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
如果您想要完整月份的确切数目,总是正的(2000-01-15,2000-02-14返回0),则考虑完整月份是当您到达下个月的同一天时(类似于年龄计算)
public static int GetMonthsBetween(DateTime from, DateTime to)
{
if (from > to) return GetMonthsBetween(to, from);
var monthDiff = Math.Abs((to.Year * 12 + (to.Month - 1)) - (from.Year * 12 + (from.Month - 1)));
if (from.AddMonths(monthDiff) > to || to.Day < from.Day)
{
return monthDiff - 1;
}
else
{
return monthDiff;
}
}
编辑原因:旧代码在某些情况下不正确,如:
new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },
Test cases I used to test the function:
var tests = new[]
{
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 1), Result = 0 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 2), Result = 0 },
new { From = new DateTime(1900, 1, 2), To = new DateTime(1900, 1, 1), Result = 0 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 2, 1), Result = 1 },
new { From = new DateTime(1900, 2, 1), To = new DateTime(1900, 1, 1), Result = 1 },
new { From = new DateTime(1900, 1, 31), To = new DateTime(1900, 2, 1), Result = 0 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 9, 30), Result = 0 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 10, 1), Result = 1 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1901, 1, 1), Result = 12 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1911, 1, 1), Result = 132 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },
};
其他回答
我在VB中检查了这个方法的用法。NET通过MSDN,它似乎有很多用途。c#中没有这样的内置方法。(即使这不是一个好主意)你可以在c#中调用VB。
将Microsoft.VisualBasic.dll添加到 你的项目作为参考 使用 Microsoft.VisualBasic.DateAndTime.DateDiff 在代码中
如果您想要完整月份的确切数目,总是正的(2000-01-15,2000-02-14返回0),则考虑完整月份是当您到达下个月的同一天时(类似于年龄计算)
public static int GetMonthsBetween(DateTime from, DateTime to)
{
if (from > to) return GetMonthsBetween(to, from);
var monthDiff = Math.Abs((to.Year * 12 + (to.Month - 1)) - (from.Year * 12 + (from.Month - 1)));
if (from.AddMonths(monthDiff) > to || to.Day < from.Day)
{
return monthDiff - 1;
}
else
{
return monthDiff;
}
}
编辑原因:旧代码在某些情况下不正确,如:
new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },
Test cases I used to test the function:
var tests = new[]
{
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 1), Result = 0 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 1, 2), Result = 0 },
new { From = new DateTime(1900, 1, 2), To = new DateTime(1900, 1, 1), Result = 0 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1900, 2, 1), Result = 1 },
new { From = new DateTime(1900, 2, 1), To = new DateTime(1900, 1, 1), Result = 1 },
new { From = new DateTime(1900, 1, 31), To = new DateTime(1900, 2, 1), Result = 0 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 9, 30), Result = 0 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1900, 10, 1), Result = 1 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1901, 1, 1), Result = 12 },
new { From = new DateTime(1900, 1, 1), To = new DateTime(1911, 1, 1), Result = 132 },
new { From = new DateTime(1900, 8, 31), To = new DateTime(1901, 8, 30), Result = 11 },
};
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;
}
我们是这样做的:
public static int MonthDiff(DateTime date1, DateTime date2)
{
if (date1.Month < date2.Month)
{
return (date2.Year - date1.Year) * 12 + date2.Month - date1.Month;
}
else
{
return (date2.Year - date1.Year - 1) * 12 + date2.Month - date1.Month + 12;
}
}
你可以使用野田时间https://nodatime.org/
LocalDate start = new LocalDate(2010, 1, 5);
LocalDate end = new LocalDate(2012, 6, 1);
Period period = Period.Between(start, end, PeriodUnits.Months);
Console.WriteLine(period.Months);