Mik*_*len 2017
一个易于理解和简单的解决方案.
// Save today's date.
var today = DateTime.Today;
// Calculate the age.
var age = today.Year - birthdate.Year;
// Go back to the year the person was born in case of a leap year
if (birthdate.Date > today.AddYears(-age)) age--;
然而,这假设您正在寻找西方的年龄观念,而不是使用东亚推算.
ScA*_*er2 983
这是一种奇怪的方法,但是如果您将日期格式化yyyymmdd为当前日期并从当前日期减去出生日期,则删除您已达到年龄的最后4位数字:)
我不知道C#,但我相信这适用于任何语言.
20080814 - 19800703 = 280111
删掉最后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;
或者可选地,没有以扩展方法的形式进行所有类型转换.错误检查省略:
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;
}
小智 377
我不知道如何接受错误的解决方案.正确的C#片段由Michael Stum撰写
这是一个测试片段:
DateTime bDay = new DateTime(2000, 2, 29);
DateTime now = new DateTime(2009, 2, 28);
MessageBox.Show(string.Format("Test {0} {1} {2}",
CalculateAgeWrong1(bDay, now), // outputs 9
CalculateAgeWrong2(bDay, now), // outputs 9
CalculateAgeCorrect(bDay, now), // outputs 8
CalculateAgeCorrect2(bDay, now))); // outputs 8
在这里你有方法:
public int CalculateAgeWrong1(DateTime birthDate, DateTime now)
{
return new DateTime(now.Subtract(birthDate).Ticks).Year - 1;
}
public int CalculateAgeWrong2(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
if (now < birthDate.AddYears(age))
age--;
return age;
}
public int CalculateAgeCorrect(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
if (now.Month < birthDate.Month || (now.Month == birthDate.Month && now.Day < birthDate.Day))
age--;
return age;
}
public int CalculateAgeCorrect2(DateTime birthDate, DateTime now)
{
int age = now.Year - birthDate.Year;
// for leap years we need this
if (birthDate > now.AddYears(-age)) age--;
// don't use:
// if (birthDate.AddYears(age) > now) age--;
return age;
}
Jam*_*sen 130
我认为迄今为止的任何答案都没有提供以不同方式计算年龄的文化.例如,参见东亚年龄估算与西方年龄相比.
任何真正的答案都必须包括本地化.在这个例子中,战略模式可能是有序的.
小智 105
对此的简单回答是应用AddYears如下所示,因为这是闰年2月29日增加年份的唯一本机方法,并获得2月28日的正确年份的正确结果.
有些人认为3月1日是闰人的生日,但是.Net和任何官方规则都不支持这一点,也没有共同逻辑解释为什么2月份出生的人应该在另一个月有75%的生日.
此外,Age方法适合作为扩展添加DateTime.通过这种方式,您可以以最简单的方式获得年龄:
int age = birthDate.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();
}
}
关键日期的例子如下:
出生日期:2000-02-29后期:2011-02-28年龄:11岁
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
}
而对于2012-02-28的后期日期:
{
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
}
Jam*_*ran 86
我的建议
int age = (int) ((DateTime.Now - bday).TotalDays/365.242199);
这似乎在正确的日期改变了一年.(我发现测试到107岁)
Mic*_*tum 72
另一个功能,不是我,而是在网上找到并稍微改进一下:
public static int GetAge(DateTime birthDate)
{
DateTime n = DateTime.Now; // To avoid a race condition around midnight
int age = n.Year - birthDate.Year;
if (n.Month < birthDate.Month || (n.Month == birthDate.Month && n.Day < birthDate.Day))
age--;
return age;
}
我想到的只有两件事:那些来自不使用格里高历的国家的人呢？DateTime.Now是我认为的特定于服务器的文化.我对亚洲日历的实际工作知之甚少,我不知道是否有一种简单的方法可以在日历之间转换日期,但万一你想知道那些来自4660年的中国人:-)
Mar*_*oth 51
2要解决的主要问题是:
1.计算确切年龄 - 以年,月,日等为单位.
2.计算一般认知的年龄 - 人们通常不关心他们到底多大年纪,他们只关心当年的生日.
1的解决方案很明显:
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的解决方案是在确定总年龄时不那么精确的解决方案,但被人们认为是精确的.当人们"手动"计算他们的年龄时,人们通常也会使用它:
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
}
2的注释:
再说一遍......我会为它创建2个静态重载方法,一个用于通用,第二个用于友好性:
public static int GetAge(DateTime bithDay, DateTime today)
{
//chosen solution method body
}
public static int GetAge(DateTime birthDay)
{
return GetAge(birthDay, DateTime.Now);
}
Sil*_*key 47
我迟到了,但这是一个单行:
int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;
Dav*_*ier 40
这是我们在这里使用的版本.它有效,而且非常简单.它与杰夫的想法相同,但我认为它更清晰一点,因为它分离出减去逻辑的逻辑,所以它更容易理解.
public static int GetAge(this DateTime dateOfBirth, DateTime dateAsAt)
{
return dateAsAt.Year - dateOfBirth.Year - (dateOfBirth.DayOfYear < dateAsAt.DayOfYear ? 0 : 1);
}
如果您认为有些事情不清楚,您可以扩展三元运算符以使其更清晰.
显然这是作为一种扩展方法完成的DateTime,但显然你可以抓住那一行代码完成工作并把它放在任何地方.这里我们有另一个传入的Extension方法的重载DateTime.Now,只是为了完整性.
Nic*_*rdi 37
我知道的最好的方法是因为闰年和一切都是:
DateTime birthDate = new DateTime(2000,3,1);
int age = (int)Math.Floor((DateTime.Now - birthDate).TotalDays / 365.25D);
希望这可以帮助.
Elm*_*mer 32
我用这个:
public static class DateTimeExtensions
{
public static int Age(this DateTime birthDate)
{
return Age(birthDate, DateTime.Now);
}
public static int Age(this DateTime birthDate, DateTime offsetDate)
{
int result=0;
result = offsetDate.Year - birthDate.Year;
if (offsetDate.DayOfYear < birthDate.DayOfYear)
{
result--;
}
return result;
}
}
Jac*_*ult 30
这为这个问题提供了"更多细节".也许这就是你要找的东西
DateTime birth = new DateTime(1974, 8, 29);
DateTime today = DateTime.Now;
TimeSpan span = today - birth;
DateTime age = DateTime.MinValue + span;
// Make adjustment due to MinValue equalling 1/1/1
int years = age.Year - 1;
int months = age.Month - 1;
int days = age.Day - 1;
// Print out not only how many years old they are but give months and days as well
Console.Write("{0} years, {1} months, {2} days", years, months, days);
use*_*601 26
我创建了一个SQL Server用户定义函数来计算某人的年龄,考虑到他们的出生日期.当您需要它作为查询的一部分时,这非常有用:
using System;
using System.Data;
using System.Data.Sql;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
public partial class UserDefinedFunctions
{
[SqlFunction(DataAccess = DataAccessKind.Read)]
public static SqlInt32 CalculateAge(string strBirthDate)
{
DateTime dtBirthDate = new DateTime();
dtBirthDate = Convert.ToDateTime(strBirthDate);
DateTime dtToday = DateTime.Now;
// get the difference in years
int years = dtToday.Year - dtBirthDate.Year;
// subtract another year if we're before the
// birth day in the current year
if (dtToday.Month < dtBirthDate.Month || (dtToday.Month == dtBirthDate.Month && dtToday.Day < dtBirthDate.Day))
years=years-1;
int intCustomerAge = years;
return intCustomerAge;
}
};
Mat*_*son 26
这是另一个答案:
public static int AgeInYears(DateTime birthday, DateTime today)
{
return ((today.Year - birthday.Year) * 372 + (today.Month - birthday.Month) * 31 + (today.Day - birthday.Day)) / 372;
}
这已经过广泛的单元测试.它确实看起来有点"神奇".如果每个月有31天,则数字372是一年中的天数.
它起作用的解释(从这里解除)是:
我们来吧
Yn = DateTime.Now.Year, Yb = birthday.Year, Mn = DateTime.Now.Month, Mb = birthday.Month, Dn = DateTime.Now.Day, Db = birthday.Day
age = Yn - Yb + (31*(Mn - Mb) + (Dn - Db)) / 372我们知道我们需要的是,
Yn-Yb如果已经达到日期,Yn-Yb-1如果没有达到.a)如果
Mn<Mb,我们有-341 <= 31*(Mn-Mb) <= -31 and -30 <= Dn-Db <= 30
-371 <= 31*(Mn - Mb) + (Dn - Db) <= -1用整数除法
(31*(Mn - Mb) + (Dn - Db)) / 372 = -1b)如果
Mn=Mb和Dn<Db,我们有31*(Mn - Mb) = 0 and -30 <= Dn-Db <= -1再次使用整数除法
(31*(Mn - Mb) + (Dn - Db)) / 372 = -1c)如果
Mn>Mb,我们有31 <= 31*(Mn-Mb) <= 341 and -30 <= Dn-Db <= 30
1 <= 31*(Mn - Mb) + (Dn - Db) <= 371用整数除法
(31*(Mn - Mb) + (Dn - Db)) / 372 = 0d)如果
Mn=Mb和Dn>Db,我们有31*(Mn - Mb) = 0 and 1 <= Dn-Db <= 30再次使用整数除法
(31*(Mn - Mb) + (Dn - Db)) / 372 = 0e)如果
Mn=Mb和Dn=Db,我们有31*(Mn - Mb) + Dn-Db = 0因此
(31*(Mn - Mb) + (Dn - Db)) / 372 = 0
小智 24
我花了一些时间研究这个并想出了这个来计算一个人的年龄,数月和数日.我已经对2月29日的问题和闰年进行了测试,似乎有效,我很感激任何反馈:
public void 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 +=
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++;
}
}
roc*_*ock 20
我们需要考虑小于1年的人吗？作为中国文化,我们将小婴儿的年龄描述为2个月或4周.
下面是我的实现,它并不像我想象的那么简单,尤其是像2/28那样处理日期.
public static string HowOld(DateTime birthday, DateTime now)
{
if (now < birthday)
throw new ArgumentOutOfRangeException("birthday must be less than now.");
TimeSpan diff = now - birthday;
int diffDays = (int)diff.TotalDays;
if (diffDays > 7)//year, month and week
{
int age = now.Year - birthday.Year;
if (birthday > now.AddYears(-age))
age--;
if (age > 0)
{
return age + (age > 1 ? " years" : " year");
}
else
{// month and week
DateTime d = birthday;
int diffMonth = 1;
while (d.AddMonths(diffMonth) <= now)
{
diffMonth++;
}
age = diffMonth-1;
if (age == 1 && d.Day > now.Day)
age--;
if (age > 0)
{
return age + (age > 1 ? " months" : " month");
}
else
{
age = diffDays / 7;
return age + (age > 1 ? " weeks" : " week");
}
}
}
else if (diffDays > 0)
{
int age = diffDays;
return age + (age > 1 ? " days" : " day");
}
else
{
int age = diffDays;
return "just born";
}
}
此实现已通过测试用例.
[TestMethod]
public void TestAge()
{
string age = HowOld(new DateTime(2011, 1, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("1 year", age);
age = HowOld(new DateTime(2011, 11, 30), new DateTime(2012, 11, 30));
Assert.AreEqual("1 year", age);
age = HowOld(new DateTime(2001, 1, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("11 years", age);
age = HowOld(new DateTime(2012, 1, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("10 months", age);
age = HowOld(new DateTime(2011, 12, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("11 months", age);
age = HowOld(new DateTime(2012, 10, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("1 month", age);
age = HowOld(new DateTime(2008, 2, 28), new DateTime(2009, 2, 28));
Assert.AreEqual("1 year", age);
age = HowOld(new DateTime(2008, 3, 28), new DateTime(2009, 2, 28));
Assert.AreEqual("11 months", age);
age = HowOld(new DateTime(2008, 3, 28), new DateTime(2009, 3, 28));
Assert.AreEqual("1 year", age);
age = HowOld(new DateTime(2009, 1, 28), new DateTime(2009, 2, 28));
Assert.AreEqual("1 month", age);
age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 3, 1));
Assert.AreEqual("1 month", age);
// NOTE.
// new DateTime(2008, 1, 31).AddMonths(1) == new DateTime(2009, 2, 28);
// new DateTime(2008, 1, 28).AddMonths(1) == new DateTime(2009, 2, 28);
age = HowOld(new DateTime(2009, 1, 31), new DateTime(2009, 2, 28));
Assert.AreEqual("4 weeks", age);
age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 2, 28));
Assert.AreEqual("3 weeks", age);
age = HowOld(new DateTime(2009, 2, 1), new DateTime(2009, 3, 1));
Assert.AreEqual("1 month", age);
age = HowOld(new DateTime(2012, 11, 5), new DateTime(2012, 11, 30));
Assert.AreEqual("3 weeks", age);
age = HowOld(new DateTime(2012, 11, 1), new DateTime(2012, 11, 30));
Assert.AreEqual("4 weeks", age);
age = HowOld(new DateTime(2012, 11, 20), new DateTime(2012, 11, 30));
Assert.AreEqual("1 week", age);
age = HowOld(new DateTime(2012, 11, 25), new DateTime(2012, 11, 30));
Assert.AreEqual("5 days", age);
age = HowOld(new DateTime(2012, 11, 29), new DateTime(2012, 11, 30));
Assert.AreEqual("1 day", age);
age = HowOld(new DateTime(2012, 11, 30), new DateTime(2012, 11, 30));
Assert.AreEqual("just born", age);
age = HowOld(new DateTime(2000, 2, 29), new DateTime(2009, 2, 28));
Assert.AreEqual("8 years", age);
age = HowOld(new DateTime(2000, 2, 29), new DateTime(2009, 3, 1));
Assert.AreEqual("9 years", age);
Exception e = null;
try
{
age = HowOld(new DateTime(2012, 12, 1), new DateTime(2012, 11, 30));
}
catch (ArgumentOutOfRangeException ex)
{
e = ex;
}
Assert.IsTrue(e != null);
}
希望它有用.
小智 19
保持简单(可能是愚蠢:)).
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.");
Nic*_*rey 18
我发现的最简单的方法就是这个.它适用于美国和西欧的地区.不能和其他地方说话,特别是像中国这样的地方.在初始计算年龄后,最多只能进行4次比较.
public int AgeInYears(DateTime birthDate, DateTime referenceDate)
{
Debug.Assert(referenceDate >= birthDate,
"birth date must be on or prior to the reference date");
DateTime birth = birthDate.Date;
DateTime reference = referenceDate.Date;
int years = (reference.Year - birth.Year);
//
// an offset of -1 is applied if the birth date has
// not yet occurred in the current year.
//
if (reference.Month > birth.Month);
else if (reference.Month < birth.Month)
--years;
else // in birth month
{
if (reference.Day < birth.Day)
--years;
}
return years ;
}
我正在寻找答案,并注意到没有人提到闰日分娩的监管/法律影响.例如,根据维基百科,如果您出生于2月29日的各个司法管辖区,那么您的非闰年生日会有所不同:
而且就我所知,在美国,法规对这个问题保持沉默,使其符合普通法以及各种监管机构如何在其规则中定义事物.
为此,改进:
public enum LeapDayRule
{
OrdinalDay = 1 ,
LastDayOfMonth = 2 ,
}
static int ComputeAgeInYears(DateTime birth, DateTime reference, LeapYearBirthdayRule ruleInEffect)
{
bool isLeapYearBirthday = CultureInfo.CurrentCulture.Calendar.IsLeapDay(birth.Year, birth.Month, birth.Day);
DateTime cutoff;
if (isLeapYearBirthday && !DateTime.IsLeapYear(reference.Year))
{
switch (ruleInEffect)
{
case LeapDayRule.OrdinalDay:
cutoff = new DateTime(reference.Year, 1, 1)
.AddDays(birth.DayOfYear - 1);
break;
case LeapDayRule.LastDayOfMonth:
cutoff = new DateTime(reference.Year, birth.Month, 1)
.AddMonths(1)
.AddDays(-1);
break;
default:
throw new InvalidOperationException();
}
}
else
{
cutoff = new DateTime(reference.Year, birth.Month, birth.Day);
}
int age = (reference.Year - birth.Year) + (reference >= cutoff ? 0 : -1);
return age < 0 ? 0 : age;
}
应该注意,此代码假定:
Dak*_*ock 18
TimeSpan diff = DateTime.Now - birthdayDateTime;
string age = String.Format("{0:%y} years, {0:%M} months, {0:%d}, days old", diff);
我不确定你是多么希望它回到你身边,所以我只是做了一个可读的字符串.
Raj*_*S P 17
这是一个解决方案.
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);
fli*_*erg 17
这不是一个直接的答案,而是从准科学的角度来看更多关于手头问题的哲学推理.
我认为这个问题没有说明衡量年龄的单位或文化,大多数答案似乎都假设一个整数年度表示.时间的SI单位是second,正确的通用答案应该是(当然假设归一化DateTime并且不考虑相对论效应):
var lifeInSeconds = (DateTime.Now.Ticks - then.Ticks)/TickFactor;
以基督徒计算年龄的方式:
var then = ... // Then, in this case the birthday
var now = DateTime.UtcNow;
int age = now.Year - then.Year;
if (now.AddYears(-age) < then) age--;
在财务方面,在计算通常被称为日计数分数的事情时存在类似的问题,该分数大致是给定时期的若干年.年龄问题实际上是衡量问题的时间.
实际/实际(计算所有天"正确")约定的示例:
DateTime start, end = .... // Whatever, assume start is before end
double startYearContribution = 1 - (double) start.DayOfYear / (double) (DateTime.IsLeapYear(start.Year) ? 366 : 365);
double endYearContribution = (double)end.DayOfYear / (double)(DateTime.IsLeapYear(end.Year) ? 366 : 365);
double middleContribution = (double) (end.Year - start.Year - 1);
double DCF = startYearContribution + endYearContribution + middleContribution;
测量时间的另一种常见方式通常是"序列化"(命名此日期约定的人必须认真地进行骚扰):
DateTime start, end = .... // Whatever, assume start is before end
int days = (end - start).Days;
我想知道我们必须走多长时间才能在几秒钟内相对论时代变得比在一生中到目前为止的地球周围太阳周期的粗略近似更有用:)或者换句话说,当一个时期必须给定一个位置或者表示动作本身有效的函数:)
mjb*_*mjb 16
与2月28日的任何一年相比,这是能够解决2月29日生日的最准确答案之一.
public int GetAge(DateTime birthDate)
{
int age = DateTime.Now.Year - birthDate.Year;
if (birthDate.DayOfYear > DateTime.Now.DayOfYear)
age--;
return age;
}
Dar*_*vil 15
我有一个自定义的方法来计算年龄,加上奖金验证消息,以防它有帮助:
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datetime值(如果服务器设置为USA locale,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
请记住,您可以按照自己喜欢的方式格式化邮件.
小智 14
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"; ;
}
AEM*_*iji 12
private int GetAge(int _year, int _month, int _day
{
DateTime yourBirthDate= new DateTime(_year, _month, _day);
DateTime todaysDateTime = DateTime.Today;
int noOfYears = todaysDateTime.Year - yourBirthDate.Year;
if (DateTime.Now.Month < yourBirthDate.Month ||
(DateTime.Now.Month == yourBirthDate.Month && DateTime.Now.Day < yourBirthDate.Day))
{
noOfYears--;
}
return noOfYears;
}
小智 10
以下方法(从.NET类DateDiff的时间段文件中提取)考虑文化信息的日历:
// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2 )
{
return YearDiff( date1, date2, DateTimeFormatInfo.CurrentInfo.Calendar );
} // YearDiff
// ----------------------------------------------------------------------
private static int YearDiff( DateTime date1, DateTime date2, Calendar calendar )
{
if ( date1.Equals( date2 ) )
{
return 0;
}
int year1 = calendar.GetYear( date1 );
int month1 = calendar.GetMonth( date1 );
int year2 = calendar.GetYear( date2 );
int month2 = calendar.GetMonth( date2 );
// find the the day to compare
int compareDay = date2.Day;
int compareDaysPerMonth = calendar.GetDaysInMonth( year1, month1 );
if ( compareDay > compareDaysPerMonth )
{
compareDay = compareDaysPerMonth;
}
// build the compare date
DateTime compareDate = new DateTime( year1, month2, compareDay,
date2.Hour, date2.Minute, date2.Second, date2.Millisecond );
if ( date2 > date1 )
{
if ( compareDate < date1 )
{
compareDate = compareDate.AddYears( 1 );
}
}
else
{
if ( compareDate > date1 )
{
compareDate = compareDate.AddYears( -1 );
}
}
return year2 - calendar.GetYear( compareDate );
} // YearDiff
用法:
// ----------------------------------------------------------------------
public void CalculateAgeSamples()
{
PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2009, 02, 28 ) );
// > Birthdate=29.02.2000, Age at 28.02.2009 is 8 years
PrintAge( new DateTime( 2000, 02, 29 ), new DateTime( 2012, 02, 28 ) );
// > Birthdate=29.02.2000, Age at 28.02.2012 is 11 years
} // CalculateAgeSamples
// ----------------------------------------------------------------------
public void PrintAge( DateTime birthDate, DateTime moment )
{
Console.WriteLine( "Birthdate={0:d}, Age at {1:d} is {2} years", birthDate, moment, YearDiff( birthDate, moment ) );
} // PrintAge
Mat*_*int 10
这个经典问题值得Noda Time解决方案.
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:
作为一个IClock而不是使用时钟传递SystemClock.Instance将提高可测试性.
目标时区可能会发生变化,因此您也需要一个DateTimeZone参数.
另请参阅我关于此主题的博客文章:处理生日和其他周年纪念日
我使用ScArcher2的解决方案来准确计算一个人的年龄,但我需要进一步计算它们的月份和日期以及年份.
public static Dictionary<string,int> CurrentAgeInYearsMonthsDays(DateTime? ndtBirthDate, DateTime? ndtReferralDate)
{
//----------------------------------------------------------------------
// Can't determine age if we don't have a dates.
//----------------------------------------------------------------------
if (ndtBirthDate == null) return null;
if (ndtReferralDate == null) return null;
DateTime dtBirthDate = Convert.ToDateTime(ndtBirthDate);
DateTime dtReferralDate = Convert.ToDateTime(ndtReferralDate);
//----------------------------------------------------------------------
// Create our Variables
//----------------------------------------------------------------------
Dictionary<string, int> dYMD = new Dictionary<string,int>();
int iNowDate, iBirthDate, iYears, iMonths, iDays;
string sDif = "";
//----------------------------------------------------------------------
// Store off current date/time and DOB into local variables
//----------------------------------------------------------------------
iNowDate = int.Parse(dtReferralDate.ToString("yyyyMMdd"));
iBirthDate = int.Parse(dtBirthDate.ToString("yyyyMMdd"));
//----------------------------------------------------------------------
// Calculate Years
//----------------------------------------------------------------------
sDif = (iNowDate - iBirthDate).ToString();
iYears = int.Parse(sDif.Substring(0, sDif.Length - 4));
//----------------------------------------------------------------------
// Store Years in Return Value
//----------------------------------------------------------------------
dYMD.Add("Years", iYears);
//----------------------------------------------------------------------
// Calculate Months
//----------------------------------------------------------------------
if (dtBirthDate.Month > dtReferralDate.Month)
iMonths = 12 - dtBirthDate.Month + dtReferralDate.Month - 1;
else
iMonths = dtBirthDate.Month - dtReferralDate.Month;
//----------------------------------------------------------------------
// Store Months in Return Value
//----------------------------------------------------------------------
dYMD.Add("Months", iMonths);
//----------------------------------------------------------------------
// Calculate Remaining Days
//----------------------------------------------------------------------
if (dtBirthDate.Day > dtReferralDate.Day)
//Logic: Figure out the days in month previous to the current month, or the admitted month.
// Subtract the birthday from the total days which will give us how many days the person has lived since their birthdate day the previous month.
// then take the referral date and simply add the number of days the person has lived this month.
//If referral date is january, we need to go back to the following year's December to get the days in that month.
if (dtReferralDate.Month == 1)
iDays = DateTime.DaysInMonth(dtReferralDate.Year - 1, 12) - dtBirthDate.Day + dtReferralDate.Day;
else
iDays = DateTime.DaysInMonth(dtReferralDate.Year, dtReferralDate.Month - 1) - dtBirthDate.Day + dtReferralDate.Day;
else
iDays = dtReferralDate.Day - dtBirthDate.Day;
//----------------------------------------------------------------------
// Store Days in Return Value
//----------------------------------------------------------------------
dYMD.Add("Days", iDays);
return dYMD;
}
小智 9
SQL版本:
declare @dd smalldatetime = '1980-04-01'
declare @age int = YEAR(GETDATE())-YEAR(@dd)
if (@dd> DATEADD(YYYY, -@age, GETDATE())) set @age = @age -1
print @age
我对Mark Soen的回答做了一个小改动:我重写了第三行,以便可以更轻松地解析表达式.
public int AgeInYears(DateTime bday)
{
DateTime now = DateTime.Today;
int age = now.Year - bday.Year;
if (bday.AddYears(age) > now)
age--;
return age;
}
为了清晰起见,我还把它变成了一个功能.
这很简单,似乎对我的需求是准确的.我为闰年的目的做出了一个假设,即无论何时选择庆祝生日,他们在技术上都不会超过一年,直到自去年生日过去整整365天(即2月28日不会使他们成为一年以上)
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;
如果您发现任何问题,请告诉我们;)
=== 俗语(从几个月到几岁) ===
如果您只是常用,请输入以下代码作为您的信息:
DateTime today = DateTime.Today;
DateTime bday = DateTime.Parse("2016-2-14");
int age = today.Year - bday.Year;
var unit = "";
if (bday > today.AddYears(-age))
{
age--;
}
if (age == 0) // Under one year old
{
age = today.Month - bday.Month;
age = age <= 0 ? (12 + age) : age; // The next year before birthday
age = today.Day - bday.Day >= 0 ? age : --age; // Before the birthday.day
unit = "month";
}
else {
unit = "year";
}
if (age > 1)
{
unit = unit + "s";
}
测试结果如下:
The birthday: 2016-2-14
2016-2-15 => age=0, unit=month;
2016-5-13 => age=2, unit=months;
2016-5-14 => age=3, unit=months;
2016-6-13 => age=3, unit=months;
2016-6-15 => age=4, unit=months;
2017-1-13 => age=10, unit=months;
2017-1-14 => age=11, unit=months;
2017-2-13 => age=11, unit=months;
2017-2-14 => age=1, unit=year;
2017-2-15 => age=1, unit=year;
2017-3-13 => age=1, unit=year;
2018-1-13 => age=1, unit=year;
2018-1-14 => age=1, unit=year;
2018-2-13 => age=1, unit=year;
2018-2-14 => age=2, unit=years;
小智 7
哇,我不得不在这里发表评论..这么简单的答案太多了
private int CalcularIdade(DateTime dtNascimento)
{
var nHoje = Convert.ToInt32(DateTime.Today.ToString("yyyyMMdd"));
var nAniversario = Convert.ToInt32(dtNascimento.ToString("yyyyMMdd"));
double diff = (nHoje - nAniversario) / 10000;
var ret = Convert.ToInt32(Math.Truncate(diff));
return ret;
}
希望它可以帮助别人,至少会让别人想到.. :)
这是在一行中回答这个问题的最简单方法.
DateTime Dob = DateTime.Parse("1985-04-24");
int Age = DateTime.MinValue.AddDays(DateTime.Now.Subtract(Dob).TotalHours/24).Year - 1;
这也适用于闰年.
小智 6
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
}
这可能有效：
public override bool IsValid(DateTime value)
{
_dateOfBirth = value;
var yearsOld = (double) (DateTime.Now.Subtract(_dateOfBirth).TotalDays/365);
if (yearsOld > 18)
return true;
return false;
}
这是一个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;
}
}
为什么不能这么简单？
int age = DateTime.Now.AddTicks(0 - dob.Ticks).Year - 1;