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);
}

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));

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);

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);
}

var now = System.DateTime.Now;

var result = now.AddDays(-((now.DayOfWeek - System.Threading.Thread.CurrentThread.CurrentCulture.DateTimeFormat.FirstDayOfWeek + 7) % 7)).Date;

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;
}