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
这是我自己的库,将返回两个日期之间的月差。
public static int MonthDiff(DateTime d1, DateTime d2)
{
int retVal = 0;
// Calculate the number of years represented and multiply by 12
// Substract the month number from the total
// Substract the difference of the second month and 12 from the total
retVal = (d1.Year - d2.Year) * 12;
retVal = retVal - d1.Month;
retVal = retVal - (12 - d2.Month);
return retVal;
}
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 },
};
我只是需要一些简单的东西来满足例如,只输入月份/年的就业日期,所以希望工作的年份和月份不同。这就是我所使用的,只是为了实用
public static YearsMonths YearMonthDiff(DateTime startDate, DateTime endDate) {
int monthDiff = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month) + 1;
int years = (int)Math.Floor((decimal) (monthDiff / 12));
int months = monthDiff % 12;
return new YearsMonths {
TotalMonths = monthDiff,
Years = years,
Months = months
};
}
net小提琴
在这个问题上没有很多明确的答案,因为你总是在假设事情。
这个解决方案在两个日期之间进行计算,假设您想保存一个月中的某一天进行比较,(这意味着在计算中考虑了这个月中的某一天)
例如,如果你的日期是2012年1月30日,2012年2月29日就不是一个月,但2013年3月1日就不是一个月。
它经过了相当彻底的测试,可能稍后我们会在使用时清理它,但这里:
private static int TotalMonthDifference(DateTime dtThis, DateTime dtOther)
{
int intReturn = 0;
bool sameMonth = false;
if (dtOther.Date < dtThis.Date) //used for an error catch in program, returns -1
intReturn--;
int dayOfMonth = dtThis.Day; //captures the month of day for when it adds a month and doesn't have that many days
int daysinMonth = 0; //used to caputre how many days are in the month
while (dtOther.Date > dtThis.Date) //while Other date is still under the other
{
dtThis = dtThis.AddMonths(1); //as we loop, we just keep adding a month for testing
daysinMonth = DateTime.DaysInMonth(dtThis.Year, dtThis.Month); //grabs the days in the current tested month
if (dtThis.Day != dayOfMonth) //Example 30 Jan 2013 will go to 28 Feb when a month is added, so when it goes to march it will be 28th and not 30th
{
if (daysinMonth < dayOfMonth) // uses day in month max if can't set back to day of month
dtThis.AddDays(daysinMonth - dtThis.Day);
else
dtThis.AddDays(dayOfMonth - dtThis.Day);
}
if (((dtOther.Year == dtThis.Year) && (dtOther.Month == dtThis.Month))) //If the loop puts it in the same month and year
{
if (dtOther.Day >= dayOfMonth) //check to see if it is the same day or later to add one to month
intReturn++;
sameMonth = true; //sets this to cancel out of the normal counting of month
}
if ((!sameMonth)&&(dtOther.Date > dtThis.Date))//so as long as it didn't reach the same month (or if i started in the same month, one month ahead, add a month)
intReturn++;
}
return intReturn; //return month
}
这是我所需要的。对我来说,一个月的哪一天并不重要,因为它总是碰巧是一个月的最后一天。
public static int MonthDiff(DateTime d1, DateTime d2){
int retVal = 0;
if (d1.Month<d2.Month)
{
retVal = (d1.Month + 12) - d2.Month;
retVal += ((d1.Year - 1) - d2.Year)*12;
}
else
{
retVal = d1.Month - d2.Month;
retVal += (d1.Year - d2.Year)*12;
}
//// Calculate the number of years represented and multiply by 12
//// Substract the month number from the total
//// Substract the difference of the second month and 12 from the total
//retVal = (d1.Year - d2.Year) * 12;
//retVal = retVal - d1.Month;
//retVal = retVal - (12 - d2.Month);
return retVal;
}