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
周一
DateTime startAtMonday = DateTime.Now.AddDays(DayOfWeek.Monday - DateTime.Now.DayOfWeek);
在周日
DateTime startAtSunday = DateTime.Now.AddDays(DayOfWeek.Sunday- DateTime.Now.DayOfWeek);
其他回答
这将是一周的第一个周日的午夜:
DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek, t.Hour, t.Minute, t.Second);
这就给了你午夜的第一个星期一:
DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek - 1, t.Hour, t.Minute, t.Second);
周一
DateTime startAtMonday = DateTime.Now.AddDays(DayOfWeek.Monday - DateTime.Now.DayOfWeek);
在周日
DateTime startAtSunday = DateTime.Now.AddDays(DayOfWeek.Sunday- DateTime.Now.DayOfWeek);
让我们结合文化安全答案和扩展方法答案:
public static class DateTimeExtensions
{
public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
{
System.Globalization.CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
DayOfWeek fdow = ci.DateTimeFormat.FirstDayOfWeek;
return DateTime.Today.AddDays(-(DateTime.Today.DayOfWeek- fdow));
}
}
我和很多学校都有合作,所以正确使用周一作为一周的第一天在这里很重要。
这里很多最简洁的答案在周日都不起作用——我们经常在周日返回明天的日期,这不利于运行关于上周活动的报告。
这是我的解决方案,上周一周日,今天周一。
// Adding 7 so remainder is always positive; Otherwise % returns -1 on Sunday.
var daysToSubtract = (7 + (int)today.DayOfWeek - (int)DayOfWeek.Monday) % 7;
var monday = today
.AddDays(-daysToSubtract)
.Date;
记住为“today”使用一个方法参数,这样它就可以进行单元测试!!
我是为周一做的,但对周日也有类似的逻辑。
public static DateTime GetStartOfWeekDate()
{
// Get today's date
DateTime today = DateTime.Today;
// Get the value for today. DayOfWeek is an enum with 0 being Sunday, 1 Monday, etc
var todayDayOfWeek = (int)today.DayOfWeek;
var dateStartOfWeek = today;
// If today is not Monday, then get the date for Monday
if (todayDayOfWeek != 1)
{
// How many days to get back to Monday from today
var daysToStartOfWeek = (todayDayOfWeek - 1);
// Subtract from today's date the number of days to get to Monday
dateStartOfWeek = today.AddDays(-daysToStartOfWeek);
}
return dateStartOfWeek;
}