Question

我的问题的逻辑是满口的，我已经提出了解决方案。我的解决方案发布在下面。我希望看看是否有人能够提出更有效/更简单的解决方案。

该方法应该返回下一个，从现在起的未来，在给定日期的工作日发生。应在输入日期和输出日期之间保留时间。

对于所有示例，今天是2015年5月8日星期五下午4:00：所有投入和产出都在2015年：

+---------------------------+---------------------------+
|           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   |
+---------------------------+---------------------------+

以下是逻辑的一些伪代码：

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

这是我提出的解决方案，避免使用循环来保持效率：

- (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;
}

在发布解决方案之前，使用上面的输入/输出表来测试您的逻辑。

Answer 1

没有理由不将逻辑放在一个循环中，因为它最多会循环两次（如果当天是星期六）。

如果当天是周末，这是一个解决方案：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];
}

然后我们可以调用此方法，检查返回的NSInteger以查看它是sat还是sun并再次运行它：

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];
}

如果你真的想要删除循环，你可以检查它是否是星期六并且增加2天值，或者如果它是Sun并且增加1天值。

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.
}

NSDate将来会发生下一个工作日

1 个答案: