計算閏年在xcode中的時間差的天數

Question

我試圖確定回到19C的應用程序在兩個不同日期之間的閏年天數 - 這裏是一個方法示例：

-(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 

} 

所以我有日期之間的總天數。現在我需要減去閏年的日子？

PS - 我知道兩閏年的天也有在這個例子中，但應用程序將回到19世紀...

Answer 1

好的，好吧，你太過於複雜。我認爲這是你想要什麼：

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;  
} 

Answer 2

一個簡單的解決方案是通過所有年份的兩個日期之間進行迭代，並調用如果是閏年，函數會增加計數器。 （維基百科）

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 

這會給你的閏年數，因而閏年日子裏，你需要減去的數量。 可能有更有效的方法，但這是我現在可以想到的最簡單的方法。

Answer 3

對於斯威夫特

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; 
}