我是为周一做的，但对周日也有类似的逻辑。

public static DateTime GetStartOfWeekDate()
{
    // Get today's date
    DateTime today = DateTime.Today;
    // Get the value for today. DayOfWeek is an enum with 0 being Sunday, 1 Monday, etc
    var todayDayOfWeek = (int)today.DayOfWeek;

    var dateStartOfWeek = today;
    // If today is not Monday, then get the date for Monday
    if (todayDayOfWeek != 1)
    {
        // How many days to get back to Monday from today
        var daysToStartOfWeek = (todayDayOfWeek - 1);
        // Subtract from today's date the number of days to get to Monday
        dateStartOfWeek = today.AddDays(-daysToStartOfWeek);
    }

    return dateStartOfWeek;

}

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));
    }
}

我是为周一做的，但对周日也有类似的逻辑。

public static DateTime GetStartOfWeekDate()
{
    // Get today's date
    DateTime today = DateTime.Today;
    // Get the value for today. DayOfWeek is an enum with 0 being Sunday, 1 Monday, etc
    var todayDayOfWeek = (int)today.DayOfWeek;

    var dateStartOfWeek = today;
    // If today is not Monday, then get the date for Monday
    if (todayDayOfWeek != 1)
    {
        // How many days to get back to Monday from today
        var daysToStartOfWeek = (todayDayOfWeek - 1);
        // Subtract from today's date the number of days to get to Monday
        dateStartOfWeek = today.AddDays(-daysToStartOfWeek);
    }

    return dateStartOfWeek;

}

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);

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;
}

谢谢你的例子。我需要总是使用“CurrentCulture”一周的第一天，对于一个数组，我需要知道确切的天数。这里是我的第一个扩展:

public static class DateTimeExtensions
{
    //http://stackoverflow.com/questions/38039/how-can-i-get-the-datetime-for-the-start-of-the-week
    //http://stackoverflow.com/questions/1788508/calculate-date-with-monday-as-dayofweek1
    public static DateTime StartOfWeek(this DateTime dt)
    {
        //difference in days
        int diff = (int)dt.DayOfWeek - (int)CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek; //sunday=always0, monday=always1, etc.

        //As a result we need to have day 0,1,2,3,4,5,6 
        if (diff < 0)
        {
            diff += 7;
        }
        return dt.AddDays(-1 * diff).Date;
    }

    public static int DayNoOfWeek(this DateTime dt)
    {
        //difference in days
        int diff = (int)dt.DayOfWeek - (int)CultureInfo.CurrentCulture.DateTimeFormat.FirstDayOfWeek; //sunday=always0, monday=always1, etc.

        //As a result we need to have day 0,1,2,3,4,5,6 
        if (diff < 0)
        {
            diff += 7;
        }
        return diff + 1; //Make it 1..7
    }
}