Question

有没有办法将这段代码汇总成1-2行？

我的目标是返回，例如，我有一个星期一的DayOfWeek，我希望在那之后（星期二）或之后的n天获得。

         switch (_RESETDAY)
        {
            case DayOfWeek.Monday:
                _STARTDAY = DayOfWeek.Tuesday;
                break;
            case DayOfWeek.Tuesday:
                _STARTDAY = DayOfWeek.Wednesday;
                break;
            case DayOfWeek.Wednesday:
                _STARTDAY = DayOfWeek.Thursday;
                break;
            case DayOfWeek.Thursday:
                _STARTDAY = DayOfWeek.Friday;
                break;
            case DayOfWeek.Friday:
                _STARTDAY = DayOfWeek.Saturday;
                break;
            case DayOfWeek.Saturday:
                _STARTDAY = DayOfWeek.Sunday;
                break;
            case DayOfWeek.Sunday:
                _STARTDAY = DayOfWeek.Monday;
                break;
            default:
                _STARTDAY = DayOfWeek.Tuesday;
                break;
        }

Answer 1

这只是一个int枚举，从星期日（0）到星期六（6），根据MSDN：

DayOfWeek枚举表示每周七天的日历中的星期几。此枚举中常量的值范围从DayOfWeek.Sunday到DayOfWeek.Saturday。如果强制转换为整数，则其值的范围为零（表示DayOfWeek.Sunday）至六（表示DayOfWeek.Saturday）。

如此简单的数学应该这样做：

DayOfWeek nextDay = (DayOfWeek)(((int)_RESETDAY + 1) % 7);

如果您需要，请将+ 1替换为+ n。

Answer 2

是

(DayOfWeek)((int)(_RESETDAY+1)%7)

Answer 3

与上面回答的加法和模数相同的结果，但更可读的imho：

day = (day == DayOfWeek.Saturday) ? DayOfWeek.Sunday : day + 1;

明显的代码意图总是更有趣。

Answer 4

    static DayOfWeek dayplus (DayOfWeek day)
    {
        if (day == DayOfWeek.Saturday)
            return DayOfWeek.Sunday;
        else
            return day + 1;
    }

例如

Console.WriteLine(dayplus(DayOfWeek.Sunday));

将返回星期一

Answer 5

一般情况解决方案使用模数arithemtics ：

  DayOfWeek _RESETDAY = ...;
  int shift = 1; // can be positive or negative

  // + 7) % 7 to support negative shift´s
  DayOfWeek result = (DayOfWeek) ((((int)_RESETDAY + shift) % 7 + 7) % 7);

可能，最好的方法是隐藏 扩展方法中的combersome公式：

 public static class DayOfWeekExtensions {
   public static DayOfWeekShift(this DayOfWeek value, int shift) {
     return (DayOfWeek) ((((int)value + shift) % 7 + 7) % 7);
   }
 }

 ...

 var result = _RESETDAY.Shift(1);

稍微减少（仅在负移位不低于-7的情况下才适用于所有情况）：

     return (DayOfWeek)(((int)value + shift + 7) % 7);

DayOfWeek获得下一个DayOfWeek（周一，周二......周日）

5 个答案: