我需要知道周数的第一天和最后一天的日期.
我得到一个开始日期和结束日期,代表给定年份中选定周的第一天和最后一天.然后我需要获取上一年同一周的开始日期和结束日期,以对某些数据进行图形比较.
我设法根据给定的开始日期和结束日期获得周数.现在我需要获取前一年同一周的第一天和最后一天的日期.我怎么能最快地做到这一点?
编辑:这是我得到周数的方式:
private int GetWeekNumber(DateTime date)
{
GregorianCalendar calendar = new GregorianCalendar(GregorianCalendarTypes.USEnglish);
return calendar.GetWeekOfYear(date, CalendarWeekRule.FirstDay, DayOfWeek.Sunday);
}
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Tim*_*ter 19
您可以使用以下两种方法根据给定年份计算给定周数的周数和开始日期:
// this method is borrowed from http://stackoverflow.com/a/11155102/284240
public static int GetIso8601WeekOfYear(DateTime time)
{
DayOfWeek day = CultureInfo.InvariantCulture.Calendar.GetDayOfWeek(time);
if (day >= DayOfWeek.Monday && day <= DayOfWeek.Wednesday)
{
time = time.AddDays(3);
}
return CultureInfo.InvariantCulture.Calendar.GetWeekOfYear(time, CalendarWeekRule.FirstFourDayWeek, DayOfWeek.Monday);
}
public static DateTime FirstDateOfWeek(int year, int weekOfYear, System.Globalization.CultureInfo ci)
{
DateTime jan1 = new DateTime(year, 1, 1);
int daysOffset = (int)ci.DateTimeFormat.FirstDayOfWeek - (int)jan1.DayOfWeek;
DateTime firstWeekDay = jan1.AddDays(daysOffset);
int firstWeek = ci.Calendar.GetWeekOfYear(jan1, ci.DateTimeFormat.CalendarWeekRule, ci.DateTimeFormat.FirstDayOfWeek);
if ((firstWeek <= 1 || firstWeek >= 52) && daysOffset >= -3)
{
weekOfYear -= 1;
}
return firstWeekDay.AddDays(weekOfYear * 7);
}
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然后您可以通过以下方式获取两个日期:
// 46
int thisWeekNumber = GetIso8601WeekOfYear(DateTime.Today);
// 11/11/2013
DateTime firstDayOfWeek= FirstDateOfWeek(2013, thisWeekNumber, CultureInfo.CurrentCulture);
// 11/12/2012
DateTime firstDayOfLastYearWeek = FirstDateOfWeek(2012, thisWeekNumber, CultureInfo.CurrentCulture);
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添加6天即可获得本周结束.
小智 12
这是我的解决方案:
public static DateTime FirstDayOfWeek(DateTime date)
{
DayOfWeek fdow = CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek;
int offset = fdow - date.DayOfWeek;
DateTime fdowDate = date.AddDays(offset);
return fdowDate;
}
public static DateTime LastDayOfWeek(DateTime date)
{
DateTime ldowDate = FirstDayOfWeek(date).AddDays(6);
return ldowDate;
}
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