namespace DateTimeExample
    {
        using System;

        public static class DateTimeExtension
        {
            public static DateTime GetMonday(this DateTime time)
            {
                if (time.DayOfWeek != DayOfWeek.Monday)
                    return GetMonday(time.AddDays(-1)); //Recursive call

                return time;
            }
        }

        internal class Program
        {
            private static void Main()
            {
                Console.WriteLine(DateTime.Now.GetMonday());
                Console.ReadLine();
            }
        }
    } 

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));
    }
}

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);

   d = DateTime.Now;
            int dayofweek =(int) d.DayOfWeek;
            if (dayofweek != 0)
            {
                d = d.AddDays(1 - dayofweek);
            }
            else { d = d.AddDays(-6); }

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);
}

下面的方法应该返回所需的DateTime。周日为true，周一为false:

private DateTime getStartOfWeek(bool useSunday)
{
    DateTime now = DateTime.Now;
    int dayOfWeek = (int)now.DayOfWeek;

    if(!useSunday)
        dayOfWeek--;

    if(dayOfWeek < 0)
    {// day of week is Sunday and we want to use Monday as the start of the week
    // Sunday is now the seventh day of the week
        dayOfWeek = 6;
    }

    return now.AddDays(-1 * (double)dayOfWeek);
}