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

试试这个解决方案,它奏效了。

int age = (Int32.Parse(DateTime.Today.ToString("yyyyMMdd")) - 
           Int32.Parse(birthday.ToString("yyyyMMdd rawrrr"))) / 10000;

其他回答

我的建议

int age = (int) ((DateTime.Now - bday).TotalDays/365.242199);

这一年似乎在正确的日期发生了变化。(我在107岁之前进行了现场测试。)

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.
}

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;
}

要使用最近的年龄计算年龄:

var ts = DateTime.Now - new DateTime(1988, 3, 19);
var age = Math.Round(ts.Days / 365.0);

简单代码

 var birthYear=1993;
 var age = DateTime.Now.AddYears(-birthYear).Year;