我如何在c#中找到一周的开始(包括周日和周一),只知道当前时间?
喜欢的东西:
DateTime.Now.StartWeek(Monday);
我如何在c#中找到一周的开始(包括周日和周一),只知道当前时间?
喜欢的东西:
DateTime.Now.StartWeek(Monday);
当前回答
我能想到的最快方法是:
var sunday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek);
如果您希望将一周中的任何一天作为您的开始日期,您所需要做的就是在结束时添加DayOfWeek值
var monday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Monday);
var tuesday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Tuesday);
其他回答
我尝试了几次,但我没有解决从星期一开始的一周的问题,导致我的下一个星期一是星期天。所以我对它做了一些修改,让它与下面的代码一起工作:
int delta = DayOfWeek.Monday - DateTime.Now.DayOfWeek;
DateTime monday = DateTime.Now.AddDays(delta == 1 ? -6 : delta);
return monday;
稍微啰嗦一点,了解一下文化:
System.Globalization.CultureInfo ci =
System.Threading.Thread.CurrentThread.CurrentCulture;
DayOfWeek fdow = ci.DateTimeFormat.FirstDayOfWeek;
DayOfWeek today = DateTime.Now.DayOfWeek;
DateTime sow = DateTime.Now.AddDays(-(today - fdow)).Date;
通过这种方式计算,您可以选择一周中的哪一天表示新一周的开始(在示例中,我选择了星期一)。
请注意,对星期一进行这个计算将得到当前的星期一,而不是以前的星期一。
//Replace with whatever input date you want
DateTime inputDate = DateTime.Now;
//For this example, weeks start on Monday
int startOfWeek = (int)DayOfWeek.Monday;
//Calculate the number of days it has been since the start of the week
int daysSinceStartOfWeek = ((int)inputDate.DayOfWeek + 7 - startOfWeek) % 7;
DateTime previousStartOfWeek = inputDate.AddDays(-daysSinceStartOfWeek);
从Compile This' answer开始,使用以下方法获取一周中任何一天的日期:
public static DateTime GetDayOfWeek(DateTime dateTime, DayOfWeek dayOfWeek)
{
var monday = dateTime.Date.AddDays((7 + (dateTime.DayOfWeek - DayOfWeek.Monday) % 7) * -1);
var diff = dayOfWeek - DayOfWeek.Monday;
if (diff == -1)
{
diff = 6;
}
return monday.AddDays(diff);
}
使用扩展方法:
public static class DateTimeExtensions
{
public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
int diff = (7 + (dt.DayOfWeek - startOfWeek)) % 7;
return dt.AddDays(-1 * diff).Date;
}
}
可以这样使用:
DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Monday);
DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Sunday);