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我正在尝试确定可以追溯到 19 世纪的应用程序的两个不同日期之间的闰年天数 - 这是一个方法示例：

-(NSInteger)leapYearDaysWithinEraFromDate:(NSDate *) startingDate toDate:(NSDate *) endingDate {

// this is for testing - it will be changed to a datepicker object
NSDateComponents *startDateComp = [[NSDateComponents alloc] init];
[startDateComp setSecond:1];
[startDateComp setMinute:0];
[startDateComp setHour:1];
[startDateComp setDay:14];
[startDateComp setMonth:4];
[startDateComp setYear:2005];

NSCalendar *GregorianCal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];

//startDate declared in .h//
startDate = [GregorianCal dateFromComponents:startDateComp];
NSLog(@"This program's start date is %@", startDate);


NSDate *today = [NSDate date];
NSUInteger unitFlags = NSDayCalendarUnit;
NSDateComponents *temporalDays = [GregorianCal components:unitFlags fromDate:startDate toDate:today options:0];

NSInteger days = [temporalDays day];

// then i will need code for the number of leap year Days

return 0;//will return the number of 2/29 days

}

所以我有日期之间的总天数。现在我需要减去闰年的天数？？？

PS - 我知道这个例子中有两个闰年，但该应用程序将追溯到 19 世纪......

4

3 回答 3

3

对于斯威夫特 

func getLeapCount(startDate : NSDate , endDate : NSDate)-> Int{
    var intialDate = startDate
    var dateComponent = NSDateComponents()
    dateComponent.day = 1
    var leapCount = 0;
    var currentCalendar = NSCalendar.currentCalendar()
        while (intialDate.compare(endDate) == NSComparisonResult.OrderedAscending) {
            intialDate = currentCalendar.dateByAddingComponents(dateComponent, toDate: intialDate, options: NSCalendarOptions.allZeros)!            
            if self.isLeapYear(startDate){
                ++leapCount
            }
    }

    return leapCount
}
func isLeapYear (year : NSDate )-> Bool{
    let cal = NSCalendar.currentCalendar()
    let year = cal.component(NSCalendarUnit.CalendarUnitYear, fromDate: year)
    return (( year%100 != 0) && (year%4 == 0)) || year%400 == 0;
}
于 2015-06-08T07:57:15.283 回答
1

一个简单的解决方案是遍历两个日期之间的所有年份，如果是闰年，则调用函数递增计数器。（来自维基百科）

if year modulo 400 is 0 then 
   is_leap_year
else if year modulo 100 is 0 then 
   not_leap_year
else if year modulo 4 is 0 then 
   is_leap_year
else
   not_leap_year

这会给你闰年的数量，因此你需要减去的闰年天数。可能有更有效的方法，但这是我现在能想到的最简单的方法。

于 2012-05-16T18:54:26.293 回答
0

好的，好吧，你把这个复杂化了。我认为这就是你想要的：

NSUInteger leapYearsInTimeFrame(NSDate *startDate, NSDate *endDate)
{
    // check to see if it's possible for a leap year (e.g. endDate - startDate > 1 year)
    if ([endDate timeIntervalSinceDate:startDate] < 31556926)
        return 0;

    // now we go year by year
    NSUInteger leapYears = 0;
    NSUInteger startYear = [[NSCalendar currentCalendar] components:NSYearCalendarUnit fromDate:startDate].year;
    NSUInteger numYears = [[NSCalendar currentCalendar] components:NSYearCalendarUnit fromDate:endDate].year - startYear;

    for (NSUInteger currentYear = startYear; currentYear <= (startYear + numYears); currentYear++) {
        if (currentYear % 400 == 0)
            // divisible by 400 is a leap year
            leapYears++;
        else if (currentYear % 100 == 0)
            /* not a leap year, divisible by 100 but not 400 isn't a leap year */ 
            continue;
        else if (currentYear % 4 == 0)
            // divisible by 4, and not by 100 is a leap year
            leapYears++;
        else 
            /* not a leap year, undivisble by 4 */
            continue;
    }

    return leapYears;    
}
于 2012-05-16T19:04:09.427 回答