DateTime.AddMonths仅添加月而不是天

Question

让我们说,我已经在2010年2月28日使用了AddMonths(1)...... 这个日期增加了一个月......
结果日期是3月28日,但不是3月31日,我想要的.
有没有办法调整一点,所以这可以工作而不添加自定义代码？

编辑:我不需要一个月的最后一天,实际上我需要添加一个月,但是当它是一个月的最后一天时,我需要找到下个月的最后一天.

Answer 1

我不知道你想要实现什么,但你可以添加一天,添加一个月并减去一天.

DateTime nextMonth = date.AddDays(1).AddMonths(1).AddDays(-1);

编辑:

正如其中一位评论者指出的那样,这有时会产生错误的结果.在阅读更新后的问题后,我认为计算所需日期的最简单方法是:

public static DateTime NextMonth(this DateTime date)
{
   if (date.Day != DateTime.DaysInMonth(date.Year, date.Month))
      return date.AddMonths(1);
   else 
      return date.AddDays(1).AddMonths(1).AddDays(-1);
}

此扩展方法返回下个月的日期.当前日期是该月的最后一天,它将返回下个月的最后一天.

Answer 2

public static DateTime NextMonth(DateTime date)
{
    DateTime nextMonth = date.AddMonths(1);

    if (date.Day != DateTime.DaysInMonth(date.Year, date.Month)) //is last day in month
    {
        //any other day then last day
        return nextMonth;
    }
    else
    {
       //last day in the month will produce the last day in the next month
       return date.AddDays(DateTime.DaysInMonth(nextMonth.Year, nextMonth.Month));
    }
}

并概括为多个月：

public static DateTime AddMonthToEndOfMonth(DateTime date, int numberOfMonths)
{
    DateTime nextMonth = date.AddMonths(numberOfMonths);

    if (date.Day != DateTime.DaysInMonth(date.Year, date.Month)) //is last day in month
    {
        //any other day then last day
        return nextMonth;
    }
    else
    {
        //if date was end of month, add remaining days
        int addDays = DateTime.DaysInMonth(nextMonth.Year, nextMonth.Month) - nextMonth.Day;
        return nextMonth.AddDays(addDays);
    }
}

该代码针对二月问题、闰年和新年过渡进行了测试。全部测试通过。

[TestMethod]
public void AddMonthTest_January()
{
    for (int i = 1; i <= 28; i++)
    {
        Assert.AreEqual(new DateTime(2015, 2, i), NextMonth(new DateTime(2015, 1, i)));
    }
    Assert.AreEqual(new DateTime(2015, 2, 28), NextMonth(new DateTime(2015, 1, 29)));
    Assert.AreEqual(new DateTime(2015, 2, 28), NextMonth(new DateTime(2015, 1, 30)));
    Assert.AreEqual(new DateTime(2015, 2, 28), NextMonth(new DateTime(2015, 1, 31)));
}

[TestMethod]
public void AddMonthTest_February()
{
    Assert.AreEqual(new DateTime(2015, 3, 31), NextMonth(new DateTime(2015, 2, 28)));

    for (int i = 1; i <= 27; i++)
    {
        Assert.AreEqual(new DateTime(2015, 3, i), NextMonth(new DateTime(2015, 2, i)));
    }            
}

[TestMethod]
public void AddMonthTest_March()
{
    Assert.AreEqual(new DateTime(2015, 4, 30), NextMonth(new DateTime(2015, 3, 31)));

    for (int i = 1; i <= 30; i++)
    {
        Assert.AreEqual(new DateTime(2015, 4, i), NextMonth(new DateTime(2015, 3, i)));
    }
}

[TestMethod]
public void AddMonthTest_December()
{            
    for (int i = 1; i <= 31; i++)
    {
        Assert.AreEqual(new DateTime(2016, 1, i), NextMonth(new DateTime(2015, 12, i)));
    }
}

[TestMethod]
public void AddMonthTest_January_LeapYear()
{
    for (int i = 1; i <= 29; i++)
    {
        Assert.AreEqual(new DateTime(2016, 2, i), NextMonth(new DateTime(2016, 1, i)));
    }            
    Assert.AreEqual(new DateTime(2016, 2, 29), NextMonth(new DateTime(2016, 1, 30)));
    Assert.AreEqual(new DateTime(2016, 2, 29), NextMonth(new DateTime(2016, 1, 31)));
}

[TestMethod]
public void AddMonthTest_February_LeapYear()
{
    Assert.AreEqual(new DateTime(2016, 3, 31), NextMonth(new DateTime(2016, 2, 29)));

    for (int i = 1; i <= 28; i++)
    {
        Assert.AreEqual(new DateTime(2016, 3, i), NextMonth(new DateTime(2016, 2, i)));
    }
}

[TestMethod]
public void AddHalfYearTest_January_LeapYear()
{        
    for (int i = 1; i <= 31; i++)
    {
        Assert.AreEqual(new DateTime(2016, 7, i), new DateTime(2016, 1, i).AddMonthToEndOfMonth(6));
    }
}

[TestMethod]
public void AddHalfYearTest_February_LeapYear()
{
    Assert.AreEqual(new DateTime(2016, 8, 31), new DateTime(2016, 2, 29).AddMonthToEndOfMonth(6));

    for (int i = 1; i <= 28; i++)
    {
        Assert.AreEqual(new DateTime(2016, 8, i), new DateTime(2016, 2, i).AddMonthToEndOfMonth(6));
    }
}

[TestMethod]
public void AddHalfYearTest_December()
{
    Assert.AreEqual(new DateTime(2016, 6, 30), new DateTime(2015, 12, 31).AddMonthToEndOfMonth(6));
    for (int i = 1; i <= 30; i++)
    {
        Assert.AreEqual(new DateTime(2016, 6, i), new DateTime(2015, 12, i).AddMonthToEndOfMonth(6));
    }
}

Answer 3

如果你的意思是结果日期应该与月末的距离相同,那么你就会进入自定义代码 - 类似于(未完全测试,特别是28/30/31个月):

class Program
{
    static void Main()
    {
        var when = DateTime.Today;
        DateTime fromEndOfNextMonth = when.AddMonthsRelativeToEndOfMonth(1);
    }

}
public static class DateTimeExtensions
{
    public static DateTime AddMonthsRelativeToEndOfMonth(
               this DateTime when, int months)
    {
        if (months == 0) return when;
        DateTime startOfNextMonth = when;
        int month = when.Month;
        while (startOfNextMonth.Month == month)
        {
            startOfNextMonth = startOfNextMonth.AddDays(1);
        }
        TimeSpan delta = startOfNextMonth - when;
        return startOfNextMonth.AddMonths(1) - delta;
    }

}