这将给你前一个星期天(我想):

DateTime t = DateTime.Now;
t -= new TimeSpan ((int) t.DayOfWeek, 0, 0, 0);

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));
    }
}

我们喜欢简单的语句:获取当前文化中每周的第一天与当前日期之间的差值，然后从当前日期减去天数:

var weekStartDate = DateTime.Now.AddDays(-((int)now.DayOfWeek - (int)DateTimeFormatInfo.CurrentInfo.FirstDayOfWeek));

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);

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);

如果您需要周六、周日或一周中的任何一天，但不超过当前一周(周六-日)，我用这段代码为您提供了支持。

public static DateTime GetDateInCurrentWeek(this DateTime date, DayOfWeek day)
{
    var temp = date;
    var limit = (int)date.DayOfWeek;
    var returnDate = DateTime.MinValue;

    if (date.DayOfWeek == day) 
        return date;

    for (int i = limit; i < 6; i++)
    {
        temp = temp.AddDays(1);

        if (day == temp.DayOfWeek)
        {
            returnDate = temp;
            break;
        }
    }
    if (returnDate == DateTime.MinValue)
    {
        for (int i = limit; i > -1; i++)
        {
            date = date.AddDays(-1);

            if (day == date.DayOfWeek)
            {
                returnDate = date;
                break;
            }
        }
    }
    return returnDate;
}