C# 用年份格式化时间跨度
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Format A TimeSpan With Years
提问by Dan
I have a class with 2 date properties: FirstDayand LastDay. LastDayis nullable. I would like to generate a string in the format of "x year(s) y day(s)". If the total years are less than 1, I would like to omit the year section. If the total days are less than 1, I would like to omit the day section. If either years or days are 0, they should say "day/year", rather than "days/years" respectively.
我有一个具有 2 个日期属性的类：FirstDay和LastDay. LastDay可以为空。我想生成一个格式为"x year(s) y day(s)". 如果总年份小于 1，我想省略年份部分。如果总天数小于 1，我想省略天部分。如果年数或天数为 0，则应分别表示“天/年”，而不是“天/年”。
Examples:
2.2 years: "2 years 73 days"
1.002738 years: "1 year 1 day"
0.2 years: "73 days"
2 years: "2 years"
示例：
2.2 年：“2 年 73 天”
1.002738 年：“1 年 1 天”
0.2 年：“73 天”
2 年：“2 年”
What I have works, but it is long:
我有什么作品，但很长：
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;
}
}
Is there a more concise way of doing this (but still readable)?
有没有更简洁的方法来做到这一点（但仍然可读）？
采纳答案by Jon Skeet
A TimeSpandoesn't have a sensible concept of "years" because it depends on the start and end point. (Months is similar - how many months are there in 29 days? Well, it depends...)
ATimeSpan没有合理的“年”概念，因为它取决于起点和终点。（月份是相似的 - 29 天内有多少个月？嗯，这取决于...）
To give a shameless plug, my Noda Timeproject makes this really simple though:
为了提供一个无耻的插件，我的Noda Time项目让这变得非常简单：
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);
}
}
Output:
输出：
Between 19 June 2010 and 11 April 2013 are 2 years and 296 days
回答by D Stanley
I wouldn't do this with a TimeSpan. Date math gets tricky as soon as you go beyond days because the number of days in a month and days in a year is no longer constant. It's likely why TimeSpandoes not contain properties for Yearsand Months. I would instead determine the number of years/months/days, etc between the two DateTimevalues and display the results accordingly.
我不会用TimeSpan. 一旦超过天数，日期数学就会变得棘手，因为一个月中的天数和一年中的天数不再是恒定的。很可能为什么TimeSpan不包含Yearsand 的属性Months。我会改为确定两个DateTime值之间的年数/月数/天数等，并相应地显示结果。
回答by Jens Borrisholt
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);
}
回答by user3469676
I think this should work:
我认为这应该有效：
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;
}
回答by Mark
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
