Question

我需要使用Objective C方法将Gregorian日期转换为Julian天，与此PHP方法（GregorianToJD）相同。

Answer 1

根据http://en.wikipedia.org/wiki/Julian_day，2000年1月1日的朱利安日数为2,451,545。因此，您可以计算日期与此之间的天数参考日期。例如（2014年1月1日）：

NSUInteger julianDayFor01012000 = 2451545;
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[cal setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0]];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2014;
comp.month = 1;
comp.day = 1;
NSDate *date = [cal dateFromComponents:comp];
comp.year = 2000;
comp.month = 1;
comp.day = 1;
NSDate *ref = [cal dateFromComponents:comp];

NSDateComponents *diff = [cal components:NSDayCalendarUnit fromDate:ref toDate:date options:0];
NSInteger julianDays = diff.day + julianDayFor01012000;
NSLog(@"%ld", (long)julianDays);
// Output: 2456659

这与http://www.php.net/manual/en/function.gregoriantojd.php的结果相同：

<?php
$jd = GregorianToJD(1, 1, 2014);
echo "$jd\n";
?>

反方向（朱利安天到格里高利年/月/日）：

NSInteger julianDays = 2456659; // From above example
NSUInteger julianDayFor01012000 = 2451545;
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
[cal setTimeZone:[NSTimeZone timeZoneForSecondsFromGMT:0]];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2000;
comp.month = 1;
comp.day = 1;
NSDate *ref = [cal dateFromComponents:comp];
NSDateComponents *diff = [[NSDateComponents alloc] init];
diff.day = julianDays - julianDayFor01012000;

NSDate *date = [cal dateByAddingComponents:diff toDate:ref options:0];
comp = [cal components:NSDayCalendarUnit|NSMonthCalendarUnit|NSYearCalendarUnit fromDate:date];
NSLog(@"%04ld-%02ld-%02ld", (long)comp.year, (long)comp.month, (long)comp.day);
// Output: 2014-01-01

更新：在评论中正确陈述了Hot Licks，使用日期更容易格式化程序采用“g”格式：

NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *comp = [[NSDateComponents alloc] init];
comp.year = 2014;
comp.month = 1;
comp.day = 1;
NSDate *date = [cal dateFromComponents:comp];
NSDateFormatter *fmt = [[NSDateFormatter alloc] init];
[fmt setDateFormat:@"g"];
NSInteger julianDays = [[fmt stringFromDate:date] integerValue];
NSLog(@"%ld", (long)julianDays);
// Output: 2456659

反方向：

NSInteger julianDays = 2456659;
NSDateFormatter *fmt = [[NSDateFormatter alloc] init];
[fmt setDateFormat:@"g"];
NSDate *date = [fmt dateFromString:[@(julianDays) stringValue]];
NSCalendar *cal = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *comp = [cal components:NSDayCalendarUnit|NSMonthCalendarUnit|NSYearCalendarUnit fromDate:date];
NSLog(@"%04ld-%02ld-%02ld", (long)comp.year, (long)comp.month, (long)comp.day);
// Output: 2014-01-01

Answer 2

精确度：在朱利安日期转换中纳入时间

这些Julian日期转换方法产生的结果与美国相同。海军天文台的 Online Julian Date Converter，比NSDateFormatter's朱利安日期转换更精确。具体来说，下面的函数包含时间（例如小时，分钟和秒），而NSDateFormatter轮到中午GMT。

Swift示例：

func jdFromDate(date : NSDate) -> Double {
    let JD_JAN_1_1970_0000GMT = 2440587.5
    return JD_JAN_1_1970_0000GMT + date.timeIntervalSince1970 / 86400
}

func dateFromJd(jd : Double) -> NSDate {
    let JD_JAN_1_1970_0000GMT = 2440587.5
    return  NSDate(timeIntervalSince1970: (jd - JD_JAN_1_1970_0000GMT) * 86400)
}

Objective-C示例：

double jdFromDate(NSDate *date) {
   double JD_JAN_1_1970_0000GMT = 2440587.5;
   return JD_JAN_1_1970_0000GMT + date.timeIntervalSince1970 / 86400;
}

NSDate dataFromJd(double jd) {
   double JD_JAN_1_1970_0000GMT = 2440587.5;
   return [[NSDate alloc] initWithTimeIntervalSince1970: (jd - JD_JAN_1_1970_0000GMT) * 86400)];        
}

注意：研究确认accepted answer将日期舍入到24小时间隔，因为它使用g格式 - NSDateFormatter的指定符，根据Apple的日期格式化API所遵循的UNICODE标准Modified Julian Day（根据Date Format Patterns）返回Date Formatting Guide。

在目标C中将格里高利日期转换为儒略日计数

2 个答案:

精确度：在朱利安日期转换中纳入时间