鉴于代表一个人生日的DateTime，我如何计算他们的年龄？
答案 0 :(得分：1928)
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--;
但是，这假设您正在寻找西方年龄的想法，而不是East Asian reckoning。
答案 1 :(得分：955)
这是一种奇怪的方法，但如果您将日期格式化为yyyymmdd并从当前日期减去出生日期，则删除您已获得年龄的最后4位数字：）
我不知道C＃，但我相信这会适用于任何语言。
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;
}
答案 2 :(得分：366)
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
这里有方法：
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;
}
答案 3 :(得分：126)
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，请参阅East Asian Age Reckoning与西方的情况。
任何真正的答案都必须包含本地化。在这个例子中，Strategy Pattern可能是有序的。
答案 4 :(得分：103)
对此的简单回答是应用AddYears，如下所示，因为这是闰年2月29日添加年份的唯一本机方法，并获得2月28日的正确结果共同的岁月。
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，也没有共同的逻辑解释为什么2月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DateTime的扩展添加。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晚期：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
}
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
}
答案 5 :(得分：80)
我的建议
int age = (int) ((DateTime.Now - bday).TotalDays/365.242199);
这似乎在正确的日期发生了变化。 （我现在测试到107岁）
答案 6 :(得分：69)
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4660年的中国人： - ）
答案 7 :(得分：47)
2要解决的主要问题是：
<强> 1。计算确切年龄 - 以年，月，日等为单位
<强> 2。计算一般感知的年龄 - 人们通常不在乎他们的年龄，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
}
注2：
再多注意一下......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);
}
答案 8 :(得分：45)
我迟到了，但这是一个单行：
int age = new DateTime(DateTime.Now.Subtract(birthday).Ticks).Year-1;
答案 9 :(得分：37)
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，因为它分离了减去逻辑的逻辑，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);
}
如果你认为那种事情不清楚，你可以扩展三元运算符以使其更清晰。
显然，这是作为DateTime上的扩展方法完成的，但很明显，您可以抓住一行代码来完成工作并将其放在任何位置。这里我们有另一个Extension方法的重载，它传入DateTime.Now，只是为了完整性。
答案 10 :(得分：32)
我知道的最好的方法是因为闰年而且一切都是：
DateTime birthDate = new DateTime(2000,3,1);
int age = (int)Math.Floor((DateTime.Now - birthDate).TotalDays / 365.25D);
希望这有帮助。
答案 11 :(得分：30)
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;
}
}
答案 12 :(得分：29)
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);
答案 13 :(得分：25)
我已经创建了一个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;
}
};
答案 14 :(得分：23)
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。
它的工作原理（lifted from here）的解释是：
我们设置
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，如果已经达到日期，我们需要
Yn-Yb，如果没有，我们需要Yn-Yb-1。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使用整数除法
(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使用整数除法
(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
答案 15 :(得分：23)
我花了一些时间研究这个并想出了这个来计算某人的年龄，月份和日期。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++;
}
}
答案 16 :(得分：19)
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。
以下是我的实现，它并不像我想象的那么简单，特别是像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。
答案 17 :(得分：18)
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.");
答案 18 :(得分：17)
我发现的最简单的方法就是这样。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，per Wikipedia，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：
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监管机构如何在其规则中定义事物。
为此，一项改进：
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：
答案 19 :(得分：17)
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复给你，所以我只是制作了一个可读的字符串。
答案 20 :(得分：16)
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);
答案 21 :(得分：16)
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和文化，大多数答案似乎都假设一个整数年度代表。时间的SI单位是second，应该是正确的通用答案（当然假设归一化DateTime并且不考虑任何相对论效应）：
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东西时会出现类似的问题，这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（计算所有天数“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;
衡量时间的另一种常用方法通常是“序列化”（命名此日期约定的家伙必须认真对待）：
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，当必须给出一个时期时代表自身有效的运动的位置或函数：）
答案 22 :(得分：15)
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;
}
答案 23 :(得分：14)
我有一个自定义方法来计算年龄，加上奖励验证信息，以防它有所帮助：
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
}
此处调用方法并传递datetime值（如果服务器设置为USA语言环境，则传递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。
答案 24 :(得分：13)
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"; ;
}
答案 25 :(得分：11)
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;
}
答案 26 :(得分：9)
以下方法（摘自Time Period Library for .NET class DateDiff ）会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
答案 27 :(得分：8)
SQL版：
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
答案 28 :(得分：8)
我使用ScArcher2的解决方案对一个人的年龄进行准确的年度计算，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;
}
答案 29 :(得分：7)
我对Mark Soen's回答了一个小改动：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，我也把它变成了一个功能。
答案 30 :(得分：7)
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：
将时钟作为IClock传递，而不是使用SystemClock.Instance，可以提高可测试性。
目标时区可能会发生变化，因此您也需要DateTimeZone参数。
另请参阅我关于此主题的博文：Handling Birthdays, and Other Anniversaries
答案 31 :(得分：6)
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。
答案 32 :(得分：6)
这很简单，看起来准确无误。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;）
答案 33 :(得分：6)
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;
}
希望它可以帮助某人，至少会让某人认为.. :)
答案 34 :(得分：5)
=== 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;
答案 35 :(得分：5)
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
}
答案 36 :(得分：4)
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;
}
}
答案 37 :(得分：4)
为什么不能这么简单？
int age = DateTime.Now.AddTicks(0 - dob.Ticks).Year - 1;
答案 38 :(得分：3)
我认为TimeSpan拥有我们所需要的一切，而不必诉诸365.25（或任何其他近似值）。扩展八月的例子：
DateTime myBD = new DateTime(1980, 10, 10);
TimeSpan difference = DateTime.Now.Subtract(myBD);
textBox1.Text = difference.Years + " years " + difference.Months + " Months " + difference.Days + " days";
答案 39 :(得分：3)
public string GetAge(this DateTime birthdate, string ageStrinFormat = null)
{
var date = DateTime.Now.AddMonths(-birthdate.Month).AddDays(-birthdate.Day);
return string.Format(ageStrinFormat ?? "{0}/{1}/{2}",
(date.Year - birthdate.Year), date.Month, date.Day);
}
答案 40 :(得分：2)
这是一个很好地为我服务的功能......没有计算，非常简单。
public static string ToAge(this DateTime dob, DateTime? toDate = null)
{
if (!toDate.HasValue)
toDate = DateTime.Now;
var now = toDate.Value;
if (now.CompareTo(dob) < 0)
return "Future date";
int years = now.Year - dob.Year;
int months = now.Month - dob.Month;
int days = now.Day - dob.Day;
if (days < 0)
{
months--;
days = DateTime.DaysInMonth(dob.Year, dob.Month) - dob.Day + now.Day;
}
if (months < 0)
{
years--;
months = 12 + months;
}
return string.Format("{0} year(s), {1} month(s), {2} days(s)",
years,
months,
days);
}
这是一个单元测试：
[Test]
public void ToAgeTests()
{
var date = new DateTime(2000, 1, 1);
Assert.AreEqual("0 year(s), 0 month(s), 1 days(s)", new DateTime(1999, 12, 31).ToAge(date));
Assert.AreEqual("0 year(s), 0 month(s), 0 days(s)", new DateTime(2000, 1, 1).ToAge(date));
Assert.AreEqual("1 year(s), 0 month(s), 0 days(s)", new DateTime(1999, 1, 1).ToAge(date));
Assert.AreEqual("0 year(s), 11 month(s), 0 days(s)", new DateTime(1999, 2, 1).ToAge(date));
Assert.AreEqual("0 year(s), 10 month(s), 25 days(s)", new DateTime(1999, 2, 4).ToAge(date));
Assert.AreEqual("0 year(s), 10 month(s), 1 days(s)", new DateTime(1999, 2, 28).ToAge(date));
date = new DateTime(2000, 2, 15);
Assert.AreEqual("0 year(s), 0 month(s), 28 days(s)", new DateTime(2000, 1, 18).ToAge(date));
}
答案 41 :(得分：2)
这里有一个针对C＃I的小代码示例，在特别是闰年的边缘情况下要小心，并不是所有上述解决方案都将它们考虑在内。将答案作为DateTime推出可能会导致问题，因为您可能最终试图将特定月份放置太多天，例如2月30日
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;
}
答案 42 :(得分：2)
var birthDate = ... // DOB
var resultDate = DateTime.Now - birthDate;
使用resultDate，无论您想要显示什么属性，都可以应用TimeSpan属性。
答案 43 :(得分：2)
我已经习惯了这个问题，我知道，它不是很优雅，但是它正在工作
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);
答案 44 :(得分：2)
我经常指望我的手指。当事情发生变化时，我需要查看日历。这就是我在代码中所做的事情：
int AgeNow(DateTime birthday)
{
return AgeAt(DateTime.Now, birthday);
}
int AgeAt(DateTime now, DateTime birthday)
{
return AgeAt(now, birthday, CultureInfo.CurrentCulture.Calendar);
}
int AgeAt(DateTime now, DateTime birthday, Calendar calendar)
{
// My age has increased on the morning of my
// birthday even though I was born in the evening.
now = now.Date;
birthday = birthday.Date;
var age = 0;
if (now <= birthday) return age; // I am zero now if I am to be born tomorrow.
while (calendar.AddYears(birthday, age + 1) <= now)
{
age++;
}
return age;
}
在LinqPad中执行此操作即可：
PASSED: someone born on 28 February 1964 is age 4 on 28 February 1968
PASSED: someone born on 29 February 1964 is age 3 on 28 February 1968
PASSED: someone born on 31 December 2016 is age 0 on 01 January 2017
LinqPad中的代码为here
答案 45 :(得分：2)
只需使用：
(DateTime.Now - myDate).TotalHours / 8766.0
当前日期 - myDate = TimeSpan，获取总小时数并除以每年的总小时数，并获得年龄/月/日... ...
答案 46 :(得分：2)
我想添加希伯来语日历计算（或其他System.Globalization日历可以以相同的方式使用），使用此线程中的重写函数：
Public Shared Function CalculateAge(BirthDate As DateTime) As Integer
Dim HebCal As New System.Globalization.HebrewCalendar ()
Dim now = DateTime.Now()
Dim iAge = HebCal.GetYear(now) - HebCal.GetYear(BirthDate)
Dim iNowMonth = HebCal.GetMonth(now), iBirthMonth = HebCal.GetMonth(BirthDate)
If iNowMonth < iBirthMonth Or (iNowMonth = iBirthMonth AndAlso HebCal.GetDayOfMonth(now) < HebCal.GetDayOfMonth(BirthDate)) Then iAge -= 1
Return iAge
End Function
答案 47 :(得分：2)
这会有用吗？
public override bool IsValid(DateTime value)
{
_dateOfBirth = value;
var yearsOld = (double) (DateTime.Now.Subtract(_dateOfBirth).TotalDays/365);
if (yearsOld > 18)
return true;
return false;
}
答案 48 :(得分：1)
仅仅因为我认为最佳答案是明确的：
public static int GetAgeByLoop(DateTime birthday)
{
var age = -1;
for (var date = birthday; date < DateTime.Today; date = date.AddYears(1))
age++;
return age;
}
答案 49 :(得分：1)
我会这样做：
DateTime birthDay = new DateTime(1990, 05, 23);
DateTime age = DateTime.Now - birthDay;
通过这种方式，您可以计算出人的确切年龄，如果需要，可以计算毫秒。
答案 50 :(得分：1)
为什么MSDN帮助没有告诉你那个？它看起来很明显：
System.DateTime birthTime = AskTheUser(myUser); // :-)
System.DateTime now = System.DateTime.Now;
System.TimeSpan age = now - birthTime; //as simple as that
double ageInDays = age.TotalDays; // will you convert to whatever you want yourself?
答案 51 :(得分：1)
转换次数少且UtcNow，此代码可以照顾2月29日闰年出生的人：
public int GetAge(DateTime DateOfBirth)
{
var Now = DateTime.UtcNow;
return Now.Year - DateOfBirth.Year -
(
(
Now.Month > DateOfBirth.Month ||
(Now.Month == DateOfBirth.Month && Now.Day >= DateOfBirth.Day)
) ? 0 : 1
);
}
答案 52 :(得分：1)
这是一个非常简单易懂的示例。
private int CalculateAge()
{
//get birthdate
DateTime dtBirth = Convert.ToDateTime(BirthDatePicker.Value);
int byear = dtBirth.Year;
int bmonth = dtBirth.Month;
int bday = dtBirth.Day;
DateTime dtToday = DateTime.Now;
int tYear = dtToday.Year;
int tmonth = dtToday.Month;
int tday = dtToday.Day;
int age = tYear - byear;
if (bmonth < tmonth)
age--;
else if (bmonth == tmonth && bday>tday)
{
age--;
}
return age;
}
答案 53 :(得分：1)
尝试这个解决方案，它正在运行。
int age = (Int32.Parse(DateTime.Today.ToString("yyyyMMdd")) -
Int32.Parse(birthday.ToString("yyyyMMdd rawrrr"))) / 10000;
答案 54 :(得分：1)
简单代码
var birthYear=1993;
var age = DateTime.Now.AddYears(-birthYear).Year;
答案 55 :(得分：1)
这是计算某人年龄的最简单方法 计算某人的年龄非常简单，这就是方法！为了使代码有效，您需要一个名为BirthDate的DateTime对象，其中包含生日。
C#
// get the difference in years
int years = DateTime.Now.Year - BirthDate.Year;
// subtract another year if we're before the
// birth day in the current year
if (DateTime.Now.Month < BirthDate.Month ||
(DateTime.Now.Month == BirthDate.Month &&
DateTime.Now.Day < BirthDate.Day))
years--;
VB.NET
' get the difference in years
Dim years As Integer = DateTime.Now.Year - BirthDate.Year
' subtract another year if we're before the
' birth day in the current year
If DateTime.Now.Month < BirthDate.Month Or (DateTime.Now.Month = BirthDate.Month And DateTime.Now.Day < BirthDate.Day) Then
years = years - 1
End If
答案 56 :(得分：1)
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.
}
答案 57 :(得分：0)
非常简单的答案
DateTime dob = new DateTime(1991, 3, 4);
DateTime now = DateTime.Now;
int dobDay = dob.Day, dobMonth = dob.Month;
int add = -1;
if (dobMonth < now.Month)
{
add = 0;
}
else if (dobMonth == now.Month)
{
if(dobDay <= now.Day)
{
add = 0;
}
else
{
add = -1;
}
}
else
{
add = -1;
}
int age = now.Year - dob.Year + add;
答案 58 :(得分：0)
<form (ngSubmit)="onSubmit(f)" #f="ngForm">
<div>
<label for='mail'>Mail </label>
<input ngControl="email" name="email" type='text' id='mail' ngModel required>
</div>
<div>
<label for='password'>password </label>
<input ngControl="password" name="password" type='text' id='password' ngModel required>
</div>
<div>
<label for='confirm-password'>Confirm Password </label>
<input ngControl="confirm-password" name="confirm-password" ngModel type='text' id='confirm-password' required>
</div>
<button type="submit"> Submit</button>
</form>
try by giving name, ngModel to inputs
答案 59 :(得分：0)
我认为可以通过这种更简单的方法来解决此问题-
类可以像-
using System;
namespace TSA
{
class BirthDay
{
double ageDay;
public BirthDay(int day, int month, int year)
{
DateTime birthDate = new DateTime(year, month, day);
ageDay = (birthDate - DateTime.Now).TotalDays; //DateTime.UtcNow
}
internal int GetAgeYear()
{
return (int)Math.Truncate(ageDay / 365);
}
internal int GetAgeMonth()
{
return (int)Math.Truncate((ageDay % 365) / 30);
}
}
}
呼叫可以像-
BirthDay b = new BirthDay(1,12,1990);
int year = b.GetAgeYear();
int month = b.GetAgeMonth();
答案 60 :(得分：0)
我对DateTime一无所知，但是我所能做的就是：
using System;
public class Program
{
public static int getAge(int month, int day, int year) {
DateTime today = DateTime.Today;
int currentDay = today.Day;
int currentYear = today.Year;
int currentMonth = today.Month;
int age = 0;
if (currentMonth < month) {
age -= 1;
} else if (currentMonth == month) {
if (currentDay < day) {
age -= 1;
}
}
currentYear -= year;
age += currentYear;
return age;
}
public static void Main()
{
int ageInYears = getAge(8, 10, 2007);
Console.WriteLine(ageInYears);
}
}
有点混乱，但是仔细看一下代码，一切都会变得有意义。
答案 61 :(得分：0)
要计算一个人的年龄，
DateTime dateOfBirth;
int ageInYears = DateTime.Now.Year - dateOfBirth.Year;
if (dateOfBirth > today.AddYears(-ageInYears )) ageInYears --;
答案 62 :(得分：-1)
一个线性答案，
DateTime dateOfBirth = Convert.ToDateTime("01/16/1990");
var age = ((DateTime.Now - dateOfBirth).Days) / 365;
答案 63 :(得分：-2)
检查出来：
TimeSpan ts = DateTime.Now.Subtract(Birthdate);
age = (byte)(ts.TotalDays / 365.25);
答案 64 :(得分：-4)
int age = DateTime.Now.Year - birthday.Year;
if (DateTime.Now.Month < birthday.Month || DateTime.Now.Month == birthday.Month
&& DateTime.Now.Day < birthday.Day) age--;
答案 65 :(得分：-4)
计算年龄最近的年龄：
var ts = DateTime.Now - new DateTime(1988, 3, 19);
var age = Math.Round(ts.Days / 365.0);
答案 66 :(得分：-6)
我不喜欢这里的许多答案，因为他们需要几行代码才能做一个非常简单的日期计算（请保存所有关于其他文化中年龄计算的评论，除非你想发表一个涵盖的答案他们）。我的一个班轮，使用c＃，sqlserver，mysql等中存在的简单日期和数学函数：
year(@today)-year(@birthDate)+floor((month(@today)-month(@birthdate)+floor((day(@today)-day(@birthdate))/31))/12)
但我也非常喜欢Mathew的回答。无论哪种方式都比这里给出的其他答案更有效。