如何在c#中计算两个日期之间的月差?
c#中是否有相当于VB的DateDiff()方法。我需要找出相隔数年的两个日期之间的月差。文档说我可以像这样使用TimeSpan:
TimeSpan ts = date1 - date2;
但这里的数据是以天为单位的。我不想把这个数字除以30,因为不是每个月都是30天,而且两个操作数的值相差很大,所以我担心除以30可能会得到错误的值。
有什么建议吗?
如何在c#中计算两个日期之间的月差?
c#中是否有相当于VB的DateDiff()方法。我需要找出相隔数年的两个日期之间的月差。文档说我可以像这样使用TimeSpan:
TimeSpan ts = date1 - date2;
但这里的数据是以天为单位的。我不想把这个数字除以30,因为不是每个月都是30天,而且两个操作数的值相差很大,所以我担心除以30可能会得到错误的值。
有什么建议吗?
当前回答
假设这个月的日期不相关(即2011.1.1和2010.12.31之间的差为1),date1 > date2为正值,date2 > date1为负值
((date1.Year - date2.Year) * 12) + date1.Month - date2.Month
或者,假设你想要两个日期之间的“平均月”的大致数字,下面的方法应该适用于所有日期,但日期差异非常大。
date1.Subtract(date2).Days / (365.25 / 12)
注意,如果您要使用后一种解决方案,那么您的单元测试应该声明应用程序设计使用的最宽日期范围,并相应地验证计算结果。
更新(感谢Gary)
如果使用“平均月份”方法,“每年平均天数”的更准确数字是365.2425。
其他回答
扩展的Kirks结构与ToString(格式)和持续时间(长ms)
public struct DateTimeSpan
{
private readonly int years;
private readonly int months;
private readonly int days;
private readonly int hours;
private readonly int minutes;
private readonly int seconds;
private readonly int milliseconds;
public DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
{
this.years = years;
this.months = months;
this.days = days;
this.hours = hours;
this.minutes = minutes;
this.seconds = seconds;
this.milliseconds = milliseconds;
}
public int Years { get { return years; } }
public int Months { get { return months; } }
public int Days { get { return days; } }
public int Hours { get { return hours; } }
public int Minutes { get { return minutes; } }
public int Seconds { get { return seconds; } }
public int Milliseconds { get { return milliseconds; } }
enum Phase { Years, Months, Days, Done }
public string ToString(string format)
{
format = format.Replace("YYYY", Years.ToString());
format = format.Replace("MM", Months.ToString());
format = format.Replace("DD", Days.ToString());
format = format.Replace("hh", Hours.ToString());
format = format.Replace("mm", Minutes.ToString());
format = format.Replace("ss", Seconds.ToString());
format = format.Replace("ms", Milliseconds.ToString());
return format;
}
public static DateTimeSpan Duration(long ms)
{
DateTime dt = new DateTime();
return CompareDates(dt, dt.AddMilliseconds(ms));
}
public static DateTimeSpan CompareDates(DateTime date1, DateTime date2)
{
if (date2 < date1)
{
var sub = date1;
date1 = date2;
date2 = sub;
}
DateTime current = date1;
int years = 0;
int months = 0;
int days = 0;
Phase phase = Phase.Years;
DateTimeSpan span = new DateTimeSpan();
while (phase != Phase.Done)
{
switch (phase)
{
case Phase.Years:
if (current.AddYears(years + 1) > date2)
{
phase = Phase.Months;
current = current.AddYears(years);
}
else
{
years++;
}
break;
case Phase.Months:
if (current.AddMonths(months + 1) > date2)
{
phase = Phase.Days;
current = current.AddMonths(months);
}
else
{
months++;
}
break;
case Phase.Days:
if (current.AddDays(days + 1) > date2)
{
current = current.AddDays(days);
var timespan = date2 - current;
span = new DateTimeSpan(years, months, days, timespan.Hours, timespan.Minutes, timespan.Seconds, timespan.Milliseconds);
phase = Phase.Done;
}
else
{
days++;
}
break;
}
}
return span;
}
}
疯狂的方法,计算所有的日子,超级精确
Helper类:
public class DaysInMonth
{
public int Days { get; set; }
public int Month { get; set; }
public int Year { get; set; }
public bool Full { get; set; }
}
功能:
public static List<DaysInMonth> MonthsDelta(DateTime start, DateTime end)
{
var dates = Enumerable.Range(0, 1 + end.Subtract(start).Days)
.Select(offset => start.AddDays(offset))
.ToArray();
DateTime? prev = null;
int days = 0;
List < DaysInMonth > list = new List<DaysInMonth>();
foreach (DateTime date in dates)
{
if (prev != null)
{
if(date.Month!=prev.GetValueOrDefault().Month)
{
DaysInMonth daysInMonth = new DaysInMonth();
daysInMonth.Days = days;
daysInMonth.Month = prev.GetValueOrDefault().Month;
daysInMonth.Year = prev.GetValueOrDefault().Year;
daysInMonth.Full = DateTime.DaysInMonth(daysInMonth.Year, daysInMonth.Month) == daysInMonth.Days;
list.Add(daysInMonth);
days = 0;
}
}
days++;
prev = date;
}
//------------------ add last
if (days > 0)
{
DaysInMonth daysInMonth = new DaysInMonth();
daysInMonth.Days = days;
daysInMonth.Month = prev.GetValueOrDefault().Month;
daysInMonth.Year = prev.GetValueOrDefault().Year;
daysInMonth.Full = DateTime.DaysInMonth(daysInMonth.Year, daysInMonth.Month) == daysInMonth.Days;
list.Add(daysInMonth);
}
return list;
}
Public Class ClassDateOperation
Private prop_DifferenceInDay As Integer
Private prop_DifferenceInMonth As Integer
Private prop_DifferenceInYear As Integer
Public Function DayMonthYearFromTwoDate(ByVal DateStart As Date, ByVal DateEnd As Date) As ClassDateOperation
Dim differenceInDay As Integer
Dim differenceInMonth As Integer
Dim differenceInYear As Integer
Dim myDate As Date
DateEnd = DateEnd.AddDays(1)
differenceInYear = DateEnd.Year - DateStart.Year
If DateStart.Month <= DateEnd.Month Then
differenceInMonth = DateEnd.Month - DateStart.Month
Else
differenceInYear -= 1
differenceInMonth = (12 - DateStart.Month) + DateEnd.Month
End If
If DateStart.Day <= DateEnd.Day Then
differenceInDay = DateEnd.Day - DateStart.Day
Else
myDate = CDate("01/" & DateStart.AddMonths(1).Month & "/" & DateStart.Year).AddDays(-1)
If differenceInMonth <> 0 Then
differenceInMonth -= 1
Else
differenceInMonth = 11
differenceInYear -= 1
End If
differenceInDay = myDate.Day - DateStart.Day + DateEnd.Day
End If
prop_DifferenceInDay = differenceInDay
prop_DifferenceInMonth = differenceInMonth
prop_DifferenceInYear = differenceInYear
Return Me
End Function
Public ReadOnly Property DifferenceInDay() As Integer
Get
Return prop_DifferenceInDay
End Get
End Property
Public ReadOnly Property DifferenceInMonth As Integer
Get
Return prop_DifferenceInMonth
End Get
End Property
Public ReadOnly Property DifferenceInYear As Integer
Get
Return prop_DifferenceInYear
End Get
End Property
End Class
这个简单的静态函数计算两个Datetimes之间的月份分数。
1.1. 到31.1。= 1.0 1.4. 到15.4。= 0.5 16.4. 到30.4。= 0.5 1.3. 到1.4。= 1 + 1/30
该函数假设第一个日期比第二个日期小。要处理负时间间隔,可以通过在开始时引入符号和变量交换来轻松地修改函数。
public static double GetDeltaMonths(DateTime t0, DateTime t1)
{
DateTime t = t0;
double months = 0;
while(t<=t1)
{
int daysInMonth = DateTime.DaysInMonth(t.Year, t.Month);
DateTime endOfMonth = new DateTime(t.Year, t.Month, daysInMonth);
int cutDay = endOfMonth <= t1 ? daysInMonth : t1.Day;
months += (cutDay - t.Day + 1) / (double) daysInMonth;
t = new DateTime(t.Year, t.Month, 1).AddMonths(1);
}
return Math.Round(months,2);
}
LINQ的解决方案,
DateTime ToDate = DateTime.Today;
DateTime FromDate = ToDate.Date.AddYears(-1).AddDays(1);
int monthCount = Enumerable.Range(0, 1 + ToDate.Subtract(FromDate).Days)
.Select(x => FromDate.AddDays(x))
.ToList<DateTime>()
.GroupBy(z => new { z.Year, z.Month })
.Count();