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

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);

其他回答

我和很多学校都有合作,所以正确使用周一作为一周的第一天在这里很重要。

这里很多最简洁的答案在周日都不起作用——我们经常在周日返回明天的日期,这不利于运行关于上周活动的报告。

这是我的解决方案,上周一周日,今天周一。

// Adding 7 so remainder is always positive; Otherwise % returns -1 on Sunday.
var daysToSubtract = (7 + (int)today.DayOfWeek - (int)DayOfWeek.Monday) % 7;

var monday = today
    .AddDays(-daysToSubtract)
    .Date;

记住为“today”使用一个方法参数,这样它就可以进行单元测试!!

这将返回一周的开始和结束日期:

    private string[] GetWeekRange(DateTime dateToCheck)
    {
        string[] result = new string[2];
        TimeSpan duration = new TimeSpan(0, 0, 0, 0); //One day 
        DateTime dateRangeBegin = dateToCheck;
        DateTime dateRangeEnd = DateTime.Today.Add(duration);

        dateRangeBegin = dateToCheck.AddDays(-(int)dateToCheck.DayOfWeek);
        dateRangeEnd = dateToCheck.AddDays(6 - (int)dateToCheck.DayOfWeek);

        result[0] = dateRangeBegin.Date.ToString();
        result[1] = dateRangeEnd.Date.ToString();
        return result;

    }

我已经在我的博客上发布了计算周、月、季度和年的开始/结束的完整代码 ZamirsBlog

把这些都放在一起,加上全球化并允许指定一周的第一天作为我们通话的一部分

public static DateTime StartOfWeek ( this DateTime dt, DayOfWeek? firstDayOfWeek )
{
    DayOfWeek fdow;

    if ( firstDayOfWeek.HasValue  )
    {
        fdow = firstDayOfWeek.Value;
    }
    else
    {
        System.Globalization.CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
        fdow = ci.DateTimeFormat.FirstDayOfWeek;
    }

    int diff = dt.DayOfWeek - fdow;

    if ( diff < 0 )
    {
        diff += 7;
    }

    return dt.AddDays( -1 * diff ).Date;

}

这是一个正确的解决办法。无论一周的第一天是周一、周日还是其他日期,下面的代码都能正常工作。

public static class DateTimeExtension
{
  public static DateTime GetFirstDayOfThisWeek(this DateTime d)
  {
    CultureInfo ci = System.Threading.Thread.CurrentThread.CurrentCulture;
    var first = (int)ci.DateTimeFormat.FirstDayOfWeek;
    var current = (int)d.DayOfWeek;

    var result = first <= current ?
      d.AddDays(-1 * (current - first)) :
      d.AddDays(first - current - 7);

    return result;
  }
}

class Program
{
  static void Main()
  {
    System.Threading.Thread.CurrentThread.CurrentCulture = CultureInfo.GetCultureInfo("en-US");
    Console.WriteLine("Current culture set to en-US");
    RunTests();
    Console.WriteLine();
    System.Threading.Thread.CurrentThread.CurrentCulture = CultureInfo.GetCultureInfo("da-DK");
    Console.WriteLine("Current culture set to da-DK");
    RunTests();
    Console.ReadLine();
  }

  static void RunTests()
  {
    Console.WriteLine("Today {1}: {0}", DateTime.Today.Date.GetFirstDayOfThisWeek(), DateTime.Today.Date.ToString("yyyy-MM-dd"));
    Console.WriteLine("Saturday 2013-03-02: {0}", new DateTime(2013, 3, 2).GetFirstDayOfThisWeek());
    Console.WriteLine("Sunday 2013-03-03: {0}", new DateTime(2013, 3, 3).GetFirstDayOfThisWeek());
    Console.WriteLine("Monday 2013-03-04: {0}", new DateTime(2013, 3, 4).GetFirstDayOfThisWeek());
  }
}

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;
}