问题
I'm running into a situation where I would like to convert from a Julian date to an java.time.Instant (if that makes sense), or some Java time that can be more easily understood. My understanding of what a Julian date is comes from reading the Wikipedia page. There are bunch of different variants, and the date I am trying to read uses a different epoch than any of these.
For example, let's say the epoch is the beginning of the Calendar (New Style) Act 1750, and the Julian date is 95906.27600694445 which in this case I believe is CE 2015 April 15 06:37:26.9 UT, how do I get an instant from this? I will need to adjust for the timezone later.
I noticed there is a class called JulianFields, but I don't know where/how to use it. Also, most of the methods I see in the package make use of int or long, not really anything for double.
So, is there a simple way to convert from a Julian date using a different epoch to a Java 8 Instant (or some other time if my thinking is wrong).
回答1:
Here is a solution using the new Java 8 classes:
public class JulianDay {
private static final double NANOS_PER_DAY = 24.0 * 60.0 * 60.0 * 1000000000.0;
// Calculate Instants for some epochs as defined in Wikipedia.
public static final Instant REDUCED_JD =
ZonedDateTime.of(1858, 11, 16, 12, 0, 0, 0, ZoneOffset.UTC).toInstant();
public static final Instant MODIFIED_JD =
ZonedDateTime.of(1858, 11, 17, 0, 0, 0, 0, ZoneOffset.UTC).toInstant();
public static final Instant JULIAN_DATE =
REDUCED_JD.minus(2400000, ChronoUnit.DAYS);
private final Instant epoch;
public JulianDay(Instant epoch) {
super();
this.epoch = epoch;
}
public Instant toInstant(double day) {
long l = (long) day;
return epoch
.plus(l, ChronoUnit.DAYS)
.plusNanos(Math.round((day - l) * NANOS_PER_DAY));
}
public static void main(String[] args) {
// Use the example values from Wikipedia for 2015-09-07 13:21 UTC.
System.out.println(new JulianDay(REDUCED_JD).toInstant(57273.05625));
// Output: 2015-09-07T13:21:00.000000126Z
System.out.println(new JulianDay(MODIFIED_JD).toInstant(57272.55625));
// Output: 2015-09-07T13:21:00.000000126Z
System.out.println(new JulianDay(JULIAN_DATE).toInstant(2457273.05625));
// Output: 2015-09-07T13:20:59.999991953Z
}
}
Regarding the JulianFields you asked about, you can define a custom formatter like this:
DateTimeFormatter formatter = new DateTimeFormatterBuilder()
.appendValue(JulianFields.MODIFIED_JULIAN_DAY)
.toFormatter().withZone(ZoneOffset.UTC);
Unfortunately it doesn't support fractions of days:
System.out.println(formatter.format(Instant.now())); // Output: 57249
System.out.println(LocalDate.from(formatter.parse("57249"))); // Output: 2015-08-15
回答2:
I am going to assume that you have a numeric timestamp that is a kind of modified Julian Day Number, i.e. a continuous count of days since a defined epoch.
For example, the definition of a "Modified Julian Day Number" is a continuous count of days since midnight on Nov 17, 1858. I believe what you are asking is:
How do I convert a continuous count of days in England since the Gregorian Calendar was officially adopted to an Instant?
I'm not certain where the Gregorian Epoch officially began after the New Style Calendar act. I will assume that it is January 1, 1752, i.e. the number 95906.276 is a continuous count of days since then.
METHOD1: Here is an algorithm for processing a day number to an integer array representation in year, month(1-12), day(1-31), hours(0-23), min(0-59), sec(0-59), millis:
private static final int YEAR = 0;
private static final int MONTH = 1;
private static final int DAY = 2;
private static final int HOURS = 3;
private static final int MINUTES = 4;
private static final int SECONDS = 5;
private static final int MILLIS = 6;
public static int[] toTimeStampArray(double yourEpochDayNumber) {
int ymd_hms[] = { -1, -1, -1, -1, -1, -1, -1 };
int a, b, c, d, e, z;
// convert from your epoch (1/1/1752) to Julian Day Number
double jd = yourEpochDayNumber + 2360965.5 + 0.5;
double f, x;
z = (int) Math.floor(jd);
f = jd - z;
if (z >= 2299161) {
int alpha = (int) Math.floor((z - 1867216.25) / 36524.25);
a = z + 1 + alpha - (int) Math.floor(alpha / 4);
} else {
a = z;
}
b = a + 1524;
c = (int) Math.floor((b - 122.1) / 365.25);
d = (int) Math.floor(365.25 * c);
e = (int) Math.floor((b - d) / 30.6001);
ymd_hms[DAY] = b - d - (int) Math.floor(30.6001 * e);
ymd_hms[MONTH] = (e < 14)
? (e - 1)
: (e - 13);
ymd_hms[YEAR] = (ymd_hms[MONTH] > 2)
? (c - 4716)
: (c - 4715);
for (int i = HOURS; i <= MILLIS; i++) {
switch(i) {
case HOURS:
f = f * 24.0;
break;
case MINUTES: case SECONDS:
f = f * 60.0;
break;
case MILLIS:
f = f * 1000.0;
break;
}
x = Math.floor(f);
ymd_hms[i] = (int) x;
f = f - x;
}
return ymd_hms;
}
Algorithm is adapted from Meeus J., Astronomical Algorithms, 2nd Ed.
From these data, you can create a LocalDateTime instance. You can combine that with a ZoneId instance to create a ZonedDateTime and get an Instant.
METHOD 2. If your day number is already reckoned in GMT/UTC and does not require any offsets for time zone or daylight savings, then you can convert directly from a day number (in your epoch) to an Instant as follows:
public Instant dayNumberToInstant(double dayNumber) {
long millisFromPosixEpoch;
final double POSIX_EPOCH_AS_DAYNUM = 79622.0
millisFromPosixEpoch = (long) ((dayNumber - POSIX_EPOCH_AS_DAYNUM) *
(86400.0 * 1000.0));
return Instant.ofEpochMillis(millisFromPosixEpoch);
}
回答3:
Epoch of 14 September 1752
My reading of the Wikipedia page on Calendar (New Style) Act 1750 indicates an epoch reference date of 1752-09-14.
If we add the integer portion of your input number of 95906.27600694445, 95_906L, we do indeed get your target date of April 15, 2015 (in modern calendar system).
long input = 95_906L;
LocalDate epochCalendarNewStyleActOf1750 = LocalDate.of ( 1752 , Month.SEPTEMBER , 14 );
LocalDate localDate = epochCalendarNewStyleActOf1750.plusDays ( input );
System.out.println ( input + " days from epoch of: " + epochCalendarNewStyleActOf1750 + " is " + localDate );
95906 days from epoch of: 1752-09-14 is 2015-04-15
Regarding the fractional number, which I presume is the fraction of the number of seconds in a generic 24-hour day.
While LocalDate is for date-only without time-of-day, we now need time-of-day which is represented by our fractional number. So in place of LocalDate, we switch to OffsetDateTime.
OffsetDateTime epochCalendarNewStyleActOf1750 = LocalDate.of ( 1752 , Month.SEPTEMBER , 14 ).atStartOfDay ().atOffset ( ZoneOffset.UTC );
We use BigDecimal as double and Double are floating-point technology that trades away accuracy for speed of execution.
String input = "95906.27600694445";
BigDecimal bd = new BigDecimal ( input );
Pull from that the number of whole days.
long days = bd.toBigInteger ().longValue ();
Work on the fraction of a day. Extract the fractional number by subtracting the integer portion.
BigDecimal fractionOfADay = bd.subtract ( new BigDecimal ( days ) ); // Extract the fractional number, separate from the integer number.
We assume this decimal fraction is a fraction of the number of seconds in a day. So we can multiply by the number of seconds is a day.
BigDecimal secondsFractional = new BigDecimal ( TimeUnit.DAYS.toSeconds ( 1 ) ).multiply ( fractionOfADay );
From that, extract the number of whole seconds. From the remainder, produce a whole number nanoseconds, the resolution of the java.time classes including OffsetDateTime and Duration.
long secondsWhole = secondsFractional.longValue ();
long nanos = secondsFractional.subtract ( new BigDecimal ( secondsWhole ) ).multiply ( new BigDecimal ( 1_000_000_000L ) ).longValue ();
Create a Duration to represent the amount of time we want to add to our epoch.
Duration duration = Duration.ofDays ( days ).plusSeconds ( secondsWhole ).plusNanos ( nanos );
Add the duration to the epoch to get our final result.
OffsetDateTime odt = epochCalendarNewStyleActOf1750.plus ( duration );
You can extract a Instant object from the OffsetDateTime.
Instant instant = odt.toInstant();
Dump to console.
System.out.println ( "bd: " + bd );
System.out.println ( "days: " + days );
System.out.println ( "fractionOfADay.toString(): " + fractionOfADay );
System.out.println ( "secondsFractional: " + secondsFractional );
System.out.println ( "secondsWhole: " + secondsWhole );
System.out.println ( "nanos: " + nanos );
System.out.println ( "duration.toString(): " + duration );
System.out.println ( "duration.toDays(): " + duration.toDays () );
System.out.println ( "odt.toString(): " + odt );
This code seems to be working properly. The result here matches the expectation stated in the Question to the second, though we disagree on the fraction of a second.
No guarantees here; this code is fresh off the top of my head, and is quite untested and unproven.
See this code run live at IdeOne.com.
bd: 95906.27600694445
days: 95906
fractionOfADay.toString(): 0.27600694445
secondsFractional: 23847.00000048000
secondsWhole: 23847
nanos: 480
duration.toString(): PT2301750H37M27.00000048S
duration.toDays(): 95906
odt.toString(): 2015-04-15T06:37:27.000000480Z
Of course this math could be simpler. But I thought it might be interesting to show the pieces. One simpler ways is to multiply the 95906.27600694445 BigDecimal by the number of seconds in a generic 24 hour day. Then separate the resulting integer from its decimal fraction, and feed each to Duration.ofSeconds and Duration::plusNanos as that fits the internal data model of Duration, a total number of seconds and a total number of nanos in a fraction of a second. We would be skipping the part where we called Duration.ofDays.
About java.time
The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.
The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.
To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.
Where to obtain the java.time classes?
- Java SE 8 and SE 9 and later
- Built-in.
- Part of the standard Java API with a bundled implementation.
- Java 9 adds some minor features and fixes.
- Java SE 6 and SE 7
- Much of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
- Android
- The ThreeTenABP project adapts ThreeTen-Backport (mentioned above) for Android specifically.
- See How to use ThreeTenABP….
The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.
回答4:
The most complete and also shortest approach is obtained by my library Time4J, see this snippet using the class JulianDay:
double customJD = 95906.27600694445;
// my comment: I have never seen Julian days with that epoch until now
HistoricCalendar hcal = // date when the new style calendar act took effect
HistoricCalendar.of(ChronoHistory.of(Locale.UK), HistoricEra.AD, 1752, 9, 14);
// last term 0.5 necessary because julian days start at noon
double value = customJD + hcal.get(EpochDays.JULIAN_DAY_NUMBER) - 0.5;
JulianDay jd = JulianDay.ofSimplifiedTime(value);
Instant instant = jd.toMoment().toTemporalAccessor();
System.out.println(instant); // 2015-04-15T06:37:27Z
However, it should be noted that the most dominating field of application of Julian days is the astronomy, see also the official recommendation of the IAU. And on that field, it is much more common to include a delta-T-correction i.e. to define the Julian days on the time scale TT (Terrestrial Time). Time4J offers the methods JulianDay.ofEphemerisTime(...) for this purpose. If you seriously consider to handle time scales such as TT then you should rather work with the class Moment instead of Instant because last one cannot understand TT, UTC, UT including leap second handling etc.
来源：https://stackoverflow.com/questions/32016368/convert-a-julian-date-to-an-instant