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
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
}
其他回答
最精确的方法是以月为单位的分数返回差值:
private double ReturnDiffereceBetweenTwoDatesInMonths(DateTime startDateTime, DateTime endDateTime)
{
double result = 0;
double days = 0;
DateTime currentDateTime = startDateTime;
while (endDateTime > currentDateTime.AddMonths(1))
{
result ++;
currentDateTime = currentDateTime.AddMonths(1);
}
if (endDateTime > currentDateTime)
{
days = endDateTime.Subtract(currentDateTime).TotalDays;
}
return result + days/endDateTime.GetMonthDays;
}
你可以这样做
if ( date1.AddMonths(x) > date2 )
似乎DateTimeSpan解决方案使许多人满意。我不知道。让我们考虑一下:
BeginDate = 1972/2/29销售= 1972/4/28。
基于DateTimeSpan的答案是:
1年(s), 2个月(s)和0天(s)
我实现了一个方法,在此基础上,答案是:
1年、1个月及28天
显然没有两个月的时间。我想说的是,因为我们在开始日期的月末,剩下的实际上是整个3月加上结束日期(4月)的月份所经过的天数,所以1个月零28天。
如果你读到这里,你有兴趣,我把方法贴在下面。我在评论中解释了我所做的假设,因为有多少个月,月份的概念是一个不断变化的目标。多次测试,看看答案是否有意义。我通常选择相邻年份的考试日期,一旦我确认了答案,我就会前后移动一两天。到目前为止,它看起来不错,我相信你会发现一些bug:D。代码可能看起来有点粗糙,但我希望它足够清楚:
static void Main(string[] args) {
DateTime EndDate = new DateTime(1973, 4, 28);
DateTime BeginDate = new DateTime(1972, 2, 29);
int years, months, days;
GetYearsMonthsDays(EndDate, BeginDate, out years, out months, out days);
Console.WriteLine($"{years} year(s), {months} month(s) and {days} day(s)");
}
/// <summary>
/// Calculates how many years, months and days are between two dates.
/// </summary>
/// <remarks>
/// The fundamental idea here is that most of the time all of us agree
/// that a month has passed today since the same day of the previous month.
/// A particular case is when both days are the last days of their respective months
/// when again we can say one month has passed.
/// In the following cases the idea of a month is a moving target.
/// - When only the beginning date is the last day of the month then we're left just with
/// a number of days from the next month equal to the day of the month that end date represent
/// - When only the end date is the last day of its respective month we clearly have a
/// whole month plus a few days after the the day of the beginning date until the end of its
/// respective months
/// In all the other cases we'll check
/// - beginingDay > endDay -> less then a month just daysToEndofBeginingMonth + dayofTheEndMonth
/// - beginingDay < endDay -> full month + (endDay - beginingDay)
/// - beginingDay == endDay -> one full month 0 days
///
/// </remarks>
///
private static void GetYearsMonthsDays(DateTime EndDate, DateTime BeginDate, out int years, out int months, out int days ) {
var beginMonthDays = DateTime.DaysInMonth(BeginDate.Year, BeginDate.Month);
var endMonthDays = DateTime.DaysInMonth(EndDate.Year, EndDate.Month);
// get the full years
years = EndDate.Year - BeginDate.Year - 1;
// how many full months in the first year
var firstYearMonths = 12 - BeginDate.Month;
// how many full months in the last year
var endYearMonths = EndDate.Month - 1;
// full months
months = firstYearMonths + endYearMonths;
days = 0;
// Particular end of month cases
if(beginMonthDays == BeginDate.Day && endMonthDays == EndDate.Day) {
months++;
}
else if(beginMonthDays == BeginDate.Day) {
days += EndDate.Day;
}
else if(endMonthDays == EndDate.Day) {
days += beginMonthDays - BeginDate.Day;
}
// For all the other cases
else if(EndDate.Day > BeginDate.Day) {
months++;
days += EndDate.Day - BeginDate.Day;
}
else if(EndDate.Day < BeginDate.Day) {
days += beginMonthDays - BeginDate.Day;
days += EndDate.Day;
}
else {
months++;
}
if(months >= 12) {
years++;
months = months - 12;
}
}
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);
}
我对两个日期之间总月差的理解有一个整数部分和一个小数部分(日期很重要)。
积分部分是整个月的差额。
对我来说,小数部分是开始月份和结束月份之间一天的百分比(到一个月的全部天数)的差值。
public static class DateTimeExtensions
{
public static double TotalMonthsDifference(this DateTime from, DateTime to)
{
//Compute full months difference between dates
var fullMonthsDiff = (to.Year - from.Year)*12 + to.Month - from.Month;
//Compute difference between the % of day to full days of each month
var fractionMonthsDiff = ((double)(to.Day-1) / (DateTime.DaysInMonth(to.Year, to.Month)-1)) -
((double)(from.Day-1)/ (DateTime.DaysInMonth(from.Year, from.Month)-1));
return fullMonthsDiff + fractionMonthsDiff;
}
}
有了这个扩展,这些是结果:
2/29/2000 TotalMonthsDifference 2/28/2001 => 12
2/28/2000 TotalMonthsDifference 2/28/2001 => 12.035714285714286
01/01/2000 TotalMonthsDifference 01/16/2000 => 0.5
01/31/2000 TotalMonthsDifference 01/01/2000 => -1.0
01/31/2000 TotalMonthsDifference 02/29/2000 => 1.0
01/31/2000 TotalMonthsDifference 02/28/2000 => 0.9642857142857143
01/31/2001 TotalMonthsDifference 02/28/2001 => 1.0