Format A TimeSpan With Years
我有一个具有两个日期属性的类:
例子:
2.2年:2年73天
1.002738年:"1年1天"
0.2年:73天
2年:
我所拥有的东西是有用的,但它是长的:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 | private const decimal DaysInAYear = 365.242M; public string LengthInYearsAndDays { get { var lastDay = this.LastDay ?? DateTime.Today; var lengthValue = lastDay - this.FirstDay; var builder = new StringBuilder(); var totalDays = (decimal)lengthValue.TotalDays; var totalYears = totalDays / DaysInAYear; var years = (int)Math.Floor(totalYears); totalDays -= (years * DaysInAYear); var days = (int)Math.Floor(totalDays); Func<int, string> sIfPlural = value => value > 1 ?"s" : string.Empty; if (years > 0) { builder.AppendFormat( CultureInfo.InvariantCulture, "{0} year{1}", years, sIfPlural(years)); if (days > 0) { builder.Append(""); } } if (days > 0) { builder.AppendFormat( CultureInfo.InvariantCulture, "{0} day{1}", days, sIfPlural(days)); } var length = builder.ToString(); return length; } } |
有没有更简洁的方法(但仍然可读)?
为了提供一个无耻的插件,我的noda时间项目使这非常简单,尽管:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 | using System; using NodaTime; public class Test { static void Main(string[] args) { LocalDate start = new LocalDate(2010, 6, 19); LocalDate end = new LocalDate(2013, 4, 11); Period period = Period.Between(start, end, PeriodUnits.Years | PeriodUnits.Days); Console.WriteLine("Between {0} and {1} are {2} years and {3} days", start, end, period.Years, period.Days); } } |
输出:
1 | Between 19 June 2010 and 11 April 2013 are 2 years and 296 days |
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 | public string GetAgeText(DateTime birthDate) { const double ApproxDaysPerMonth = 30.4375; const double ApproxDaysPerYear = 365.25; /* The above are the average days per month/year over a normal 4 year period We use these approximations as they are more accurate for the next century or so After that you may want to switch over to these 400 year approximations ApproxDaysPerMonth = 30.436875 ApproxDaysPerYear = 365.2425 How to get theese numbers: The are 365 days in a year, unless it is a leepyear. Leepyear is every forth year if Year % 4 = 0 unless year % 100 == 1 unless if year % 400 == 0 then it is a leep year. This gives us 97 leep years in 400 years. So 400 * 365 + 97 = 146097 days. 146097 / 400 = 365.2425 146097 / 400 / 12 = 30,436875 Due to the nature of the leap year calculation, on this side of the year 2100 you can assume every 4th year is a leap year and use the other approximatiotions */ //Calculate the span in days int iDays = (DateTime.Now - birthDate).Days; //Calculate years as an integer division int iYear = (int)(iDays / ApproxDaysPerYear); //Decrease remaing days iDays -= (int)(iYear * ApproxDaysPerYear); //Calculate months as an integer division int iMonths = (int)(iDays / ApproxDaysPerMonth); //Decrease remaing days iDays -= (int)(iMonths * ApproxDaysPerMonth); //Return the result as an string return string.Format("{0} years, {1} months, {2} days", iYear, iMonths, iDays); } |
我不会用一个
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | Public Function TimeYMDBetween(StartDate As DateTime, EndDate As DateTime) As String Dim Years As Integer = EndDate.Year - StartDate.Year Dim Months As Integer = EndDate.Month - StartDate.Month Dim Days As Integer = EndDate.Day - StartDate.Day Dim DaysLastMonth As Integer 'figure out how many days were in last month If EndDate.Month = 1 Then DaysLastMonth = DateTime.DaysInMonth(EndDate.Year - 1, 12) Else DaysLastMonth = DateTime.DaysInMonth(EndDate.Year, EndDate.Month - 1) End If 'adjust for negative days If Days < 0 Then Months = Months - 1 Days = Days + DaysLastMonth 'borrowing from last month End If 'adjust for negative Months If Months < 0 Then 'startdate hasn't happend this year yet Years = Years - 1 Months = Months + 12 End If Return Years.ToString() +" Years," + Months.ToString() +" Months and" + Days.ToString() +" Days" End Function |
我认为这应该有效:
1 2 3 4 5 6 | public static int DiffYears(DateTime dateValue1, DateTime dateValue2) { var intToCompare1 = Convert.ToInt32(dateValue1.ToString("yyyyMMdd")); var intToCompare2 = Convert.ToInt32(dateValue2.ToString("yyyyMMdd")); return (intToCompare2 - intToCompare1) / 10000; } |