I know this has been asked on here many times previously, but I'm haven't been able to find anything specific to my case. I'm trying to find the difference between the current datetime and a previous datetime, each with the format yyyy-MM-dd HH:mm:ss.s. Based on the answer given here, I've come up with the following code:

SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.s");
String earliestRunTime = "2017-12-16 01:30:08.0";
Date currentDate = new Date();
log.info("Current Date: {}", format.format(currentDate));

try {
    Date earliestDate = format.parse(earliestRunTime);

    long diff = currentDate.getTime() - earliestDate.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) % 30;
    long diffMonths = diff / (30 * 24 * 60 * 60 * 1000) % 12;
    long diffYears = diff / (12 * 30 * 24 * 60 * 60 * 1000);

    return String.format("%s years, %s months, %s days, %s hours, %s minutes, %s seconds",
            diffYears, diffMonths, diffDays, diffHours, diffMinutes, diffSeconds);


    }
catch (Exception e) {
    e.printStackTrace();
    return e.getMessage();
}

When I run the code, the JSON returns the following result:

lifetime: "41 years, -1 months, 14 days, 9 hours, 42 minutes, 37 seconds"

I have two questions here:

  1. Where am I going wrong in my calculations 41 years and a negative number?
  2. Is there a better way for me to do this? My current setup does not consider leap years or a 365 day year, and I need to take these into account.

3 Answers

0
KevinThePepper23 On Best Solutions

Using the same approach you did, you need to explicitly identify the denominator as long values. Currently, it assumes them to be integers, which causes a numeric overflow - meaning the value computed is too large for a integer. This would explain why you get negative/arbitrary values. Fix is simple:

SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss.s");
String earliestRunTime = "2017-12-16 01:30:08.0";
Date currentDate = new Date();
log.info("Current Date: {}" + format.format(currentDate));

try {
    Date earliestDate = format.parse(earliestRunTime);

    long diff = currentDate.getTime() - earliestDate.getTime();

    long diffSeconds = diff / 1000L % 60L;
    long diffMinutes = diff / (60L * 1000L) % 60L;
    long diffHours = diff / (60L * 60L * 1000L) % 24L;
    long diffDays = diff / (24L * 60L * 60L * 1000L) % 30L;
    long diffMonths = diff / (30L * 24L * 60L * 60L * 1000L) % 12L;
    long diffYears = diff / (12L * 30L * 24L * 60L * 60L * 1000L);

    return String.format("%s years, %s months, %s days, %s hours, %s minutes, %s seconds",
            diffYears, diffMonths, diffDays, diffHours, diffMinutes, diffSeconds);


}
catch (Exception e) {
    e.printStackTrace();
    return e.getMessage();
}
1
NeplatnyUdaj On

Where am I going wrong in my calculations 41 years and a negative number?

Because the denominator will overflow. You need to use Long:

long diffMonths = diff / (30 * 24 * 60 * 60 * 1000L) % 12; //Notice the L at the end
long diffYears = diff / (12 * 30 * 24 * 60 * 60 * 1000L); //Notice the L at the end

Also note that 12 * 30 is a really bad approximation of the number of days in a year.

Is there a better way for me to do this?

Yes. Use Duration api from Java 8. https://www.mkyong.com/java8/java-8-period-and-duration-examples/

It's hard to give precise answer, because the question is a bit vague. For example - If one of the year was a leap year and you were comparing dates 2020/03/28 and 2021/03/28, what should be the result? 1 year or 1 years, 1 days? (2020 is a leap year so after 03/28, there's also 03/29)

1
Ole V.V. On
  1. Where am I going wrong in my calculations 41 years and a negative number?

Apart from using the notoriously troublesome and long outdated SimpleDateFormat class and the just as outdated Date there are the following bugs in your code:

  • You are parsing 08.0 as 8 seconds 0 seconds. On my JDK-11 SimpleDateFormat opts for the 0 seconds and discards the 8 seconds that I think are correct. SimpleDateFormat cannot parse one decimal on the seconds (only exactly three decimals), so the solution to this bug is discarding SimpleDateFormat altogether.
  • As others have said you have an int overflow in your multiplications. For example, 30 * 24 * 60 * 60 * 1000 should give 2 592 000 000, but since an int cannot hold this number, you get -1 702 967 296 instead. Since this is a negative number, the following division gives you a negative number of months.
  • As Solomon Slow pointed out in a comment, a month may be 28, 29, 30 or 31 days. When setting all months to 30 days you risk incorrect numbers of days and months and in the end also years. When I ran your code today, the correct answer would have been 1 year, 4 months, 13 days, but I got 19 days instead, 6 days too much.
  • You are not taking summer time (DST) and other time anomalies into account. These may cause a day to be for example 23 or 25 hours, giving an error.

Or to sum up: Your error was that you tried to do the calculation “by hand”. Date and time math is too complex and error-prone to do this. You should always leave it to well-proven library classes instead.

  1. Is there a better way for me to do this? My current setup does not consider leap years or a 365 day year, and I need to take these into account.

Yes, there is a much better way. The best way may be to use the PeriodDuration class from the ThreeTen Extra project, see the link below. I am not going to install that library in my computer right now, so I will just show the good and modern solution using built-in classes:

    DateTimeFormatter dtf = DateTimeFormatter.ofPattern("uuuu-MM-dd HH:mm:ss.S");
    LocalDateTime currentDateTime = LocalDateTime.now(ZoneId.of("Australia/Sydney"));
    String earliestRunTime = "2017-12-16 01:30:08.0";
    LocalDateTime earliestDateTime = LocalDateTime.parse(earliestRunTime, dtf);

    // We want to find a period (years, months, days) and a duration (hours, minutes, seconds).
    // To do that we cut at the greatest possible whole number of days
    // and then measure the period before the cut and the duration after it.
    LocalDateTime cut = earliestDateTime.plusDays(
            ChronoUnit.DAYS.between(earliestDateTime, currentDateTime));

    Period p = Period.between(earliestDateTime.toLocalDate(), cut.toLocalDate());
    Duration d = Duration.between(cut, currentDateTime);

    String result = String.format("%s years, %s months, %s days, %s hours, %s minutes, %s seconds",
            p.getYears(), p.getMonths(), p.getDays(),
            d.toHours(), d.toMinutesPart(), d.toSecondsPart());

    System.out.println(result);

When I ran the code just now I got:

1 years, 4 months, 13 days, 19 hours, 26 minutes, 7 seconds

In java.time, the modern Java date and time API, a Period is an amount of years, months and days, and a Duration is an amount of hours, minutes, seconds and fraction of second (down to nanoseconds). Since you wanted both, I am using both classes.

The toXxxPart methods of Duration I am using were introduced in Java 9. If you are using Java 8 (or the ThreeTen Backport) printing the minutes and seconds is a little bit more complicated. Search for java format duration or similar to learn how.

I am still not taking summer time into account. To do that we would need to know the time zone of the earliest run time string and then use ZonedDateTime instead of LocalDateTime. The code would otherwise be very similar.

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