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

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;
    }

其他回答

public static int PayableMonthsInDuration(DateTime StartDate, DateTime EndDate)
{
    int sy = StartDate.Year; int sm = StartDate.Month; int count = 0;
    do
    {
        count++;if ((sy == EndDate.Year) && (sm >= EndDate.Month)) { break; }
        sm++;if (sm == 13) { sm = 1; sy++; }
    } while ((EndDate.Year >= sy) || (EndDate.Month >= sm));
    return (count);
}

这个解决方案是用于租金/订阅计算的,其中的差异并不意味着减法,它意味着这两个日期之间的跨度。

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;
    }

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;
}

你可以有一个这样的函数。

例如,从2012/12/27到2012/12/29变成3天。同样,从2012/12/15到2013/01/15变成了2个月,因为到2013/01/14是1个月。从15号开始是第二个月。

如果您不想在计算中包括这两天,则可以删除第二个if条件中的“=”。即从2012/12/15到2013/01/15为1个月。

public int GetMonths(DateTime startDate, DateTime endDate)
{
    if (startDate > endDate)
    {
        throw new Exception("Start Date is greater than the End Date");
    }

    int months = ((endDate.Year * 12) + endDate.Month) - ((startDate.Year * 12) + startDate.Month);

    if (endDate.Day >= startDate.Day)
    {
        months++;
    }

    return months;
}

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);
 }