import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
public class test {
/*
* Calculate the difference between two date/times *
*
*/
private static long dateDiff(Date toDate, Date fromDate) {
Calendar cal = Calendar.getInstance();
cal.setTime(toDate);
long ms = cal.getTimeInMillis();
cal.setTime(fromDate);
ms -= cal.getTimeInMillis();
return ms;
}
public static void main(String[] args) {
SimpleDateFormat sdf = new SimpleDateFormat("hh:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = sdf.parse("11:00:00");
d2 = sdf.parse("10:00:00");
} catch (Exception e) {
e.printStackTrace();
}
long result = dateDiff(d1, d2);
Date time = new Date(result);
System.out.println(time);
}
}
When I run it I get this result :
Thu Jan 01 02:00:00 CET 1970
I would expect 1 hour difference ?! again a problem with Timezone??
Any idea how I can fix it.
thx all
I don't know what you expect this to do, but what you are actually doing is outputting the date corresponding to one hour after midnight on Jan 1 1970, using the default timezone.
You seem to want to Date to represent a duration (i.e. a number of seconds). It doesn't do that, and neither will the Date formatters render a Date as a duration.
I need the time difference between two Date fields and then put it in MySql (time format)
For what you are trying to do, you need calculate the duration value as a long, then use the java.sql.Time(long) constructor to create a Time object. You can either serialize this object using its toString() method or use it as a parameter in a JDBC prepared statement.
It turns out that my advice above is incorrect too.
Your real problem is that the SQL Time type is for representing times ... not durations. In fact, SQL does not have a dedicated duration type, so the best you can do is represent the duration as an integer number of seconds or milliseconds or whatever.
(For the more general case, the Joda Time libraries are generally thought to provide the best APIs for manipulating dates, times and related temporal values. But for this simple case, the standard J2SE libraries should suffice ... provided that you use them correctly.)
The problem is that a difference of two Date types can not be represented by another Date type.
Why don't you just take the milliseconds of both dates and substract them from each other?
Calendar cal = Calendar.getInstance();
cal.setTime(d1);
long d1ms = cal.getTimeInMillis();
cal.setTime(d2);
long d2ms = cal.getTimeInMillis();
long diffMs = d1ms - d2ms;
long diffHour = diffMs * 1000 * 60 * 60;
Hi try this setting timezone to GMT. Remove day, month in words in the resultant time difference. This method does nothing but assumes these many milliseconds since starting of time counter in java, which is 1st Jan 1970. So if your result says 3rd Jan 1970 means 3 days have passed since time counter started, which is perfect. You just need to interpret it properly, but formatting your answer
...
long result = dateDiff(d1, d2); //This is your code in main([])
SimpleDateFormat dateFormat = new SimpleDateFormat("HH:mm:ss SSS");
dateFormat.setTimeZone(TimeZone.getTimeZone("GMT"));
System.out.println(dateFormat.format(new Date(result )));
Related
This question already has answers here:
Calculate number of weekdays between two dates in Java
(20 answers)
Closed 1 year ago.
Am very beginner and new to Java platform. I have the below 3 simple Java date difference calculation functions. I wanted to exclude weekends on the below calculations in all the 3 methods. Can anyone please help how to exclude weekends for the below dateDiff calculations?
public static String getDatesDiff(String date1, String date2) {
String timeDiff = "";
SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(date1);
d2 = format.parse(date2);
long diff = d2.getTime() - d1.getTime();
timeDiff = ""+diff;
} catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}
return timeDiff;
}
public static String getDatesDiffAsDays(String date1, String date2) {
String timeDiff = "";
SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(date1);
d2 = format.parse(date2);
long diff = d2.getTime() - d1.getTime();
String days = ""+(diff / (24 * 60 * 60 * 1000));
timeDiff = days;
timeDiff = timeDiff.replaceAll("-", "");
timeDiff = timeDiff+" days";
} catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}
return timeDiff;
}
public static String getDatesDiffAsDate(String date1, String date2) {
String timeDiff = "";
SimpleDateFormat format = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
Date d1 = null;
Date d2 = null;
try {
d1 = format.parse(date1);
d2 = format.parse(date2);
long diff = d2.getTime() - d1.getTime();
String days = (diff / (24 * 60 * 60 * 1000))+" days";
String hours = (diff / (60 * 60 * 1000) % 24)+"h";
String minutes = (diff / 1000 % 60)+"mts";
String seconds = (diff / (60 * 1000) % 60)+"sec";
timeDiff = days;
timeDiff = timeDiff.replaceAll("-", "");
} catch (Exception e) {
// TODO: handle exception
e.printStackTrace();
}
return timeDiff;
}
This code is fundamentally broken. java.util.Date doesn't represent a date, it represents a timestamp. But if you're working with moments in time, you have a problem: not all days are exactly 24 hours long. For example, daylight savings exists, making some days 25 or 23 hours. At specific moments in time in specific places on the planet, entire days were skipped, such as when a place switches which side of the international date line it is on, or when Russia was the last to switch from Julian to Gregorian (the famed October Revolution? Yeah, that happened in November actually!)
Use LocalDate which represents an actual date, not a timestamp. Do not use Date, or SimpleDateFormat – these are outdated and mostly broken takes on dates and times. The java.time package is properly thought through.
When is 'the weekend'? In some places, Friday and Saturday are considered the weekend, not Saturday and Sunday.
If you're excluding weekends, presumably you'd also want to exclude mandated holidays. Many countries state that Jan 1st, regardless of what day that is, counts as a Sunday, e.g. for the purposes of government buildings and services being open or not.
Lessons you need to take away from this:
Dates are incredibly complicated, and as a consequence, are a horrible idea for teaching basic principles.
Do not use java.util.Date, Calendar, GregorianCalendar, or SimpleDateFormat, ever. Use the stuff in java.time instead.
If you're writing math like this, you're probably doing it wrong – e.g. ChronoUnit.DAYS.between(date1, date2) does all that math for you.
You should probably just start at start date, and start looping: check if that date counts as a working day or not (and if it is, increment a counter), then go to the next day. Keep going until the day is equal to the end date, and then return that counter. Yes, this is 'slow', but a computer will happily knock through 2 million days (that covers over 5000 years worth) in a heartbeat for you. The advantage is that you can calculate whether or not a day counts as a 'working day' (which can get incredibly complicated. For example, most mainland European countries and I think the US too mandates that Easter is a public holiday. Go look up and how to know when Easter is. Make some coffee first, though).
If you really insist on going formulaic and defining weekends as Saturday and Sunday, it's better to separately calculate how many full weeks are between the two dates and multiply that by 5, and then add separately the half-week 'on the front of the range' and the half-week at the back. This will be fast even if you ask for a hypothetical range of a million years.
That is not how you handle exceptions. Add throws X if you don't want to deal with it right now, or, put throw new RuntimeException("unhandled", e); in your catch blocks. Not this, this is horrible. It logs half of the error and does blindly keeps going, with invalid state.
Almost all interesting questions, such as 'is this date a holiday?' are not answerable without knowing which culture/locale you're in. This includes seemingly obvious constants such as 'is Saturday a weekend day?'.
rzwitserloot has already brought up many valid points about problems in your code.
This is an example of how you could count the working days:
LocalDate startDate = ...;
LocalDate endDateExclusive = ...;
long days = startDate.datesUntil(endDateExclusive)
.filter(date -> isWorkingDay(date))
.count();
And, of course, you need to implement the isWorkingDay method. An example would be this:
public static boolean isWorkingDay(LocalDate date) {
DayOfWeek dow = date.getDayOfWeek();
return (dow != DayOfWeek.SATURDAY && dow != DayOfWeek.SUNDAY);
}
I used LocalDate to illustrate the example. LocalDate fits well if you are working with concepts like weekend days and holidays. However, if you want to also include the time component, then you should also take clock adjustments like DST into account; otherwise a "difference" does not make sense.
I assume the user to input an object representing some datetime value, not a String. The parsing of a string does not belong to this method, but should be handled elsewhere.
Already been said, but I repeat: don't use Date, Calendar and SimpleDateFormat. They're troublesome. Here are some reasons why.
If you want to take the time into consideration, it'll get a little more complex. For instance, ChronoUnit.DAYS.between(date1, date2) only supports a single, contiguous timespan. Gaps in the timespan, like excluding certain periods of time, is not. Then you have to walk over each date and get the associated duration of that portion of date.
First, we could create a LocalTimeRange class, which represents a time span at a certain day.
public record LocalTimeRange(LocalTime start, LocalTime endExclusive) {
public static final LocalTimeRange EMPTY = new LocalTimeRange(null, null);
public Duration toDuration(LocalDate date, ZoneId zone) {
if (this.equals(EMPTY)) {
return Duration.ZERO;
}
var s = ZonedDateTime.of(date, Objects.requireNonNullElse(start, LocalTime.MIN), zone);
var e = (endExclusive != null ? ZonedDateTime.of(date, endExclusive, zone) : ZonedDateTime.of(date.plusDays(1), LocalTime.MIN, zone));
return Duration.between(s, e);
}
}
Calculations are not done immediately, because the duration in between the two wall clock times, depends on the date and timezone. The toDuration method calculates this.
Then we'll create a method which defines what times on each day are counted as a non-weekend day. In this example, I have defined a weekend to be from Friday, 12:00 (noon) until Sunday, 23:59 (midnight).
private static Duration nonWeekendHours(LocalDate date, ZoneId zone) {
var result = switch (date.getDayOfWeek()) {
case MONDAY,
TUESDAY,
WEDNESDAY,
THURSDAY -> new LocalTimeRange(LocalTime.MIDNIGHT, null);
case FRIDAY -> new LocalTimeRange(LocalTime.MIDNIGHT, LocalTime.NOON);
case SATURDAY,
SUNDAY -> new LocalTimeRange(null, null);
};
return result.toDuration(date, zone);
}
The LocalTimeRange::toDuration method is called with the passed LocalDate and ZoneId arguments.
Note that passing null as LocalTimeRange's second argument means 'until the end of the day'.
At last we could stream over all dates of a certain period and calculate how much time are the non-weekend hours for each day, and then reduce them to get the total amount of time:
LocalDate startDate = ...;
LocalDate endDate = ...;
ZoneId zone = ...;
Duration result = startDate.datesUntil(endDate)
.map(date -> nonWeekendHours(date, zone))
.reduce(Duration.ZERO, Duration::plus);
With the retrieved Duration instance, you can easily get the time parts with the get<unit>Part() methods,
Online demo
I have a map of string values which represent down times for different components.
dependencyMap.put ("sut", "14:26:12,14:27:19,00:01:07;15:01:54,15:02:54,00:01:00;15:44:30,15:46:30,00:02:00;16:10:30,16:11:30,00:01:00");
dependencyMap.put ("jms", "14:26:12,14:28:12,00:02:00;15:10:50,15:12:55,00:02:05;15:42:30,15:43:30,00:01:00;16:25:30,16:27:30,00:02:00");
The strings represent the start, end and duration of down times.
(start)14:26:12,(end)14:27:19,(duration)00:01:07
I read the values in, then add them to a list of DependencyDownTime objects which hold the Long values startTime, endTime and duration.
jArray.forEach (dependency ->{
String downTimeValues = knownDowntimesMap.get(dependency);
final String[] downtime = downTimeValues.split (";");
for (final String str : downtime) {
final DependencyDownTime depDownTime = new DependencyDownTime ();
final String[] strings = str.split (",");
if (strings.length == 3) {
final DateFormat dateFormat = new SimpleDateFormat ("HH:mm:ss");
try {
depDownTime.setStartTime(dateFormat.parse (strings[0]).getTime ());
depDownTime.setEndTime (dateFormat.parse (strings[1]).getTime ());
depDownTime.setDuration (dateFormat.parse (strings[2]).getTime ());
downTimes.add (depDownTime);
} catch (final ParseException e) {
//logger.warn (e.getMessage (), e);
}
} else {
//logger.warn ("");
}
}
I then perform simple arithmetic on the values, which calculates the total down time for each component.
// sort the list by start time
Collections.sort(downTimes, Comparator.comparing (DependencyDownTime::getStartTime));
int i = 1;
Long duration = 0L;
for(DependencyDownTime dts: downTimes){
Long curStart = dts.getStartTime ();
Long curEnd = dts.getEndTime();
Long nextStart = downTimes.get(i).getStartTime ();
Long nextEnd = downTimes.get(i).getEndTime ();
if(duration == 0){
duration = dts.getDuration();
}
if(curStart.equals(nextStart) && curEnd < nextEnd){
duration += (nextEnd - curEnd);
}
else if(nextStart > curEnd){
duration += downTimes.get(i).getDuration();
}
else if( curStart < nextStart && curEnd > nextStart){
duration += (nextEnd - curEnd);
}
else if(curEnd == nextStart){
duration += downTimes.get(i).getDuration();
}
i++;
if(i == downTimes.size ()){
componentDTimeMap.put (application, duration);
return;
}
The expected values should be something like 1970-01-01T 00:14:35 .000+0100, a matter of minutes. The actual result is usually extremely high off by a matter of hours in the difference 1969-12-31T 15:13:35 .000+0100
I have 2 questions.
Am I parsing the values correctly?
If my calculations are a little off when adding and subtracting the long values. When I convert the values back to Date format will there be a drastic difference in the expected value?
As explained in your other question, don't mistake those 2 different concepts:
a time of the day: it represents a specific point of a day, such as 10 AM or 14:45:50
a duration: it represents an amount of time, such as "1 hour and 10 minutes" or "2 years, 3 months and 4 days". The duration doesn't tell you when it starts or ends ("1 hour and 10 minutes" relative to what?), it's not attached to a chronology, it doesn't correspond to a specific point in the timeline. It's just the amount of time, by itself.
In your input, you have:
(start)14:26:12,(end)14:27:19,(duration)00:01:07
The start and end represents times of the day, and the duration represents the amount of time. SimpleDateFormat is designed to work with dates and times of the day, but not with durations. Treating the duration as a time of the day might work, but it's a hack as explained in this answer.
Another problem is that when SimpleDateFormat parses only a time, it defaults the day to January 1st 1970 at the JVM default timezone, leading to all the strange results you see. Unfortunately there's no way to avoid that, as java.util.Date works with full timestamps. A better alternative is to use the new date/time API.
As in your other question you're using Java 8, I'm assuming you can also use it here (but if you're using Java <= 7, you can use the ThreeTen Backport, a great backport for Java 8's new date/time classes. The only difference is the package names (in Java 8 is java.time and in ThreeTen Backport (or Android's ThreeTenABP) is org.threeten.bp), but the classes and methods names are the same).
As you're working only with times, there's no need to consider date fields (day/month/year), we can use a LocalTime instead. You can parse the strings directly, because they are in ISO861 compliant format:
LocalTime start = LocalTime.parse("14:26:12");
LocalTime end = LocalTime.parse("14:27:19");
Unfortunately there are no built-in parsers for a duration, so you'll have to parse it manually:
// parse the duration manually
String[] parts = "00:01:07".split(":");
Duration d = Duration
// get hours
.ofHours(Long.parseLong(parts[0]))
// plus minutes
.plusMinutes(Long.parseLong(parts[1]))
// plus seconds
.plusSeconds(Long.parseLong(parts[2]));
Another alternative is to remove the durations from your input (or ignore them) and calculate it using the start and end:
Duration d = Duration.between(start, end);
Both will give you a duration of 1 minute and 7 seconds.
My suggestion is to change the DependencyDownTime to store start and end as LocalTime objects, and the duration as a Duration object. With this, your algorithm would be like this:
Duration total = Duration.ZERO;
for (...) {
LocalTime curStart = ...
LocalTime curEnd = ...
LocalTime nextStart = ...
LocalTime nextEnd = ...
if (total.toMillis() == 0) {
duration = dts.getDuration();
}
if (curStart.equals(nextStart) && curEnd.isBefore(nextEnd)) {
total = total.plus(Duration.between(curEnd, nextEnd));
} else if (nextStart.isAfter(curEnd)) {
total = total.plus(downTimes.get(i).getDuration());
} else if (curStart.isBefore(nextStart) && curEnd.isAfter(nextStart)) {
total = total.plus(Duration.between(curEnd, nextEnd));
} else if (curEnd.equals(nextStart)) {
total = total.plus(downTimes.get(i).getDuration());
}
i++;
if (i == downTimes.size()) {
// assuming you want the duration as a total of milliseconds
componentDTimeMap.put(application, total.toMillis());
return;
}
}
You can either store the Duration object, or the respective value of milliseconds. Don't try to transform it to a Date, because a date is not designed nor supposed to work with durations. You can adapt this code to format a duration if you want (unfortunately there are no native formatters for durations).
Limitations
The code above assumes that all start and end times are in the same day. But if you have start at 23:50 and end at 00:10, should the duration be 20 minutes?
If that's the case, it's a little bit trickier, because LocalTime is not aware of the date (so it considers 23:50 > 00:10 and the duration between them is "minus 23 hours and 40 minutes").
In this case, you could do a trick and assume the dates are all at the current date, but when start is greater than end, it means that end time is in the next day:
LocalTime start = LocalTime.parse("23:50");
LocalTime end = LocalTime.parse("00:10");
// calculate duration
Duration d;
if (start.isAfter(end)) {
// start is after end, it means end is in the next day
// current date
LocalDate now = LocalDate.now();
// start is at the current day
LocalDateTime startDt = now.atTime(start);
// end is at the next day
LocalDateTime endDt = now.plusDays(1).atTime(end);
d = Duration.between(startDt, endDt);
} else {
// both start and end are in the same day
// just calculate the duration in the usual way
d = Duration.between(start, end);
}
In the code above, the result will be a Duration of 20 minutes.
Don't format dates as durations
Here are some examples of why SimpleDateFormat and Date aren't good to handle durations of time.
Suppose I have a duration of 10 seconds. If I try to transform it to a java.util.Date using the value 10 to a date (AKA treating a duration as a date):
// a 10 second duration (10000 milliseconds), treated as a date
Date date = new Date(10 * 1000);
System.out.println(date);
This will get a date that corresponds to "10000 milliseconds after unix epoch (1970-01-01T00:00Z)", which is 1970-01-01T00:00:10Z. But when I print the date object, the toString() method is implicity called (as explained here). And this method converts this millis value to the JVM default timezone.
In the JVM I'm using, the default timezone is America/Sao_Paulo, so the code above outputs:
Wed Dec 31 21:00:10 BRT 1969
Which is not what is expected: the UTC instant 1970-01-01T00:00:10Z corresponds to December 31st 1969 at 9 PM in São Paulo timezone.
This happens because I'm erroneously treating the duration as a date (and the output will be different, depending on the default timezone configured in the JVM).
A java.util.Date can't (must not) be used to work with durations. Actually, now that we have better API's, it should be avoided whenever possible. There are too many problems and design issues with this, just don't use it if you can.
SimpleDateFormat also won't work properly if you handle the durations as dates. In this code:
SimpleDateFormat dateFormat = new SimpleDateFormat("HH:mm:ss");
Date d = dateFormat.parse("10:00:00");
The input has only time fields (hour, minute and second), so SimpleDateFormat sets the date to January 1st 1970 at the JVM default timezone. If I System.out.println this date, the result will be:
Thu Jan 01 10:00:00 BRT 1970
That's January 1st 1970 at 10 AM in São Paulo timezone, which in UTC is equivalent to 1970-01-01T13:00:00Z - so d.getTime() returns 46800000.
If I change the JVM default timezone to Europe/London, it will create a date that corresponds to January 1st 1970 at 10 AM in London (or UTC 1970-01-01T09:00:00Z) - and d.getTime() now returns 32400000 (because 10 AM in London and 10 AM in São Paulo happened at different instants).
SimpleDateFormat isn't the right tool to work with durations - it isn't even the best tool to work with dates, actually.
This question already has answers here:
Get java.util.Calendar from days since epoch
(3 answers)
Closed 6 years ago.
Epoch time is the number of milliseconds that have passed since 1st January 1970, so if i want to add x days to that time, it seems natural to add milliseconds equivalent to x days to get the result
Date date = new Date();
System.out.println(date);
// Adding 30 days to current time
long longDate = date.getTime() + 30*24*60*60*1000;
System.out.println(new Date(longDate));
it gives the following output
Mon Dec 26 06:07:19 GMT 2016
Tue Dec 06 13:04:32 GMT 2016
I know i can use Calendar class to solve this issue, but just wanted to understand about this behaviour
Its Because JVM is treating Value of multiplication 30*24*60*1000 as Int And multiplication result is out of range of Integer it will give result : -1702967296 intends of 2592000000 so its giving date smaller then current date
Try Below code :
public class Test {
public static void main(final String[] args) {
Date date = new Date();
System.out.println(date);
// Adding 30 days to current time
System.out.println(30 * 24 * 60 * 60 * 1000); // it will print -1702967296
long longDate = (date.getTime() + TimeUnit.DAYS.toMillis(30));
System.out.println(TimeUnit.DAYS.toMillis(30));
date = new Date(longDate);
System.out.println(date);
}
}
You're hitting integer overflow with your number. 30*24*60*60*1000 = 2,592,000,000, which is bigger than a signed, 32-bit integer can hold (2,147,483,647).
Use longs instead, by appending L onto any of the numbers: 1000L, for instance.
Note that if you want to deal with daylight savings time (to say nothing of leap seconds!), this still won't be enough. But if you're willing to assume that a day is always exactly 24 hours, using longs will fix your problem. (Time is a complicated thing, and I would suggest using a library like joda or Java 8's classes to handle it for you!)
Edit your code as follows:
long longDate = date.getTime() + 30*24*60*60*1000L;
It will work for sure.
Try this:
DateTime timePlusTwoDays = new DateTime().plusDays(days);
long longDate = timePlusTwoDays.getTime();
System.out.println(new Date(longDate));
Dateime class has:
public DateTime plusDays(int days) {
if (days == 0) {
return this;
}
long instant = getChronology().days().add(getMillis(), days);
return withMillis(instant);
}
I want get the durations gap of two videos contain milliseconds.
A : 01:18:19.92, B : 01:18:19.57
public String calTime(String A, String B) {
String sFormat = "HH:mm:ss.SSS";
SimpleDateFormat dFormat = new SimpleDateFormat(sFormat);
try {
Date A1 = dFormat.parse(A);
Date A2 = dFormat.parse(B);
//A1 = Thu Jan 01 01:18:19 KST 1970
//A1.getTime() = -27700943
//A2 = Thu Jan 01 01:18:19 KST 1970
//A2.getTime() = -27701000
System.out.println(A1.getTime() - A2.getTime());
} catch (ParseException ex) {}
}
Why are the values from getTime() negative? Cause I didn't define yy-MM-dd the Date method?
Can I get and calculation durations like HH:mm:ss.SSS using Date method? I don't need year, month, day, etc.
how can I normal gap of two durations?
Why appear minus front of getTime() values? Cause I didn't define yy-MM-dd the Date method?
Because the dates are before Jan 1st 1970 at midnight GMT. You're clearly in a GMT+XX:XX timezone, so although they're after 1970 in that timezone, they aren't in GMT.
Cause I didn't define yy-MM-dd the Date method?
Yes, indirectly. SimpleDateFormat assumes midnight Jan 1st 1970 GMT if there is no date portion.
Can I get and calculation durations like HH:mm:ss.SSS using Date method?
Java's original Date object is not very useful. JDK 8 adds the java.time package with more useful classes in it, including Duration.
how can I normal gap of two durations?
That's what your code is currently doing. It's getting the interval (gap) in milliseconds. It's a positive number because you're doing A - B, and A is later than B. You can turn that number into a Duration via Duration.ofMillis if desired:
long interval = A1.getTime() - A2.getTime();
System.out.println("interval in ms: " + interval);
Duration d = Duration.ofMillis(interval);
System.out.println("duration: " + d);
Live Example
...though just for formatting it doesn't buy you much, I'm not immediately seeing anything in java.time.format that formats Durations or TemporalAmounts.
java.time.Duration
Since Java 8:
Why not using Duration.between(Temporal startInclusive, Temporal endExclusive)
Duration tempDuration = Duration.between(A1.toInstant(), A2.toInstant());
System.out.println(tempDuration.toMillis());
https://docs.oracle.com/javase/8/docs/api/java/time/Duration.html
The result can be negative if the first date is after the second.
Can also use LocalTime instead of date
I was looking thru java Date libraries
1. java.util.Date,
2. Date4J
3. Joda-time
to find out whether I can perform time subtraction to two Date Objects, to the precision of milliseconds.
I receive 2011-05-29T22:50:12.692 as a String, and convert it into a Date object by parsing it with SimpleDateFormat.
The other Date object will also be received a String. and I want to subtract from two Date objects.
Any ideas?
Basically i want to get an interval between the two Date objects, to the precision of milliseconds.
Try using Date.getTime()
long timeBetweenInMillis = d2.getTime() - d1.getTime();
EDIT
This assumes that d2 is chronologically after d1.
in java.util.Date there is a method called Date.getTime() that you can use to get the time in milies.
Do you mean like this?
Date d1 =
Date d2 =
long intervalInMillis = d1.getTime() - d2.getTime();
The internal representation of a Date() object is the number of milliseconds past the epoch. Just get those values with the getTime() method and do arithmetic on them. You can then construct a new Date object based on that value and you are done.
Just get the millisecond difference:
long millis = date1.getTime() - date2.getTime();
You could leverage XMLGregorianCalendar instead of SimpleDateFormat (Note SimpleDateFormat is not thread safe: http://www.codefutures.com/weblog/andygrove/2007/10/simpledateformat-and-thread-safety.html):
import javax.xml.datatype.DatatypeFactory;
import javax.xml.datatype.XMLGregorianCalendar;
public class Demo {
public static void main(String[] args) throws Exception {
DatatypeFactory df = DatatypeFactory.newInstance();
XMLGregorianCalendar date2 = df.newXMLGregorianCalendar("2011-05-29T22:50:12.692");
XMLGregorianCalendar date1 = df.newXMLGregorianCalendar("2011-03-29T22:50:12.692");
System.out.println(date2.toGregorianCalendar().getTimeInMillis() - date1.toGregorianCalendar().getTimeInMillis());
}
}
This is what you want:
Date result = new Date(d2.getTime() - d1.getTime());
Explanation:
d2.getTime() - d1.getTime()
Returns millisecond difference between the dates
new Date(milliseconds);
Creates a new Date object set to number of milliseconds since epoch.
Edit
IF you only want the positive difference, change it to this. This way it doesn't matter what order you put in inputs:
Date result = new Date(Math.abs(d2.getTime() - d1.getTime()));