使用扩展方法:

public static class DateTimeExtensions
{
    public static DateTime StartOfWeek(this DateTime dt, DayOfWeek startOfWeek)
    {
        int diff = (7 + (dt.DayOfWeek - startOfWeek)) % 7;
        return dt.AddDays(-1 * diff).Date;
    }
}

可以这样使用:

DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Monday);
DateTime dt = DateTime.Now.StartOfWeek(DayOfWeek.Sunday);

using System;
using System.Globalization;

namespace MySpace
{
    public static class DateTimeExtention
    {
        // ToDo: Need to provide culturaly neutral versions.

        public static DateTime GetStartOfWeek(this DateTime dt)
        {
            DateTime ndt = dt.Subtract(TimeSpan.FromDays((int)dt.DayOfWeek));
            return new DateTime(ndt.Year, ndt.Month, ndt.Day, 0, 0, 0, 0);
        }

        public static DateTime GetEndOfWeek(this DateTime dt)
        {
            DateTime ndt = dt.GetStartOfWeek().AddDays(6);
            return new DateTime(ndt.Year, ndt.Month, ndt.Day, 23, 59, 59, 999);
        }

        public static DateTime GetStartOfWeek(this DateTime dt, int year, int week)
        {
            DateTime dayInWeek = new DateTime(year, 1, 1).AddDays((week - 1) * 7);
            return dayInWeek.GetStartOfWeek();
        }

        public static DateTime GetEndOfWeek(this DateTime dt, int year, int week)
        {
            DateTime dayInWeek = new DateTime(year, 1, 1).AddDays((week - 1) * 7);
            return dayInWeek.GetEndOfWeek();
        }
    }
}

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;
}

这可能有点黑，但你可以将. dayofweek属性转换为int(它是一个枚举，因为它的底层数据类型没有改变，它默认为int)，并使用它来确定一周的前一周。

它显示DayOfWeek枚举中指定的周从周日开始，因此如果将这个值减去1，就等于周一离当前日期有多少天。我们还需要将周日(0)映射为7，因此给定1 - 7 = -6，周日将映射到前一个周一:-

DateTime now = DateTime.Now;
int dayOfWeek = (int)now.DayOfWeek;
dayOfWeek = dayOfWeek == 0 ? 7 : dayOfWeek;
DateTime startOfWeek = now.AddDays(1 - (int)now.DayOfWeek);

前一个星期天的代码更简单，因为我们不需要做这样的调整:-

DateTime now = DateTime.Now;
int dayOfWeek = (int)now.DayOfWeek;
DateTime startOfWeek = now.AddDays(-(int)now.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));
}