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
基于上面出色的DateTimeSpan工作,我将代码规范化了一些;这似乎很有效:
public class DateTimeSpan
{
private DateTimeSpan() { }
private DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
{
Years = years;
Months = months;
Days = days;
Hours = hours;
Minutes = minutes;
Seconds = seconds;
Milliseconds = milliseconds;
}
public int Years { get; private set; } = 0;
public int Months { get; private set; } = 0;
public int Days { get; private set; } = 0;
public int Hours { get; private set; } = 0;
public int Minutes { get; private set; } = 0;
public int Seconds { get; private set; } = 0;
public int Milliseconds { get; private set; } = 0;
public static DateTimeSpan CompareDates(DateTime StartDate, DateTime EndDate)
{
if (StartDate.Equals(EndDate)) return new DateTimeSpan();
DateTimeSpan R = new DateTimeSpan();
bool Later;
if (Later = StartDate > EndDate)
{
DateTime D = StartDate;
StartDate = EndDate;
EndDate = D;
}
// Calculate Date Stuff
for (DateTime D = StartDate.AddYears(1); D < EndDate; D = D.AddYears(1), R.Years++) ;
if (R.Years > 0) StartDate = StartDate.AddYears(R.Years);
for (DateTime D = StartDate.AddMonths(1); D < EndDate; D = D.AddMonths(1), R.Months++) ;
if (R.Months > 0) StartDate = StartDate.AddMonths(R.Months);
for (DateTime D = StartDate.AddDays(1); D < EndDate; D = D.AddDays(1), R.Days++) ;
if (R.Days > 0) StartDate = StartDate.AddDays(R.Days);
// Calculate Time Stuff
TimeSpan T1 = EndDate - StartDate;
R.Hours = T1.Hours;
R.Minutes = T1.Minutes;
R.Seconds = T1.Seconds;
R.Milliseconds = T1.Milliseconds;
// Return answer. Negate values if the Start Date was later than the End Date
if (Later)
return new DateTimeSpan(-R.Years, -R.Months, -R.Days, -R.Hours, -R.Minutes, -R.Seconds, -R.Milliseconds);
return R;
}
}
其他回答
这里有一个简单的解决方案,至少对我来说是有效的。它可能不是最快的,因为它在循环中使用了很酷的DateTime的AddMonth功能:
public static int GetMonthsDiff(DateTime start, DateTime end)
{
if (start > end)
return GetMonthsDiff(end, start);
int months = 0;
do
{
start = start.AddMonths(1);
if (start > end)
return months;
months++;
}
while (true);
}
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
基于上面出色的DateTimeSpan工作,我将代码规范化了一些;这似乎很有效:
public class DateTimeSpan
{
private DateTimeSpan() { }
private DateTimeSpan(int years, int months, int days, int hours, int minutes, int seconds, int milliseconds)
{
Years = years;
Months = months;
Days = days;
Hours = hours;
Minutes = minutes;
Seconds = seconds;
Milliseconds = milliseconds;
}
public int Years { get; private set; } = 0;
public int Months { get; private set; } = 0;
public int Days { get; private set; } = 0;
public int Hours { get; private set; } = 0;
public int Minutes { get; private set; } = 0;
public int Seconds { get; private set; } = 0;
public int Milliseconds { get; private set; } = 0;
public static DateTimeSpan CompareDates(DateTime StartDate, DateTime EndDate)
{
if (StartDate.Equals(EndDate)) return new DateTimeSpan();
DateTimeSpan R = new DateTimeSpan();
bool Later;
if (Later = StartDate > EndDate)
{
DateTime D = StartDate;
StartDate = EndDate;
EndDate = D;
}
// Calculate Date Stuff
for (DateTime D = StartDate.AddYears(1); D < EndDate; D = D.AddYears(1), R.Years++) ;
if (R.Years > 0) StartDate = StartDate.AddYears(R.Years);
for (DateTime D = StartDate.AddMonths(1); D < EndDate; D = D.AddMonths(1), R.Months++) ;
if (R.Months > 0) StartDate = StartDate.AddMonths(R.Months);
for (DateTime D = StartDate.AddDays(1); D < EndDate; D = D.AddDays(1), R.Days++) ;
if (R.Days > 0) StartDate = StartDate.AddDays(R.Days);
// Calculate Time Stuff
TimeSpan T1 = EndDate - StartDate;
R.Hours = T1.Hours;
R.Minutes = T1.Minutes;
R.Seconds = T1.Seconds;
R.Milliseconds = T1.Milliseconds;
// Return answer. Negate values if the Start Date was later than the End Date
if (Later)
return new DateTimeSpan(-R.Years, -R.Months, -R.Days, -R.Hours, -R.Minutes, -R.Seconds, -R.Milliseconds);
return R;
}
}
简单快速的解决方案,计算2个日期之间的总月份。 如果你只想得到不同的月份,而不计算From date中的月份-只需从代码中删除+1。
public static int GetTotalMonths(DateTime From, DateTime Till)
{
int MonthDiff = 0;
for (int i = 0; i < 12; i++)
{
if (From.AddMonths(i).Month == Till.Month)
{
MonthDiff = i + 1;
break;
}
}
return MonthDiff;
}
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 },
};