Calculate date/time difference in java

Question

I want to calculate difference between 2 dates in hours/minutes/seconds.

I have a slight problem with my code here it is :

String dateStart = "11/03/14 09:29:58";
String dateStop = "11/03/14 09:33:43";

// Custom date format
SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");  

Date d1 = null;
Date d2 = null;
try {
    d1 = format.parse(dateStart);
    d2 = format.parse(dateStop);
} catch (ParseException e) {
    e.printStackTrace();
}    

// Get msec from each, and subtract.
long diff = d2.getTime() - d1.getTime();
long diffSeconds = diff / 1000;         
long diffMinutes = diff / (60 * 1000);         
long diffHours = diff / (60 * 60 * 1000);                      
System.out.println("Time in seconds: " + diffSeconds + " seconds.");         
System.out.println("Time in minutes: " + diffMinutes + " minutes.");         
System.out.println("Time in hours: " + diffHours + " hours.");

This should produce :

Time in seconds: 45 seconds.
Time in minutes: 3 minutes.
Time in hours: 0 hours.

However I get this result :

Time in seconds: 225 seconds.
Time in minutes: 3 minutes.
Time in hours: 0 hours.

Can anyone see what I'm doing wrong here ?

A similar question found: http://stackoverflow.com/questions/625433/how-to-convert-milliseconds-to-x-mins-x-seconds-in-java — Yves Martin, Feb 07 '12 at 13:43
There is a better way to do it: http://stackoverflow.com/a/15541322/562769 — Martin Thoma, Mar 21 '13 at 07:52
For Java 8+ there is a real simple way: https://stackoverflow.com/a/23176621/1548776 — ᴠɪɴᴄᴇɴᴛ, Aug 07 '18 at 15:09

score 235 · Answer 1 · answered Nov 15 '13 at 10:10

235

I would prefer to use suggested java.util.concurrent.TimeUnit class.

long diff = d2.getTime() - d1.getTime();//as given

long seconds = TimeUnit.MILLISECONDS.toSeconds(diff);
long minutes = TimeUnit.MILLISECONDS.toMinutes(diff);

answered Nov 15 '13 at 10:10

Admit

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@Mark I disagree. You missed the point of the question. Despite the fact that he uses the phrase "Time in seconds" in the rest of the post he makes it clear that he doesn't actually want a plain conversion, he wants remainders. This is less efficient than the accepted answer (employing a method call, which even in JVM bytecode is a few instructions), less clear (it's longer, and frankly if something finds "1000" or "60" to be magic numbers in that context, they're not working with a full deck) and, pivotally, it doesn't do what the OP wanted. – Parthian Shot Aug 08 '14 at 17:25
There is a way to write a generic method for that. See http://stackoverflow.com/a/10650881/82609 – Sebastien Lorber Jan 29 '16 at 10:20
1

if i want, diference for weeks, months, years? – jose920405 Sep 15 '16 at 16:51
@jose920405 in that case, you need Joda time – John Henckel Jun 29 '17 at 21:13
1

This will give the answer the OP has specified they don't want. – Peter Lawrey Nov 03 '18 at 15:33
1

It may not have answered the OP's question, but it answered everybody else's, including mine. +1 – bertmoog Mar 04 '19 at 23:29

score 108 · Accepted Answer · edited Nov 09 '18 at 13:52

108

try

long diffSeconds = diff / 1000 % 60;  
long diffMinutes = diff / (60 * 1000) % 60; 
long diffHours = diff / (60 * 60 * 1000);

NOTE: this assumes that diff is non-negative.

edited Nov 09 '18 at 13:52

vels4j

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answered Mar 18 '11 at 11:46

Peter Lawrey

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@vels4j I was assuming the third line from the original code which I have just added. Thank you for pointing that out. – Peter Lawrey Nov 03 '18 at 11:43
1

but still the `diffMinutes` is wrong – vels4j Nov 03 '18 at 12:10
1

@vels4j in what way? – Peter Lawrey Nov 03 '18 at 15:32
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I've updated the answer, pl check it. discard my edit after clarification – vels4j Nov 08 '18 at 06:05
1

@vels4j which one are you saying is wrong because 6 appears to be the right answer as you also have hours in the OPs question. – Peter Lawrey Nov 08 '18 at 09:47
ok, misunderstood with below answer. – vels4j Nov 09 '18 at 13:53

score 46 · Answer 3 · edited Sep 18 '16 at 22:05

46

If you are able to use external libraries I would recommend you to use Joda-Time, noting that:

Joda-Time is the de facto standard date and time library for Java prior to Java SE 8. Users are now asked to migrate to java.time (JSR-310).

Example for between calculation:

Seconds.between(startDate, endDate);
Days.between(startDate, endDate);

edited Sep 18 '16 at 22:05

Dave Jarvis

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answered Mar 18 '11 at 11:51

heldt

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Please note that if you are dealing with 2 java.util.Date objects, then you'd need to use Days.daysBetween(LocalDate.fromDateFields(startDate), LocalDate.fromDateFields(endDate)); – user799188 Oct 09 '12 at 03:00
1

I've created an example how to use Joda time: http://stackoverflow.com/a/15541322/562769 – Martin Thoma Mar 21 '13 at 07:53
4

What is the current way to do it? (with JSR-310) – static_rtti Mar 29 '18 at 10:27

score 19 · Answer 4 · edited Apr 14 '16 at 13:57

Try this for a friendly representation of time differences (in milliseconds):

String friendlyTimeDiff(long timeDifferenceMilliseconds) {
    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 = timeDifferenceMilliseconds / ((long)60 * 60 * 1000 * 24 * 365);

    if (diffSeconds < 1) {
        return "less than a second";
    } else if (diffMinutes < 1) {
        return diffSeconds + " seconds";
    } else if (diffHours < 1) {
        return diffMinutes + " minutes";
    } else if (diffDays < 1) {
        return diffHours + " hours";
    } else if (diffWeeks < 1) {
        return diffDays + " days";
    } else if (diffMonths < 1) {
        return diffWeeks + " weeks";
    } else if (diffYears < 1) {
        return diffMonths + " months";
    } else {
        return diffYears + " years";
    }
}

on line long diffYears = (long) (timeDifferenceMilliseconds / (60 * 60 * 1000 * 24 * 365)); you need to cast long little later, otherwise the divisor is int and overflowing :diffYears = (timeDifferenceMilliseconds / ((long)60 * 60 * 1000 * 24 * 365)); — Pavel Niedoba, Apr 14 '16 at 13:38

Yves Martin · Answer 5 · 2016-05-24T15:13:44.500

19

Since Java 5, you can use java.util.concurrent.TimeUnit to avoid the use of Magic Numbers like 1000 and 60 in your code.

By the way, you should take care to leap seconds in your computation: the last minute of a year may have an additional leap second so it indeed lasts 61 seconds instead of expected 60 seconds. The ISO specification even plan for possibly 61 seconds. You can find detail in java.util.Date javadoc.

edited May 24 '16 at 15:13

answered Feb 07 '12 at 13:40

Yves Martin

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Unless there's a good business justification for including those stray leap seconds, you're pretty safe treating them as a fun, but immaterial, scientific curiosity. – phatfingers Apr 16 '13 at 02:44
1

I agree the "leap second" is a minor trick. But what about day-light saving or time zone differences ? – Yves Martin May 03 '13 at 10:28
1

@YvesMartin "time zone differences" Look at his date format. There is no place for time zone. If he did want this to work across timezones he'd have bigger problems. Daylight savings is an issue, though. – Parthian Shot Aug 08 '14 at 17:29
2

«it indeed lasts 60 seconds instead of expected 59 seconds» You mean it lasts 61 seconds (`0..60`) instead of the expected 60 seconds (`0..59`). ;) – Andrea Lazzarotto May 24 '16 at 11:45
1

Fixed my comment... a classical picket / spacing issue ! Thanks for pointing it – Yves Martin May 24 '16 at 15:15

score 11 · Answer 6 · answered Mar 13 '16 at 18:02

Here is a suggestion, using TimeUnit, to obtain each time part and format them.

private static String formatDuration(long duration) {
    long hours = TimeUnit.MILLISECONDS.toHours(duration);
    long minutes = TimeUnit.MILLISECONDS.toMinutes(duration) % 60;
    long seconds = TimeUnit.MILLISECONDS.toSeconds(duration) % 60;
    long milliseconds = duration % 1000;
    return String.format("%02d:%02d:%02d,%03d", hours, minutes, seconds, milliseconds);
}

SimpleDateFormat sdf = new SimpleDateFormat("HH:mm:ss,SSS");
Date startTime = sdf.parse("01:00:22,427");
Date now = sdf.parse("02:06:38,355");
long duration = now.getTime() - startTime.getTime();
System.out.println(formatDuration(duration));

The result is: 01:06:15,928

score 8 · Answer 7 · answered Mar 18 '11 at 12:07

This is more of a maths problem than a java problem basically.

The result you receive is correct. This because 225 seconds is 3 minutes (when doing an integral division). What you want is the this:

divide by 1000 to get the number of seconds -> rest is milliseconds
divide that by 60 to get number of minutes -> rest are seconds
divide that by 60 to get number of hours -> rest are minutes

or in java:

int millis = diff % 1000;
diff/=1000;
int seconds = diff % 60;
diff/=60;
int minutes = diff % 60;
diff/=60;
hours = diff;

score 7 · Answer 8 · answered Mar 01 '16 at 00:02

I know this is an old question, but I ended up doing something slightly different from the accepted answer. People talk about the TimeUnit class, but there were no answers using this in the way OP wanted it.

So here's another solution, should someone come by missing it ;-)

public class DateTesting {
    public static void main(String[] args) {
        String dateStart = "11/03/14 09:29:58";
        String dateStop = "11/03/14 09:33:43";

        // Custom date format
        SimpleDateFormat format = new SimpleDateFormat("yy/MM/dd HH:mm:ss");  

        Date d1 = null;
        Date d2 = null;
        try {
            d1 = format.parse(dateStart);
            d2 = format.parse(dateStop);
        } catch (ParseException e) {
            e.printStackTrace();
        }    

        // Get msec from each, and subtract.
        long diff = d2.getTime() - d1.getTime();

        long days = TimeUnit.MILLISECONDS.toDays(diff);
        long remainingHoursInMillis = diff - TimeUnit.DAYS.toMillis(days);
        long hours = TimeUnit.MILLISECONDS.toHours(remainingHoursInMillis);
        long remainingMinutesInMillis = remainingHoursInMillis - TimeUnit.HOURS.toMillis(hours);
        long minutes = TimeUnit.MILLISECONDS.toMinutes(remainingMinutesInMillis);
        long remainingSecondsInMillis = remainingMinutesInMillis - TimeUnit.MINUTES.toMillis(minutes);
        long seconds = TimeUnit.MILLISECONDS.toSeconds(remainingSecondsInMillis);

        System.out.println("Days: " + days + ", hours: " + hours + ", minutes: " + minutes + ", seconds: " + seconds);
    }
}

Although just calculating the difference yourself can be done, it's not very meaningful to do it like that and I think TimeUnit is a highly overlooked class.

Thanks! I thought in the same way, but you speeded up it :) – Akbolat SSS Nov 23 '19 at 16:56 — Akbolat SSS, Nov 23 '19 at 16:56

bobbyrne01 · Answer 9 · 2013-04-16T00:32:24.330

5

Create a Date object using the diffence between your times as a constructor,
then use Calendar methods to get values ..

Date diff = new Date(d2.getTime() - d1.getTime());

Calendar calendar = Calendar.getInstance();
calendar.setTime(diff);
int hours = calendar.get(Calendar.HOUR_OF_DAY);
int minutes = calendar.get(Calendar.MINUTE);
int seconds = calendar.get(Calendar.SECOND);

edited Apr 16 '13 at 00:32

answered Jan 08 '13 at 23:07

bobbyrne01

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Those methods are deprecated. – djechlin Feb 14 '13 at 23:15
2

works fine, but you must not forgett to set TimeZone: `Calendar calendar = Calendar.getInstance(TimeZone.getTimeZone("UTC"));` – amra Jul 11 '13 at 13:19

score 4 · Answer 10 · answered Mar 18 '11 at 12:00

difference-between-two-dates-in-java

Extracted the code from the link

public class TimeDiff {
    /**
     * (For testing purposes)
     *
     */
    public static void main(String[] args) {
        Date d1 = new Date();
        try { Thread.sleep(750); } catch(InterruptedException e) { /* ignore */ }      
        Date d0 = new Date(System.currentTimeMillis() - (1000*60*60*24*3)); // About 3 days ago
        long[] diff = TimeDiff.getTimeDifference(d0, d1);

        System.out.printf("Time difference is %d day(s), %d hour(s), %d minute(s), %d second(s) and %d millisecond(s)\n",
                diff[0], diff[1], diff[2], diff[3], diff[4]);
        System.out.printf("Just the number of days = %d\n",
                TimeDiff.getTimeDifference(d0, d1, TimeDiff.TimeField.DAY));
    }

    /**
     * Calculate the absolute difference between two Date without
     * regard for time offsets
     *
     * @param d1 Date one
     * @param d2 Date two
     * @param field The field we're interested in out of
     * day, hour, minute, second, millisecond
     *
     * @return The value of the required field
     */
    public static long getTimeDifference(Date d1, Date d2, TimeField field) {
        return TimeDiff.getTimeDifference(d1, d2)[field.ordinal()];
    }

    /**
     * Calculate the absolute difference between two Date without
     * regard for time offsets
     *
     * @param d1 Date one
     * @param d2 Date two
     * @return The fields day, hour, minute, second and millisecond
     */
    public static long[] getTimeDifference(Date d1, Date d2) {
        long[] result = new long[5];
        Calendar cal = Calendar.getInstance();
        cal.setTimeZone(TimeZone.getTimeZone("UTC"));
        cal.setTime(d1);

        long t1 = cal.getTimeInMillis();
        cal.setTime(d2);

        long diff = Math.abs(cal.getTimeInMillis() - t1);
        final int ONE_DAY = 1000 * 60 * 60 * 24;
        final int ONE_HOUR = ONE_DAY / 24;
        final int ONE_MINUTE = ONE_HOUR / 60;
        final int ONE_SECOND = ONE_MINUTE / 60;

        long d = diff / ONE_DAY;
        diff %= ONE_DAY;

        long h = diff / ONE_HOUR;
        diff %= ONE_HOUR;

        long m = diff / ONE_MINUTE;
        diff %= ONE_MINUTE;

        long s = diff / ONE_SECOND;
        long ms = diff % ONE_SECOND;
        result[0] = d;
        result[1] = h;
        result[2] = m;
        result[3] = s;
        result[4] = ms;

        return result;
    }

    public static void printDiffs(long[] diffs) {
        System.out.printf("Days:         %3d\n", diffs[0]);
        System.out.printf("Hours:        %3d\n", diffs[1]);
        System.out.printf("Minutes:      %3d\n", diffs[2]);
        System.out.printf("Seconds:      %3d\n", diffs[3]);
        System.out.printf("Milliseconds: %3d\n", diffs[4]);
    }

    public static enum TimeField {DAY,
        HOUR,
        MINUTE,
        SECOND,
        MILLISECOND;
    }
}

Good. Time offset is important - even if difference is computed based on the same offset, but also any day-light saving difference between two "calendar date" has also an impact on result. You should have used java.util.concurrent.TimeUnit for your constants. — Yves Martin, May 03 '13 at 10:26

score 3 · Answer 11 · answered Dec 28 '13 at 05:44

// d1, d2 are dates
long diff = d2.getTime() - d1.getTime();

long diffSeconds = diff / 1000 % 60;
long diffMinutes = diff / (60 * 1000) % 60;
long diffHours = diff / (60 * 60 * 1000) % 24;
long diffDays = diff / (24 * 60 * 60 * 1000);

System.out.print(diffDays + " days, ");
System.out.print(diffHours + " hours, ");
System.out.print(diffMinutes + " minutes, ");
System.out.print(diffSeconds + " seconds.");

Basil Bourque · Answer 12 · 2015-03-15T20:47:54.537

Joda-Time

Joda-Time 2.3 library offers already-debugged code for this chore.

Joad-Time includes three classes to represent a span of time: Period, Interval, and Duration. Period tracks a span as a number of months, days, hours, etc. (not tied to the timeline).

// © 2013 Basil Bourque. This source code may be used freely forever by anyone taking full responsibility for doing so.

// Specify a time zone rather than rely on default.
// Necessary to handle Daylight Saving Time (DST) and other anomalies.
DateTimeZone timeZone = DateTimeZone.forID( "America/Montreal" );

DateTimeFormatter formatter = DateTimeFormat.forPattern( "yy/MM/dd HH:mm:ss" ).withZone( timeZone ); 

DateTime dateTimeStart = formatter.parseDateTime( "11/03/14 09:29:58" );
DateTime dateTimeStop = formatter.parseDateTime( "11/03/14 09:33:43" );
Period period = new Period( dateTimeStart, dateTimeStop );

PeriodFormatter periodFormatter = PeriodFormat.getDefault();
String output = periodFormatter.print( period );

System.out.println( "output: " + output );

When run…

output: 3 minutes and 45 seconds

score 2 · Answer 13 · edited Apr 29 '13 at 19:43

Here is my code.

import java.util.Date;

// to calculate difference between two days
public class DateDifference {

// to calculate difference between two dates in milliseconds
public long getDateDiffInMsec(Date da, Date db) {
    long diffMSec = 0;
    diffMSec = db.getTime() - da.getTime();
    return diffMSec;
}

// to convert Milliseconds into DD HH:MM:SS format.
public String getDateFromMsec(long diffMSec) {
    int left = 0;
    int ss = 0;
    int mm = 0;
    int hh = 0;
    int dd = 0;
    left = (int) (diffMSec / 1000);
    ss = left % 60;
    left = (int) left / 60;
    if (left > 0) {
        mm = left % 60;
        left = (int) left / 60;
        if (left > 0) {
            hh = left % 24;
            left = (int) left / 24;
            if (left > 0) {
                dd = left;
            }
        }
    }
    String diff = Integer.toString(dd) + " " + Integer.toString(hh) + ":"
            + Integer.toString(mm) + ":" + Integer.toString(ss);
    return diff;

}
}

krems · Answer 14 · 2013-12-20T19:19:26.107

Well, I'll try yet another code sample:

/**
 * Calculates the number of FULL days between to dates
 * @param startDate must be before endDate
 * @param endDate must be after startDate
 * @return number of day between startDate and endDate
 */
public static int daysBetween(Calendar startDate, Calendar endDate) {
    long start = startDate.getTimeInMillis();
    long end = endDate.getTimeInMillis();
    // It's only approximation due to several bugs (@see java.util.Date) and different precision in Calendar chosen
    // by user (ex. day is time-quantum).
    int presumedDays = (int) TimeUnit.MILLISECONDS.toDays(end - start);
    startDate.add(Calendar.DAY_OF_MONTH, presumedDays);
    // if we still didn't reach endDate try it with the step of one day
    if (startDate.before(endDate)) {
        startDate.add(Calendar.DAY_OF_MONTH, 1);
        ++presumedDays;
    }
    // if we crossed endDate then we must go back, because the boundary day haven't completed yet
    if (startDate.after(endDate)) {
        --presumedDays;
    }
    return presumedDays;
}

MD. Mohiuddin Ahmed · Answer 15 · 2014-08-10T19:04:11.847

Date startTime = new Date();
//...
//... lengthy jobs
//...
Date endTime = new Date();
long diff = endTime.getTime() - startTime.getTime();
String hrDateText = DurationFormatUtils.formatDuration(diff, "d 'day(s)' H 'hour(s)' m 'minute(s)' s 'second(s)' ");
System.out.println("Duration : " + hrDateText);

You can use Apache Commons Duration Format Utils. It formats like SimpleDateFormatter

Output:

0 days(s) 0 hour(s) 0 minute(s) 1 second(s)

score 1 · Answer 16 · edited Dec 01 '14 at 14:20

As said before - think this is a good answer

/**
 * @param d2 the later date 
 * @param d1 the earlier date
 * @param timeUnit - Example Calendar.HOUR_OF_DAY
 * @return
 */
public static int getTimeDifference(Date d2,Date d1, int timeUnit) {
     Date diff = new Date(d2.getTime() - d1.getTime());

     Calendar calendar = Calendar.getInstance();
     calendar.setTime(diff);
     int hours = calendar.get(Calendar.HOUR_OF_DAY);
     int minutes = calendar.get(Calendar.MINUTE);
     int seconds = calendar.get(Calendar.SECOND);
     if(timeUnit==Calendar.HOUR_OF_DAY)
         return hours;
     if(timeUnit==Calendar.MINUTE)
         return minutes;
     return seconds;
 }

this will fail with timeunit minute and time diff more than hour — Pavel Niedoba, Feb 26 '16 at 23:29

score 1 · Answer 17 · answered Mar 18 '11 at 11:54

1

long diffSeconds = (diff / 1000)%60;
try this and let me know if it works correctly...

answered Mar 18 '11 at 11:54

Akshatha

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Calculate date/time difference in java

17 Answers17

Joda-Time

Linked

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