问题
I\'m using Java\'s java.util.Date class in Scala and want to compare a Date object and the current time. I know I can calculate the delta by using getTime():
(new java.util.Date()).getTime() - oldDate.getTime()
However, this just leaves me with a long representing milliseconds. Is there any simpler, nicer way to get a time delta?
回答1:
The JDK Date API is horribly broken unfortunately. I recommend using Joda Time library.
Joda Time has a concept of time Interval:
Interval interval = new Interval(oldTime, new Instant());
EDIT: By the way, Joda has two concepts: Interval for representing an interval of time between two time instants (represent time between 8am and 10am), and a Duration that represents a length of time without the actual time boundaries (e.g. represent two hours!)
If you only care about time comparisions, most Date implementations (including the JDK one) implements Comparable interface which allows you to use the Comparable.compareTo()
回答2:
Simple diff (without lib)
/**
* Get a diff between two dates
* @param date1 the oldest date
* @param date2 the newest date
* @param timeUnit the unit in which you want the diff
* @return the diff value, in the provided unit
*/
public static long getDateDiff(Date date1, Date date2, TimeUnit timeUnit) {
long diffInMillies = date2.getTime() - date1.getTime();
return timeUnit.convert(diffInMillies,TimeUnit.MILLISECONDS);
}
And then can you call:
getDateDiff(date1,date2,TimeUnit.MINUTES);
to get the diff of the 2 dates in minutes unit.
TimeUnit is java.util.concurrent.TimeUnit, a standard Java enum going from nanos to days.
Human readable diff (without lib)
public static Map<TimeUnit,Long> computeDiff(Date date1, Date date2) {
long diffInMillies = date2.getTime() - date1.getTime();
//create the list
List<TimeUnit> units = new ArrayList<TimeUnit>(EnumSet.allOf(TimeUnit.class));
Collections.reverse(units);
//create the result map of TimeUnit and difference
Map<TimeUnit,Long> result = new LinkedHashMap<TimeUnit,Long>();
long milliesRest = diffInMillies;
for ( TimeUnit unit : units ) {
//calculate difference in millisecond
long diff = unit.convert(milliesRest,TimeUnit.MILLISECONDS);
long diffInMilliesForUnit = unit.toMillis(diff);
milliesRest = milliesRest - diffInMilliesForUnit;
//put the result in the map
result.put(unit,diff);
}
return result;
}
http://ideone.com/5dXeu6
The output is something like Map:{DAYS=1, HOURS=3, MINUTES=46, SECONDS=40, MILLISECONDS=0, MICROSECONDS=0, NANOSECONDS=0}, with the units ordered.
You just have to convert that map to an user-friendly string.
Warning
The above code snippets compute a simple diff between 2 instants. It can cause problems during a daylight saving switch, like explained in this post. This means if you compute the diff between dates with no time you may have a missing day/hour.
In my opinion the date diff is kind of subjective, especially on days. You may:
count the number of 24h elapsed time: day+1 - day = 1 day = 24h
count the number of elapsed time, taking care of daylight savings: day+1 - day = 1 = 24h (but using midnight time and daylight savings it could be 0 day and 23h)
count the number of
day switches, which means day+1 1pm - day 11am = 1 day, even if the elapsed time is just 2h (or 1h if there is a daylight saving :p)
My answer is valid if your definition of date diff on days match the 1st case
With JodaTime
If you are using JodaTime you can get the diff for 2 instants (millies backed ReadableInstant) dates with:
Interval interval = new Interval(oldInstant, new Instant());
But you can also get the diff for Local dates/times:
// returns 4 because of the leap year of 366 days
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5), PeriodType.years()).getYears()
// this time it returns 5
new Period(LocalDate.now(), LocalDate.now().plusDays(365*5+1), PeriodType.years()).getYears()
// And you can also use these static methods
Years.yearsBetween(LocalDate.now(), LocalDate.now().plusDays(365*5)).getYears()
回答3:
int diffInDays = (int)( (newerDate.getTime() - olderDate.getTime())
/ (1000 * 60 * 60 * 24) )
Note that this works with UTC dates, so the difference may be a day off if you look at local dates. And getting it to work correctly with local dates requires a completely different approach due to daylight savings time.
回答4:
You need to define your problem more clearly. You could just take the number of milliseconds between the two Date objects and divide by the number of milliseconds in 24 hours, for example... but:
- This won't take time zones into consideration -
Dateis always in UTC - This won't take daylight saving time into consideration (where there can be days which are only 23 hours long, for example)
- Even within UTC, how many days are there in August 16th 11pm to August 18th 2am? It's only 27 hours, so does that mean one day? Or should it be three days because it covers three dates?
回答5:
Using the java.time framework built into Java 8+:
ZonedDateTime now = ZonedDateTime.now();
ZonedDateTime oldDate = now.minusDays(1).minusMinutes(10);
Duration duration = Duration.between(oldDate, now);
System.out.println("ISO-8601: " + duration);
System.out.println("Minutes: " + duration.toMinutes());
Output:
ISO-8601: PT24H10M
Minutes: 1450
For more info, see the Oracle Tutorial and the ISO 8601 standard.
回答6:
tl;dr
Convert your obsolete java.util.Date objects to their replacement, java.time.Instant. Then calculate the elapsed time as a Duration.
Duration d =
Duration.between( // Calculate the span of time between two moments as a number of hours, minutes, and seconds.
myJavaUtilDate.toInstant() , // Convert legacy class to modern class by calling new method added to the old class.
Instant.now() // Capture the current moment in UTC. About two and a half hours later in this example.
)
;
d.toString(): PT2H34M56S
d.toMinutes(): 154
d.toMinutesPart(): 34
ISO 8601 Format: PnYnMnDTnHnMnS
The sensible standard ISO 8601 defines a concise textual representation of a span of time as a number of years, months, days, hours, etc. The standard calls such such a span a duration. The format is PnYnMnDTnHnMnS where the P means "Period", the T separates the date portion from the time portion, and in between are numbers followed by a letter.
Examples:
P3Y6M4DT12H30M5S
three years, six months, four days, twelve hours, thirty minutes, and five secondsPT4H30M
Four and a half hours
java.time
The java.time framework built into Java 8 and later supplants the troublesome old java.util.Date/java.util.Calendar classes. The new classes are inspired by the highly successful Joda-Time framework, intended as its successor, similar in concept but re-architected. Defined by JSR 310. Extended by the ThreeTen-Extra project. See the Tutorial.
Moment
The Instant class represents a moment on the timeline in UTC with a resolution of nanoseconds (up to nine (9) digits of a decimal fraction).
Instant instant = Instant.now() ; // Capture current moment in UTC.
Best to avoid the legacy classes such as Date/Calendar. But if you must inter-operate with old code not yet updated to java.time, convert back and forth. Call new conversion methods added to the old classes. For moving from a java.util.Date to an Instant, call Date::toInstant.
Instant instant = myJavaUtilDate.toInstant() ; // Convert from legacy `java.util.Date` class to modern `java.time.Instant` class.
Span of time
The java.time classes have split this idea of representing a span of time as a number of years, months, days, hours, minutes, seconds into two halves:
- Period for years, months, days
- Duration for days, hours, minutes, seconds
Here is an example.
ZoneId zoneId = ZoneId.of ( "America/Montreal" );
ZonedDateTime now = ZonedDateTime.now ( zoneId );
ZonedDateTime future = now.plusMinutes ( 63 );
Duration duration = Duration.between ( now , future );
Dump to console.
Both Period and Duration use the ISO 8601 standard for generating a String representation of their value.
System.out.println ( "now: " + now + " to future: " + now + " = " + duration );
now: 2015-11-26T00:46:48.016-05:00[America/Montreal] to future: 2015-11-26T00:46:48.016-05:00[America/Montreal] = PT1H3M
Java 9 adds methods to Duration to get the days part, hours part, minutes part, and seconds part.
You can get the total number of days or hours or minutes or seconds or milliseconds or nanoseconds in the entire Duration.
long totalHours = duration.toHours();
In Java 9 the Duration class gets new methods for returning the various parts of days, hours, minutes, seconds, milliseconds/nanoseconds. Call the to…Part methods: toDaysPart(), toHoursPart(), and so on.
ChronoUnit
If you only care about a simpler larger granularity of time, such as “number of days elapsed”, use the ChronoUnit enum.
long daysElapsed = ChronoUnit.DAYS.between( earlier , later );
Another example.
Instant now = Instant.now();
Instant later = now.plus( Duration.ofHours( 2 ) );
…
long minutesElapsed = ChronoUnit.MINUTES.between( now , later );
120
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 java.time.
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….
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.
Joda-Time
UPDATE: The Joda-Time project is now in maintenance mode, with the team advising migration to the java.time classes. I leave this section intact for history.
The Joda-Time library uses ISO 8601 for its defaults. Its Period class parses and generates these PnYnMnDTnHnMnS strings.
DateTime now = DateTime.now(); // Caveat: Ignoring the important issue of time zones.
Period period = new Period( now, now.plusHours( 4 ).plusMinutes( 30));
System.out.println( "period: " + period );
Renders:
period: PT4H30M
回答7:
Days d = Days.daysBetween(startDate, endDate);
int days = d.getDays();
https://www.joda.org/joda-time/faq.html#datediff
回答8:
A slightly simpler alternative:
System.currentTimeMillis() - oldDate.getTime()
As for "nicer": well, what exactly do you need? The problem with representing time durations as a number of hours and days etc. is that it may lead to inaccuracies and wrong expectations due to the complexity of dates (e.g. days can have 23 or 25 hours due to daylight savings time).
回答9:
Using millisecond approach can cause problems in some locales.
Lets take, for example, the difference between the two dates 03/24/2007 and 03/25/2007 should be 1 day;
However, using the millisecond route, you'll get 0 days, if you run this in the UK!
/** Manual Method - YIELDS INCORRECT RESULTS - DO NOT USE**/
/* This method is used to find the no of days between the given dates */
public long calculateDays(Date dateEarly, Date dateLater) {
return (dateLater.getTime() - dateEarly.getTime()) / (24 * 60 * 60 * 1000);
}
Better way to implement this is to use java.util.Calendar
/** Using Calendar - THE CORRECT WAY**/
public static long daysBetween(Calendar startDate, Calendar endDate) {
Calendar date = (Calendar) startDate.clone();
long daysBetween = 0;
while (date.before(endDate)) {
date.add(Calendar.DAY_OF_MONTH, 1);
daysBetween++;
}
return daysBetween;
}
回答10:
There are many ways you can find the difference between dates & times. One of the simplest ways that I know of would be:
Calendar calendar1 = Calendar.getInstance();
Calendar calendar2 = Calendar.getInstance();
calendar1.set(2012, 04, 02);
calendar2.set(2012, 04, 04);
long milsecs1= calendar1.getTimeInMillis();
long milsecs2 = calendar2.getTimeInMillis();
long diff = milsecs2 - milsecs1;
long dsecs = diff / 1000;
long dminutes = diff / (60 * 1000);
long dhours = diff / (60 * 60 * 1000);
long ddays = diff / (24 * 60 * 60 * 1000);
System.out.println("Your Day Difference="+ddays);
The print statement is just an example - you can format it, the way you like.
回答11:
Since all the answers here are correct but use legacy java or 3rd party libs like joda or similar, I will just drop another way using new java.time classes in Java 8 and later. See Oracle Tutorial.
Use LocalDate and ChronoUnit:
LocalDate d1 = LocalDate.of(2017, 5, 1);
LocalDate d2 = LocalDate.of(2017, 5, 18);
long days = ChronoUnit.DAYS.between(d1, d2);
System.out.println( days );
回答12:
Subtracting the dates in milliseconds works (as described in another post), but you have to use HOUR_OF_DAY and not HOUR when clearing the time parts of your dates:
public static final long MSPERDAY = 60 * 60 * 24 * 1000;
...
final Calendar dateStartCal = Calendar.getInstance();
dateStartCal.setTime(dateStart);
dateStartCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateStartCal.set(Calendar.MINUTE, 0);
dateStartCal.set(Calendar.SECOND, 0);
dateStartCal.set(Calendar.MILLISECOND, 0);
final Calendar dateEndCal = Calendar.getInstance();
dateEndCal.setTime(dateEnd);
dateEndCal.set(Calendar.HOUR_OF_DAY, 0); // Crucial.
dateEndCal.set(Calendar.MINUTE, 0);
dateEndCal.set(Calendar.SECOND, 0);
dateEndCal.set(Calendar.MILLISECOND, 0);
final long dateDifferenceInDays = ( dateStartCal.getTimeInMillis()
- dateEndCal.getTimeInMillis()
) / MSPERDAY;
if (dateDifferenceInDays > 15) {
// Do something if difference > 15 days
}
回答13:
If you don't want to use JodaTime or similar, the best solution is probably this:
final static long MILLIS_PER_DAY = 24 * 3600 * 1000;
long msDiff= date1.getTime() - date2.getTime();
long daysDiff = Math.round(msDiff / ((double)MILLIS_PER_DAY));
The number of ms per day is not always the same (because of daylight saving time and leap seconds), but it's very close, and at least deviations due to daylight saving time cancel out over longer periods. Therefore dividing and then rounding will give a correct result (at least as long as the local calendar used does not contain weird time jumps other than DST and leap seconds).
Note that this still assumes that date1 and date2 are set to the same time of day. For different times of day, you'd first have to define what "date difference" means, as pointed out by Jon Skeet.
回答14:
Take a look at Joda Time, which is an improved Date/Time API for Java and should work fine with Scala.
回答15:
Let me show difference between Joda Interval and Days:
DateTime start = new DateTime(2012, 2, 6, 10, 44, 51, 0);
DateTime end = new DateTime(2012, 2, 6, 11, 39, 47, 1);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
System.out.println(period.getYears() + " years, " + period.getMonths() + " months, " + period.getWeeks() + " weeks, " + period.getDays() + " days");
System.out.println(period.getHours() + " hours, " + period.getMinutes() + " minutes, " + period.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *1 weeks, 1 days*
//0 hours, 54 minutes, 56 seconds
//Period can set PeriodType,such as PeriodType.yearMonthDay(),PeriodType.yearDayTime()...
Period p = new Period(start, end, PeriodType.yearMonthDayTime());
System.out.println(p.getYears() + " years, " + p.getMonths() + " months, " + p.getWeeks() + " weeks, " + p.getDays() + "days");
System.out.println(p.getHours() + " hours, " + p.getMinutes() + " minutes, " + p.getSeconds() + " seconds ");
//Result is:
//0 years, 0 months, *0 weeks, 8 days*
//0 hours, 54 minutes, 56 seconds
回答16:
If you need a formatted return String like "2 Days 03h 42m 07s", try this:
public String fill2(int value)
{
String ret = String.valueOf(value);
if (ret.length() < 2)
ret = "0" + ret;
return ret;
}
public String get_duration(Date date1, Date date2)
{
TimeUnit timeUnit = TimeUnit.SECONDS;
long diffInMilli = date2.getTime() - date1.getTime();
long s = timeUnit.convert(diffInMilli, TimeUnit.MILLISECONDS);
long days = s / (24 * 60 * 60);
long rest = s - (days * 24 * 60 * 60);
long hrs = rest / (60 * 60);
long rest1 = rest - (hrs * 60 * 60);
long min = rest1 / 60;
long sec = s % 60;
String dates = "";
if (days > 0) dates = days + " Days ";
dates += fill2((int) hrs) + "h ";
dates += fill2((int) min) + "m ";
dates += fill2((int) sec) + "s ";
return dates;
}
回答17:
Note: startDate and endDates are -> java.util.Date
import org.joda.time.Duration;
import org.joda.time.Interval;
// Use .getTime() unless it is a joda DateTime object
Interval interval = new Interval(startDate.getTime(), endDate.getTime());
Duration period = interval.toDuration();
//gives the number of days elapsed between start
period.getStandardDays();
and end date
Similar to days, you can also get hours, minutes and seconds
period.getStandardHours();
period.getStandardMinutes();
period.getStandardSeconds();
回答18:
Check example here http://www.roseindia.net/java/beginners/DateDifferent.shtml This example give you difference in days, hours, minutes, secs and milli sec's :).
import java.util.Calendar;
import java.util.Date;
public class DateDifferent {
public static void main(String[] args) {
Date date1 = new Date(2009, 01, 10);
Date date2 = new Date(2009, 07, 01);
Calendar calendar1 = Calendar.getInstance();
Calendar calendar2 = Calendar.getInstance();
calendar1.setTime(date1);
calendar2.setTime(date2);
long milliseconds1 = calendar1.getTimeInMillis();
long milliseconds2 = calendar2.getTimeInMillis();
long diff = milliseconds2 - milliseconds1;
long diffSeconds = diff / 1000;
long diffMinutes = diff / (60 * 1000);
long diffHours = diff / (60 * 60 * 1000);
long diffDays = diff / (24 * 60 * 60 * 1000);
System.out.println("\nThe Date Different Example");
System.out.println("Time in milliseconds: " + diff + " milliseconds.");
System.out.println("Time in seconds: " + diffSeconds + " seconds.");
System.out.println("Time in minutes: " + diffMinutes + " minutes.");
System.out.println("Time in hours: " + diffHours + " hours.");
System.out.println("Time in days: " + diffDays + " days.");
}
}
回答19:
Use GMT time zone to get an instance of the Calendar, set the time using the set method of Calendar class. The GMT timezone has 0 offset (not really important) and daylight saving time flag set to false.
final Calendar cal = Calendar.getInstance(TimeZone.getTimeZone("GMT"));
cal.set(Calendar.YEAR, 2011);
cal.set(Calendar.MONTH, 9);
cal.set(Calendar.DAY_OF_MONTH, 29);
cal.set(Calendar.HOUR, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
final Date startDate = cal.getTime();
cal.set(Calendar.YEAR, 2011);
cal.set(Calendar.MONTH, 12);
cal.set(Calendar.DAY_OF_MONTH, 21);
cal.set(Calendar.HOUR, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
final Date endDate = cal.getTime();
System.out.println((endDate.getTime() - startDate.getTime()) % (1000l * 60l * 60l * 24l));
回答20:
Following code can give you the desired output:
String startDate = "Jan 01 2015";
DateTimeFormatter formatter = DateTimeFormatter.ofPattern("MMM dd yyyy");
LocalDate date = LocalDate.parse(startDate, formatter);
String currentDate = "Feb 11 2015";
LocalDate date1 = LocalDate.parse(currentDate, formatter);
System.out.println(date1.toEpochDay() - date.toEpochDay());
回答21:
public static String getDifferenceBtwTime(Date dateTime) {
long timeDifferenceMilliseconds = new Date().getTime() - dateTime.getTime();
long diffSeconds = timeDifferenceMilliseconds / 1000;
long diffMinutes = timeDifferenceMilliseconds / (60 * 1000);
long diffHours = timeDifferenceMilliseconds / (60 * 60 * 1000);
long diffDays = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24);
long diffWeeks = timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 7);
long diffMonths = (long) (timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 30.41666666));
long diffYears = (long)(timeDifferenceMilliseconds / (1000 * 60 * 60 * 24 * 365));
if (diffSeconds < 1) {
return "one sec ago";
} else if (diffMinutes < 1) {
return diffSeconds + " seconds ago";
} else if (diffHours < 1) {
return diffMinutes + " minutes ago";
} else if (diffDays < 1) {
return diffHours + " hours ago";
} else if (diffWeeks < 1) {
return diffDays + " days ago";
} else if (diffMonths < 1) {
return diffWeeks + " weeks ago";
} else if (diffYears < 12) {
return diffMonths + " months ago";
} else {
return diffYears + " years ago";
}
}
回答22:
int daysDiff = (date1.getTime() - date2.getTime()) / MILLIS_PER_DAY;
回答23:
Best thing to do is
(Date1-Date2)/86 400 000
That number is the number of milliseconds in a day.
One date-other date gives you difference in milliseconds.
Collect the answer in a double variable.
回答24:
That's probably the most straightforward way to do it - perhaps it's because I've been coding in Java (with its admittedly clunky date and time libraries) for a while now, but that code looks "simple and nice" to me!
Are you happy with the result being returned in milliseconds, or is part of your question that you would prefer to have it returned in some alternative format?
回答25:
Not using the standard API, no. You can roll your own doing something like this:
class Duration {
private final TimeUnit unit;
private final long length;
// ...
}
Or you can use Joda:
DateTime a = ..., b = ...;
Duration d = new Duration(a, b);
回答26:
Just to answer the initial question:
Put the following code in a Function like Long getAge(){}
Date dahora = new Date();
long MillisToYearsByDiv = 1000l *60l * 60l * 24l * 365l;
long javaOffsetInMillis = 1990l * MillisToYearsByDiv;
long realNowInMillis = dahora.getTime() + javaOffsetInMillis;
long realBirthDayInMillis = this.getFechaNac().getTime() + javaOffsetInMillis;
long ageInMillis = realNowInMillis - realBirthDayInMillis;
return ageInMillis / MillisToYearsByDiv;
The most important here is to work with long numbers when multiplying and dividing. And of course, the offset that Java applies in its calculus of Dates.
:)
回答27:
Since the question is tagged with Scala,
import scala.concurrent.duration._
val diff = (System.currentTimeMillis() - oldDate.getTime).milliseconds
val diffSeconds = diff.toSeconds
val diffMinutes = diff.toMinutes
val diffHours = diff.toHours
val diffDays = diff.toDays
回答28:
Here's a correct Java 7 solution in O(1) without any dependencies.
public static int countDaysBetween(Date date1, Date date2) {
Calendar c1 = removeTime(from(date1));
Calendar c2 = removeTime(from(date2));
if (c1.get(YEAR) == c2.get(YEAR)) {
return Math.abs(c1.get(DAY_OF_YEAR) - c2.get(DAY_OF_YEAR)) + 1;
}
// ensure c1 <= c2
if (c1.get(YEAR) > c2.get(YEAR)) {
Calendar c = c1;
c1 = c2;
c2 = c;
}
int y1 = c1.get(YEAR);
int y2 = c2.get(YEAR);
int d1 = c1.get(DAY_OF_YEAR);
int d2 = c2.get(DAY_OF_YEAR);
return d2 + ((y2 - y1) * 365) - d1 + countLeapYearsBetween(y1, y2) + 1;
}
private static int countLeapYearsBetween(int y1, int y2) {
if (y1 < 1 || y2 < 1) {
throw new IllegalArgumentException("Year must be > 0.");
}
// ensure y1 <= y2
if (y1 > y2) {
int i = y1;
y1 = y2;
y2 = i;
}
int diff = 0;
int firstDivisibleBy4 = y1;
if (firstDivisibleBy4 % 4 != 0) {
firstDivisibleBy4 += 4 - (y1 % 4);
}
diff = y2 - firstDivisibleBy4 - 1;
int divisibleBy4 = diff < 0 ? 0 : diff / 4 + 1;
int firstDivisibleBy100 = y1;
if (firstDivisibleBy100 % 100 != 0) {
firstDivisibleBy100 += 100 - (firstDivisibleBy100 % 100);
}
diff = y2 - firstDivisibleBy100 - 1;
int divisibleBy100 = diff < 0 ? 0 : diff / 100 + 1;
int firstDivisibleBy400 = y1;
if (firstDivisibleBy400 % 400 != 0) {
firstDivisibleBy400 += 400 - (y1 % 400);
}
diff = y2 - firstDivisibleBy400 - 1;
int divisibleBy400 = diff < 0 ? 0 : diff / 400 + 1;
return divisibleBy4 - divisibleBy100 + divisibleBy400;
}
public static Calendar from(Date date) {
Calendar c = Calendar.getInstance();
c.setTime(date);
return c;
}
public static Calendar removeTime(Calendar c) {
c.set(HOUR_OF_DAY, 0);
c.set(MINUTE, 0);
c.set(SECOND, 0);
c.set(MILLISECOND, 0);
return c;
}
回答29:
After wading through all the other answers, to keep the Java 7 Date type but be more precise/standard with the Java 8 diff approach,
public static long daysBetweenDates(Date d1, Date d2) {
Instant instant1 = d1.toInstant();
Instant instant2 = d2.toInstant();
long diff = ChronoUnit.DAYS.between(instant1, instant2);
return diff;
}
回答30:
try this:
int epoch = (int) (new java.text.SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse("01/01/1970 00:00:00").getTime() / 1000);
you can edit the string in the parse() methods param.
来源：https://stackoverflow.com/questions/1555262/calculating-the-difference-between-two-java-date-instances