Later date: 2011-02-27 Age: 10
Birth date: 2000-02-29 Later date: 2011-02-28 Age: 11
Birth date: 2000-02-29 Later date: 2011-03-01 Age: 11
Birth date: 2000-03-01 Later date: 2011-02-27 Age: 10
Birth date: 2000-03-01 Later date: 2011-02-28 Age: 10
Birth date: 2000-03-01 Later date: 2011-03-01 Age: 11
}
2012年2月28日晚些时候:
{
Birth date: 2000-02-28 Later date: 2012-02-28 Age: 12
Birth date: 2000-02-28 Later date: 2012-02-29 Age: 12
Birth date: 2000-02-28 Later date: 2012-03-01 Age: 12
Birth date: 2000-02-29 Later date: 2012-02-28 Age: 11
Birth date: 2000-02-29 Later date: 2012-02-29 Age: 12
Birth date: 2000-02-29 Later date: 2012-03-01 Age: 12
Birth date: 2000-03-01 Later date: 2012-02-28 Age: 11
Birth date: 2000-03-01 Later date: 2012-02-29 Age: 11
Birth date: 2000-03-01 Later date: 2012-03-01 Age: 12
}
其他回答
这很简单,似乎符合我的需要。我为闰年的目的做了一个假设,即无论一个人选择什么时候庆祝生日,从技术上讲,他们都不会比自己大一岁,直到他们的上一个生日过去365天(即2月28日不会使他们大一岁)。
DateTime now = DateTime.Today;
DateTime birthday = new DateTime(1991, 02, 03);//3rd feb
int age = now.Year - birthday.Year;
if (now.Month < birthday.Month || (now.Month == birthday.Month && now.Day < birthday.Day))//not had bday this year yet
age--;
return age;
我在这个问题上使用了以下内容。我知道它不太优雅,但它很管用。
DateTime zeroTime = new DateTime(1, 1, 1);
var date1 = new DateTime(1983, 03, 04);
var date2 = DateTime.Now;
var dif = date2 - date1;
int years = (zeroTime + dif).Year - 1;
Log.DebugFormat("Years -->{0}", years);
这里有一个解决方案。
DateTime dateOfBirth = new DateTime(2000, 4, 18);
DateTime currentDate = DateTime.Now;
int ageInYears = 0;
int ageInMonths = 0;
int ageInDays = 0;
ageInDays = currentDate.Day - dateOfBirth.Day;
ageInMonths = currentDate.Month - dateOfBirth.Month;
ageInYears = currentDate.Year - dateOfBirth.Year;
if (ageInDays < 0)
{
ageInDays += DateTime.DaysInMonth(currentDate.Year, currentDate.Month);
ageInMonths = ageInMonths--;
if (ageInMonths < 0)
{
ageInMonths += 12;
ageInYears--;
}
}
if (ageInMonths < 0)
{
ageInMonths += 12;
ageInYears--;
}
Console.WriteLine("{0}, {1}, {2}", ageInYears, ageInMonths, ageInDays);
只是因为我认为最重要的答案不是那么明确:
public static int GetAgeByLoop(DateTime birthday)
{
var age = -1;
for (var date = birthday; date < DateTime.Today; date = date.AddYears(1))
{
age++;
}
return age;
}
我有一个定制的计算年龄的方法,加上一条奖金验证消息,以防有帮助:
public void GetAge(DateTime dob, DateTime now, out int years, out int months, out int days)
{
years = 0;
months = 0;
days = 0;
DateTime tmpdob = new DateTime(dob.Year, dob.Month, 1);
DateTime tmpnow = new DateTime(now.Year, now.Month, 1);
while (tmpdob.AddYears(years).AddMonths(months) < tmpnow)
{
months++;
if (months > 12)
{
years++;
months = months - 12;
}
}
if (now.Day >= dob.Day)
days = days + now.Day - dob.Day;
else
{
months--;
if (months < 0)
{
years--;
months = months + 12;
}
days += DateTime.DaysInMonth(now.AddMonths(-1).Year, now.AddMonths(-1).Month) + now.Day - dob.Day;
}
if (DateTime.IsLeapYear(dob.Year) && dob.Month == 2 && dob.Day == 29 && now >= new DateTime(now.Year, 3, 1))
days++;
}
private string ValidateDate(DateTime dob) //This method will validate the date
{
int Years = 0; int Months = 0; int Days = 0;
GetAge(dob, DateTime.Now, out Years, out Months, out Days);
if (Years < 18)
message = Years + " is too young. Please try again on your 18th birthday.";
else if (Years >= 65)
message = Years + " is too old. Date of Birth must not be 65 or older.";
else
return null; //Denotes validation passed
}
方法调用此处并传递日期时间值(如果服务器设置为美国语言环境,则为MM/dd/yyyy)。将其替换为消息框或要显示的任何容器:
DateTime dob = DateTime.Parse("03/10/1982");
string message = ValidateDate(dob);
lbldatemessage.Visible = !StringIsNullOrWhitespace(message);
lbldatemessage.Text = message ?? ""; //Ternary if message is null then default to empty string
记住,您可以按任何方式格式化邮件。