我有一个DayOfWeek
,我需要检查这一天是否在另外两个DayOfWeek
变量之间。
例如:
DayOfWeek monday = DayOfWeek.Monday;
DayOfWeek friday= DayOfWeek.Friday;
DayOfWeek today = DateTime.Today.DayOfWeek;
if (today is between monday and friday)
{
...
}
注意:这些日子将是包容性的。在这种情况下,如果日期星期一,星期二,星期三,星期四和星期五,则该日期有效。
我唯一能想到的是用不同的方法做一个大量的if语句,也许是扩展方法,但这不是很优雅。
修改
以下是我的要求示例:
public enum RecurringModes
{
Minutes,
Hours
}
public RecurringModes RecurringMode { get; set; }
public int RecurringValue { get; set; }
...
public IEnumerable<DateTime> AllDueDatesToday()
{
//Get the current date (starting at 00:00)
DateTime current = DateTime.Today;
//Get today and tomorrow's day of week.
DayOfWeek today = current.DayOfWeek;
DayOfWeek tomorrow = current.AddDays(1).DayOfWeek;
//If it isn't in the date range, then return nothing for today.
if (!IsInDateRange(today, StartingOn, EndingOn))
yield break;
while (current.DayOfWeek != tomorrow)
{
//Check the selected recurring mode
switch (RecurringMode)
{
//If it's minutes, then add the desired minutes
case RecurringModes.Minutes:
current = current.AddMinutes(RecurringValue);
break;
//If it's hours, then add the desired hours.
case RecurringModes.Hours:
current = current.AddHours(RecurringValue);
break;
}
//Add the calculated date to the collection.
yield return current;
}
}
public bool IsInDateRange(DayOfWeek day, DayOfWeek start, DayOfWeek end)
{
//if they are all the same date
if (start == end && start == day)
return true;
//This if statement is where the problem lies.
if ((start <= end && (day >= start && day <= end)) ||
(start > end && (day <= start && day >= end)))
return true;
else return false;
}
有效地,方法AllDueDatesToday()
将返回DateTime
的列表,该列表代表今天的时间表。
答案 0 :(得分:4)
您可以将枚举比较为数字:
if (today >= monday && today <= friday) {
正如@Tyrsius指出的那样,这只能起作用monday < friday
。所以,从技术上讲,你需要先检查一下:
if ((monday <= friday && (today >= monday && today <= friday)) ||
(monday > friday && (today <= monday && today >= friday))) {
请注意,.NET周从星期日开始:DayOfWeek.Sunday为0.
如果你希望你的星期在星期一开始,你必须进行一些算术运算。
var lowLimit = ((int)monday + 6) % 7;
var highLimit = ((int)friday + 6) % 7;
var valueToCheck = ((int)today + 6) % 7;
if ((lowLimit <= highLimit && (valueToCheck >= lowLimit && valueToCheck <= highLimit)) ||
(lowLimit > highLimit && (valueToCheck <= lowLimit && valueToCheck >= highLimit))) {
答案 1 :(得分:2)
如上所述,您可以在逻辑比较中使用枚举,但是存在值不会环绕的问题。并且终点的顺序很重要。例如“星期六和星期二之间的星期一”应该返回true
而“星期二和星期六之间的星期一”应该返回false
。
public static bool IsBetween(this DayOfWeek weekday, DayOfWeek inclusiveStart, DayOfWeek inclusiveEnd)
{
if (inclusiveStart <= inclusiveEnd)
{
return (weekday >= inclusiveStart) && (weekday <= inclusiveEnd);
}
return (weekday >= inclusiveStart) || (weekday <= inclusiveEnd);
}
这应该有效。还有其他方法可以做到,但这是一种方式。