使用Fluent DateTime:

var monday = DateTime.Now.Previous(DayOfWeek.Monday);
var sunday = DateTime.Now.Previous(DayOfWeek.Sunday);

使用Fluent DateTime:

var monday = DateTime.Now.Previous(DayOfWeek.Monday);
var sunday = DateTime.Now.Previous(DayOfWeek.Sunday);

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;
}

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);
}

周末也一样(以Compile This's answer的方式):

    public static DateTime EndOfWeek(this DateTime dt)
    {
        int diff = 7 - (int)dt.DayOfWeek;

        diff = diff == 7 ? 0 : diff;

        DateTime eow = dt.AddDays(diff).Date;

        return new DateTime(eow.Year, eow.Month, eow.Day, 23, 59, 59, 999) { };
    }

c#中的模数对-1 mod 7(它应该是6，但c#返回-1)效果很差 所以…一个“一行程序”的解决方案是这样的:

private static DateTime GetFirstDayOfWeek(DateTime date)
{
    return date.AddDays(
        -(((int)date.DayOfWeek - 1) -
          (int)Math.Floor((double)((int)date.DayOfWeek - 1) / 7) * 7));
}