NSDate得到下一個發生在未來的工作日

Question

我的問題的邏輯是一口，我已經想出了一個解決方案。我的解決方案發布如下。我期待看看是否有人能夠提出更高效/更簡單的解決方案。

該方法應該返回從現在開始的下一個未來，在給定日期發生的工作日。時間應該保留在輸入日期和輸出日期之間。

對於所有的例子，今天是星期五2015年5月8日，4:00 PM： 所有輸入和輸出都在2015：

+---------------------------+---------------------------+ 
|   Input   |   Output   | 
+---------------------------+---------------------------+ 
| Tuesday April 7, 3:00 PM | Monday May 11, 3:00 PM | 
| Thursday May 7, 3:00 PM | Monday May 11, 3:00 PM | 
| Thursday May 7, 5:00 PM | Friday May 8, 5:00 PM  | 
| Tuesday May 12, 3:00 PM | Wednesday May 13, 3:00 PM | 
| Saturday June 20, 3:00 PM | Monday June 22, 3:00 PM | 
+---------------------------+---------------------------+ 

下面是一些邏輯僞代碼：

do { 
    date += 1 day; 
} while(date.isWeekend || date.isInThePast) 

這裏是我想出瞭解決方案，避免了使用循環保持效率記住：

- (NSDate *)nextWeekDayInFuture { 
    NSDate *now = [NSDate date]; 
    NSDate *nextWeekDaydate; 
    NSCalendar *calendar = [NSCalendar currentCalendar]; 
    nextWeekDaydate = [self dateByAddingDays:1]; // custom method, adds 1 day 

    if ([nextWeekDaydate isLessThan:now]) { 
     NSDateComponents *nowComponents = [calendar components:(NSCalendarUnitYear | NSCalendarUnitMonth | NSCalendarUnitDay) fromDate:now]; 
     NSDateComponents *dateComponents = [calendar components:(NSCalendarUnitHour | NSCalendarUnitMinute | NSCalendarUnitSecond) fromDate:nextWeekDaydate]; 

     [nowComponents setHour:dateComponents.hour]; 
     [nowComponents setMinute:dateComponents.minute]; 
     [nowComponents setSecond:dateComponents.second]; 

     nextWeekDaydate = [calendar dateFromComponents:nowComponents]; 
     if ([nextWeekDaydate isLessThan:now]) { 
      nextWeekDaydate = [nextWeekDaydate dateByAddingDays:1]; 
     } 
    } 

    NSDateComponents *components = [calendar components:NSCalendarUnitWeekday fromDate:nextWeekDaydate]; 
    if (components.weekday == Saturday) { 
     nextWeekDaydate = [nextWeekDaydate dateByAddingDays:2]; 
    } else if (components.weekday == Sunday) { 
     nextWeekDaydate = [nextWeekDaydate dateByAddingDays:1]; 
    } 

    return nextWeekDaydate; 
} 

在發佈解決方案之前，使用上面的輸入/輸出表來測試您的邏輯。

Answer 1

沒有理由不把邏輯放在循環中，因爲最多它會循環兩次（如果當前日期是星期六）。

這裏是爲制定一個解決方案，如果當天是週末：How to find weekday from today's date using NSDate?

- (NSInteger)isWeekend:(NSDate*)inDate 
{ 
    NSCalendar* cal = [NSCalendar currentCalendar]; 
    NSDateComponents* comp = [cal components:kCFCalendarUnitWeekday fromDate:inDate]; 
    return [comp weekday]; 
} 

然後，我們可以調用這個方法，檢查返回NSInteger的，看它是否是坐着或太陽再次運行：

NSDate *now = [NSDate date]; 
now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth. 

NSInteger curDay = [self isWeekend:now]; 

while ((curDay == 6) || (curDay == 1)) // Sat or Sund. 
{ 
    now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth. 
    curDay = [self isWeekend:now]; 
} 

如果你真的要刪除的循環，你可以檢查它是否是星期六和加2天的，或者如果它的陽光加1天的。

NSDate *now = [NSDate date]; 
now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth. 
NSInteger curDay = [self isWeekend:now]; 

if (curDay == 6) // Sat 
{ 
    now = [now dateByAddingTimeInterval:60*60*24*2]; // Add 2 day's worth. 
} 
else if (curDay == 1) // Sun 
{ 
    now = [now dateByAddingTimeInterval:60*60*24]; // Add 1 day's worth. 
}