Reputation: 21962
Calculating the difference between two Java date instances
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?
Upvotes: 508
Views: 810670
Answers (30)
Reputation: 33
This will perfectly works with joda api
import java.util.Calendar;
import java.util.Date;
import java.util.GregorianCalendar;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import org.joda.time.DateTime;
import org.joda.time.Months;
import org.joda.time.Weeks;
import org.joda.time.Years;
import java.util.LinkedHashMap;
public Map<String,Integer> Datedifference(String fromDate,String toDate) throws ParseException {
Map<String,Integer> dateUtilMap = new LinkedHashMap<String, Integer>();
Date date1=new SimpleDateFormat("dd-MM-yyyy").parse(fromDate);
Date date2=new SimpleDateFormat("dd-MM-yyyy").parse(toDate);
Calendar fromDay = new GregorianCalendar();
fromDay.setTime(date1);
Calendar toDay = new GregorianCalendar();
toDay.setTime(date2);
int days = (int) (1+(toDay.getTimeInMillis() -fromDay.getTimeInMillis()) / 1000 / 60 / 60 / 24);
if(days >1) {
DateTime dateTime1 = new DateTime(fromDay);
DateTime dateTime2 = new DateTime(toDay);
int weeks = Weeks.weeksBetween(dateTime1.dayOfWeek().withMinimumValue().minusDays(1),
dateTime2.dayOfWeek().withMaximumValue().plusDays(1)).getWeeks();
int months = Months.monthsBetween(dateTime1.dayOfMonth().withMinimumValue().minusDays(1),
dateTime2.dayOfMonth().withMaximumValue().plusDays(1)).getMonths();
int years =1+Years.yearsBetween(dateTime1, dateTime2).getYears();
dateUtilMap.put("days", days);
dateUtilMap.put("weeks", weeks);
dateUtilMap.put("months", months);
dateUtilMap.put("years", years);
System.out.println(dateUtilMap);
return dateUtilMap;
} else {
System.out.println("From Date Must be samller than to Date");
return null;
}
}
Upvotes: -1
Reputation: 92112
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 you can 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;
}
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 a 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()
Upvotes: 619
Reputation: 4096
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 and end date.
period.getStandardDays();
Similar to days, you can also get hours, minutes and seconds
period.getStandardHours();
period.getStandardMinutes();
period.getStandardSeconds();
Upvotes: 5
Reputation: 593
Since dates can contain hours and minutes, final result will be rounded down, which will result in incorrect value. For example, you calculate difference between today at 22:00 p.m and day after tomorrow 00:00 a.m, so the final result will be 1, because in reality it was 1.08 or smth difference, then it gets rounded down when calling TimeUnit.MILLISECONDS.toDays(..). That's why you need to take that in account, so in my solution I subtract the remainder of milliseconds from milliseconds in a day. Additionally, if you want to count the end date, you need to +1 it.
import java.util.Date;
import java.util.concurrent.TimeUnit;
public static long getDaysBetween(Date date1, Date date2, boolean includeEndDate) {
long millisInDay = 60 * 60 * 24 * 1000;
long difference = Math.abs(date1.getTime() - date2.getTime());
long add = millisInDay - (difference % millisInDay);//is used to calculate true number of days, because by default hours, minutes are also counted
return TimeUnit.MILLISECONDS.toDays(difference + add) + (includeEndDate ? 1 : 0);
}
Tests:
Date date1 = new Date(121, Calendar.NOVEMBER, 27); //2021 Nov 27
Date date2 = new Date(121, Calendar.DECEMBER, 29); //2021 Dec 29
System.out.println( getDaysBetween(date1, date2, false) ); //32 days difference
System.out.println( getDaysBetween(date1, date2, true) ); //33 days difference
Upvotes: -1
Reputation: 153
If you want to fix the issue for date ranges that cross daylight savings time boundary (e.g. one date in summer time and the other one in winter time), you can use this to get the difference in days:
public static long calculateDifferenceInDays(Date start, Date end, Locale locale) {
Calendar cal = Calendar.getInstance(locale);
cal.setTime(start);
cal.set(Calendar.HOUR_OF_DAY, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
cal.set(Calendar.MILLISECOND, 0);
long startTime = cal.getTimeInMillis();
cal.setTime(end);
cal.set(Calendar.HOUR_OF_DAY, 0);
cal.set(Calendar.MINUTE, 0);
cal.set(Calendar.SECOND, 0);
cal.set(Calendar.MILLISECOND, 0);
long endTime = cal.getTimeInMillis();
// calculate the offset if one of the dates is in summer time and the other one in winter time
TimeZone timezone = cal.getTimeZone();
int offsetStart = timezone.getOffset(startTime);
int offsetEnd = timezone.getOffset(endTime);
int offset = offsetEnd - offsetStart;
return TimeUnit.MILLISECONDS.toDays(endTime - startTime + offset);
}
Upvotes: 2
Reputation: 14952
Take a look at Joda Time, which is an improved Date/Time API for Java and should work fine with Scala.
Upvotes: 8
Reputation: 59299
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()
Upvotes: 215
Reputation: 1154
Days d = Days.daysBetween(startDate, endDate);
int days = d.getDays();
https://www.joda.org/joda-time/faq.html#datediff
Upvotes: 39
Reputation: 11974
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;
}
Upvotes: 5
Reputation: 338181
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:
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
Upvotes: 69
Reputation: 6525
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.
Upvotes: 23
Reputation: 48258
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 );
Upvotes: 25
Reputation: 121
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;
}
Upvotes: 3
Reputation: 10236
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
Upvotes: 2
Reputation: 12063
Another pure Java variation:
public boolean isWithin30Days(Calendar queryCalendar) {
// 1. Take the date you are checking, and roll it back N days
Calendar queryCalMinus30Days = Calendar.getInstance();
queryCalMinus30Days.setTime(queryCalendar.getTime());
queryCalMinus30Days.add(Calendar.DATE, -30); // subtract 30 days from the calendar
// 2. Get respective milliseconds for the two Calendars: now & queryCal minus N days
long nowL = Calendar.getInstance().getTimeInMillis();
long queryCalMinus30DaysL = queryCalMinus30Days.getTimeInMillis();
// 3. if nowL is still less than the queryCalMinus30DaysL, it means queryCalendar is more than 30 days into future
boolean isWithin30Days = nowL >= queryCalMinus30DaysL;
return isWithin30Days;
}
thanks to starter code here: https://stackoverflow.com/a/30207726/2162226
Upvotes: 0
Reputation: 3124
Earlier versions of Java you can try.
public static String daysBetween(Date createdDate, Date expiryDate) {
Calendar createdDateCal = Calendar.getInstance();
createdDateCal.clear();
createdDateCal.setTime(createdDate);
Calendar expiryDateCal = Calendar.getInstance();
expiryDateCal.clear();
expiryDateCal.setTime(expiryDate);
long daysBetween = 0;
while (createdDateCal.before(expiryDateCal)) {
createdDateCal.add(Calendar.DAY_OF_MONTH, 1);
daysBetween++;
}
return daysBetween+"";
}
Upvotes: 1
Reputation: 114400
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.
Upvotes: 83
Reputation: 354
After having read many answers and comments to this question, I was left with the impression that one either has to use the Joda time or else take into account some peculiarities with the daylight saving time etc. Since I didn't want to do either of these, I ended up writing a few lines of code to calculate the difference between two dates without using any date or time related Java classes.
In the code below the numbers of year, month and day are the same as in real life. For example in December 24th 2015, the year = 2015, the month = 12 and the day = 24.
I want to share this code in case someone else wants to use it. There are 3 methods: 1) A method to find out whether a given year is a leap year 2) A method to calculate the number of a given day in relation to January 1st of a given year 3) A method to calculate the number of days between any two dates using the method 2 (number of the end date minus number of the start date).
Here are the methods:
1)
public static boolean isLeapYear (int year) {
//Every 4. year is a leap year, except if the year is divisible by 100 and not by 400
//For example 1900 is not a leap year but 2000 is
boolean result = false;
if (year % 4 == 0) {
result = true;
}
if (year % 100 == 0) {
result = false;
}
if (year % 400 == 0) {
result = true;
}
return result;
}
2)
public static int daysGoneSince (int yearZero, int year, int month, int day) {
//Calculates the day number of the given date; day 1 = January 1st in the yearZero
//Validate the input
if (year < yearZero || month < 1 || month > 12 || day < 1 || day > 31) {
//Throw an exception
throw new IllegalArgumentException("Too many or too few days in month or months in year or the year is smaller than year zero");
}
else if (month == 4 || month == 6 || month == 9 || month == 11) {//Months with 30 days
if (day == 31) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
else if (month == 2) {//February 28 or 29
if (isLeapYear(year)) {
if (day > 29) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
else if (day > 28) {
//Throw an exception
throw new IllegalArgumentException("Too many days in month");
}
}
//Start counting days
int days = 0;
//Days in the target month until the target day
days = days + day;
//Days in the earlier months in the target year
for (int i = 1; i < month; i++) {
switch (i) {
case 1: case 3: case 5:
case 7: case 8: case 10:
case 12:
days = days + 31;
break;
case 2:
days = days + 28;
if (isLeapYear(year)) {
days = days + 1;
}
break;
case 4: case 6: case 9: case 11:
days = days + 30;
break;
}
}
//Days in the earlier years
for (int i = yearZero; i < year; i++) {
days = days + 365;
if (isLeapYear(i)) {
days = days + 1;
}
}
return days;
}
3)
public static int dateDiff (int startYear, int startMonth, int startDay, int endYear, int endMonth, int endDay) {
int yearZero;
//daysGoneSince presupposes that the first argument be smaller or equal to the second argument
if (10000 * startYear + 100 * startMonth + startDay > 10000 * endYear + 100 * endMonth + endDay) {//If the end date is earlier than the start date
yearZero = endYear;
}
else {
yearZero = startYear;
}
return daysGoneSince(yearZero, endYear, endMonth, endDay) - daysGoneSince(yearZero, startYear, startMonth, startDay);
}
Upvotes: 1
Reputation: 4617
There is an easy way to do this in java
//create an utility method
public long getDaysBetweenDates(Date d1, Date d2){
return TimeUnit.MILLISECONDS.toDays(d1.getTime() - d2.getTime());
}
This method will return the number of days between the 2 dates. Either you can use the default java date format or can easily convert from any date format.
Upvotes: 0
Reputation: 2534
Just use below method with two Date objects. If you want to pass current date, just pass new Date() as a second parameter as it is initialised with current time.
public String getDateDiffString(Date dateOne, Date dateTwo)
{
long timeOne = dateOne.getTime();
long timeTwo = dateTwo.getTime();
long oneDay = 1000 * 60 * 60 * 24;
long delta = (timeTwo - timeOne) / oneDay;
if (delta > 0) {
return "dateTwo is " + delta + " days after dateOne";
}
else {
delta *= -1;
return "dateTwo is " + delta + " days before dateOne";
}
}
Also, apart from from number of days, if, you want other parameter difference too, use below snippet,
int year = delta / 365;
int rest = delta % 365;
int month = rest / 30;
rest = rest % 30;
int weeks = rest / 7;
int days = rest % 7;
P.S Code is entirely taken from an SO answer.
Upvotes: 1
Reputation: 3448
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";
}
}
Upvotes: 4
Reputation: 31
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.
:)
Upvotes: 2
Reputation: 110
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());
Upvotes: 4
Reputation: 72
This is another sample. Which basically works for user-defined pattern.
public static LinkedHashMap<String, Object> checkDateDiff(DateTimeFormatter dtfObj, String startDate, String endDate)
{
Map<String, Object> dateDiffMap = new HashMap<String, Object>();
DateTime start = DateTime.parse(startDate,dtfObj);
DateTime end = DateTime.parse(endDate,dtfObj);
Interval interval = new Interval(start, end);
Period period = interval.toPeriod();
dateDiffMap.put("ISO-8601_PERIOD_FORMAT", period);
dateDiffMap.put("YEAR", period.getYears());
dateDiffMap.put("MONTH", period.getMonths());
dateDiffMap.put("WEEK", period.getWeeks());
dateDiffMap.put("DAY", period.getWeeks());
dateDiffMap.put("HOUR", period.getHours());
dateDiffMap.put("MINUTE", period.getMinutes());
dateDiffMap.put("SECOND", period.getSeconds());
return dateDiffMap;
}
Upvotes: 0
Reputation: 5381
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;
}
Upvotes: 5
Reputation: 1474
I liked the TimeUnit-based approach until I found out that it only covers the trivial cases where the number of how many units of one timeunit are in the next higher unit is fixed. This breaks apart when you want to know how many months, year, etc are in between.
here is a counting approach, not as efficient as some others, but it seems to work for me and takes into account DST, too.
public static String getOffsetAsString( Calendar cNow, Calendar cThen) {
Calendar cBefore;
Calendar cAfter;
if ( cNow.getTimeInMillis() < cThen.getTimeInMillis()) {
cBefore = ( Calendar) cNow.clone();
cAfter = cThen;
} else {
cBefore = ( Calendar) cThen.clone();
cAfter = cNow;
}
// compute diff
Map<Integer, Long> diffMap = new HashMap<Integer, Long>();
int[] calFields = { Calendar.YEAR, Calendar.MONTH, Calendar.DAY_OF_MONTH, Calendar.HOUR_OF_DAY, Calendar.MINUTE, Calendar.SECOND, Calendar.MILLISECOND};
for ( int i = 0; i < calFields.length; i++) {
int field = calFields[ i];
long d = computeDist( cAfter, cBefore, field);
diffMap.put( field, d);
}
final String result = String.format( "%dY %02dM %dT %02d:%02d:%02d.%03d",
diffMap.get( Calendar.YEAR), diffMap.get( Calendar.MONTH), diffMap.get( Calendar.DAY_OF_MONTH), diffMap.get( Calendar.HOUR_OF_DAY), diffMap.get( Calendar.MINUTE), diffMap.get( Calendar.SECOND), diffMap.get( Calendar.MILLISECOND));
return result;
}
private static int computeDist( Calendar cAfter, Calendar cBefore, int field) {
cBefore.setLenient( true);
System.out.print( "D " + new Date( cBefore.getTimeInMillis()) + " --- " + new Date( cAfter.getTimeInMillis()) + ": ");
int count = 0;
if ( cAfter.getTimeInMillis() > cBefore.getTimeInMillis()) {
int fVal = cBefore.get( field);
while ( cAfter.getTimeInMillis() >= cBefore.getTimeInMillis()) {
count++;
fVal = cBefore.get( field);
cBefore.set( field, fVal + 1);
System.out.print( count + "/" + ( fVal + 1) + ": " + new Date( cBefore.getTimeInMillis()) + " ] ");
}
int result = count - 1;
cBefore.set( field, fVal);
System.out.println( "" + result + " at: " + field + " cb = " + new Date( cBefore.getTimeInMillis()));
return result;
}
return 0;
}
Upvotes: 0
Reputation: 1
public static void main(String[] args) {
String dateStart = "01/14/2012 09:29:58";
String dateStop = "01/14/2012 10:31:48";
SimpleDateFormat format = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(dateStart);
d2 = format.parse(dateStop);
DateTime date11 = new DateTime(d1);
DateTime date22 = new DateTime(d2);
int days = Days.daysBetween(date11.withTimeAtStartOfDay(), date22.withTimeAtStartOfDay()).getDays();
int hours = Hours.hoursBetween(date11, date22).getHours() % 24;
int minutes = Minutes.minutesBetween(date11, date22).getMinutes() % 60;
int seconds = Seconds.secondsBetween(date11, date22).getSeconds() % 60;
if (hours > 0 || minutes > 0 || seconds > 0) {
days = days + 1;
}
System.out.println(days);
} catch (Exception e) {
e.printStackTrace();
}
}
This will give date diff for the same day as well
Upvotes: 0
Reputation: 837
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.
Upvotes: 3
Reputation: 2095
Since you are using Scala, there is a very good Scala library Lamma. With Lamma you can minus date directly with - operator
scala> Date(2015, 5, 5) - 2 // minus days by int
res1: io.lamma.Date = Date(2015,5,3)
scala> Date(2015, 5, 15) - Date(2015, 5, 8) // minus two days => difference between two days
res2: Int = 7
Upvotes: 1
Reputation: 1092
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
}
Upvotes: 10
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