我如何在c#中找到一周的开始(包括周日和周一),只知道当前时间?

喜欢的东西:

DateTime.Now.StartWeek(Monday);

当前回答

public static System.DateTime getstartweek()
{
    System.DateTime dt = System.DateTime.Now;
    System.DayOfWeek dmon = System.DayOfWeek.Monday;
    int span = dt.DayOfWeek - dmon;
    dt = dt.AddDays(-span);
    return dt;
}

其他回答

我能想到的最快方法是:

var sunday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek);

如果您希望将一周中的任何一天作为您的开始日期,您所需要做的就是在结束时添加DayOfWeek值

var monday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Monday);

var tuesday = DateTime.Today.AddDays(-(int)DateTime.Today.DayOfWeek + (int)DayOfWeek.Tuesday);

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;

}

我尝试了几次,但我没有解决从星期一开始的一周的问题,导致我的下一个星期一是星期天。所以我对它做了一些修改,让它与下面的代码一起工作:

int delta = DayOfWeek.Monday - DateTime.Now.DayOfWeek;
DateTime monday = DateTime.Now.AddDays(delta == 1 ? -6 : delta);
return monday;

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);