Develop Reference

How to check the number of fixed time segments between two dates?

Two Dates are given:
Let's say,
Date dt1 = 22 June 2013 8:00 PM
Date dt2 = 24 June 2013 6:00 AM
Given the two dates, I want to determine that how many segments from 1 am to 5 am are between these two dates.
For above, there are two segments:
23 June 1 am to 5am
24 June 1am to 5am
So the answer should be 2.
I can get the difference between the two times,
var time1 = new Date(dt1).getTime();
var time2 = new Date(dt2).getTime();
var diff = new Date(time1 - time2);
And the number of hours, min and seconds,
var hours = diff.getHours();
var minutes = diff.getMinutes();
var seconds = diff.getMinutes();
But this only gives difference as expected.
What approach is needed to do so ?

Like everything else in computer science: break the problem down into a series of smaller problems that you're able to solve.
For example, in this problem you might simply determine if there's at least one of your "segments" in the time span. If there is, you might remove the first 24 hours from the full time span, then repeat the process for as long as there exists 24 hours to remove. Remember to count along the way.

Another approach might be to check
if time1.getHours() before or equal to 1am then result=1
if time2.getHours() after 5am then resutl++
result+= diff.getDays() - 1

Determining if a time stamp is an exact hour

I need to determine if a time stamp is an exact hour (i.e. it represents a time with no minutes, seconds or milliseconds components) - using primitives only. The code is called several hundred times per second and I don't want the overhead of a Calendar or other object.
I've tried this but rounding errors cause the comparison to fail.
float hours = (time / 3600000f);
boolean isExactHour = hours == (int)hours;
For example, if time = 1373763600470, hours = 381601.03125. That time stamp represents 01:00:00:000 GMT today and hours should be 381601.
Any ideas for a quick and simple way to do this? Thanks!
[EDIT]
It seems that this is more complex than at first sight (when is it not? :)
For clarity, I don't care about time zones, nor leap seconds. The time stamp is generated by a method which returns the previous midnight - i.e. 13 July 2013 00:00:00:000 for today. I am then calculating, for any time stamp in an array of longs which is always this initial time stamp plus/minus an exact multiple of 5 minutes. My aim to to determine if a given stamp is "top of the hour". I might have edge cases where the multiple of 5 minutes overlaps a year end but I can live with those.

(time % 36000000) == 0
Surely this is obvious?
EDIT
To get accuracy w.r.t. leap seconds, assuming a lookup table of leap seconds indexed by year since (say) 1970, something like:
((time-leapSeconds[(int)(time/(1000*60*60*24*365.2425))-1970]*1000) % 3600000) == 0
Programmers needing this level of accuracy should also see all of Einar's comments below.

need to find every X minutes between to hours

I have two hours lets say 12:00 and 14:00 (the input will always be two hours of "whatever" day) second time will always be greater than first time. So the input is simple : two given times in a day. I need to find every 30 minutes between those two times.
I need the following output (the ellapsed time between each output will always be 30 minutes in my case:
12:00
12:30
13:00
13:30
14:00
14:30
I am discovering jodatime but I am a bit confused with how the determine "startTime" and "end Time"

It's pretty easy, actually:
DateTime current = start;
while(true){
System.out.println(current);
current=current.plusMinutes(30);
if(current.isAfter(stop))break;
}
Responding to comment:
to parse a String, you need a DateTimeFormatter. Here's one of several ways to acquire one:
private static final DateTimeFormatter DATE_TIME_FORMATTER =
DateTimeFormat.forPattern("HH:mm");
Now you can do:
DateTime startDate = DATE_TIME_FORMATTER.parseDateTime(startString);

Java date jump to the next 10th minute

I need a fast implementation of a "Date jump" algorithm that is to be included in an event management system.
An event is triggered and sets a date (in a synchronized method) to the next 10th minute.
For instance
Event occurs at "2010-01-05 13:10:12" and sets the
next date to be "2010-01-05 13:20:00"
and if an event occurs exactly (supposedly) at a 10th minute, the next one must be set
Event occurs at "2010-01-05 13:30:00" and sets the
next date to be "2010-01-05 13:40:00"
(unlikely since the date goes down to the 1/1000th of a second, but just in case...).
My first idea would be to get the current Date() and work directly with the ms from the getTime() method, via integer (long) division, like ((time / 10mn)+1)*10mn.
Since it has to be fast, and also reliable, I thought I'll ask my fellow OSers prior to the implementation.

You can use / adapt my answer to a very similar question:
How to round time to the nearest quarter in java?
Something like this:
int unroundedMinutes = calendar.get(Calendar.MINUTE);
int mod = unroundedMinutes % 10;
calendar.add(Calendar.MINUTE, mod == 0 ? 10 : 10 - mod);
calendar.set(Calendar.SECOND, 0);
calendar.set(Calendar.MILLISECOND, 0);

Your approach with the epoch time in ms will not work for arbitrary time zones, since some time zones have a N*15 minutes offset from GMT. Within these time zones, your code would move the interval 5-14 to 15, 15-24 to 25 and so on.
Have you actually tested the performance when manipulating the appropriate fields in the GregorianCalendar class and concluded that the performance is insufficient, or are you trying to reinvent a wheel just for the fun of it?

calculate frequency on certain range

I have maths problem ... (at the moment i solved it using manual iteration which is pretty slow) ...
For example if an employee got paid weekly (it can be fortnightly / every 2 weeks and monthly) with certain date (let's call the employee got paid every tuesday and for monthly the employee paid on certain date).
I have date range between 10th August 2009- 31 December 2009, now how to get frequency the employee got paid ?
is it possible to calculate this using jodatime ?
Example to make this question clear:
I have date range between Friday 14 August - Monday 14 Sept 2009 (31 days)
the employee got paid on every Tuesday
so he got paid on 18 & 25 August, 1 & 8 August we got 4 times payment
(frequency)
another example:
with the same date range Friday 14 August - Monday 14 Sept 2009 (31 days)
but different pay date .. for example on Sunday
so he got paid on : 15, 22 & 29 August , 5 & 12 September ... we got 5 times payment.
same date range but different pay day .. will result different.
So my question is, are there any formula to solve this case ?
at the moment I calculate using manual iterator .. which is very slow (because the range could be some years or months)
thank you
ps: I am using groovy .. any solutions using java or groovy or just algorithm are welcome :)

Oftentimes pay periods are on the 15th and the end of every month, so in that case you'd count the number of months and multiply by 2, checking the end conditions (if start is before the 15th, subtract one pay period; if end is after end of the month subtract one pay period).
It's possible to get counts of days, weeks, and months, but you'll have to add in the logic to handle the dodgy end conditions. It's probably not a simple formula, as the case I described demonstrates.

abosolutely, using the Weeks class is very simple:
DateTime start = new LocalDate(2009, 8, 10).toDateTimeAtStartOfDay();
DateTime end = new LocalDate(2009, 12, 31).toDateTimeAtStartOfDay();
int numberOfWeeks = Weeks.weeksBetween(start, end).getWeeks();
this code give 20 as result. It is right?
EDIT
maybe this is better:
DateMidnight start = new DateMidnight(2009, 8, 10);
DateMidnight end = new DateMidnight(2009, 12, 31);
int numberOfWeeks = Weeks.weeksBetween(start, end).getWeeks();
System.out.println(numberOfWeeks);

Subtracting one date from the other to get the "number of days" (or weeks) is generally the wrong way to go for these kinds of calculations. For example, if someone is 365 days old, they are exactly one year old, unless there was a February 29 during that time. In any (modern) 7-day period, there is always exactly one Tuesday; but for 8 days, it's either one or two. The calendar often figures into the calculations.
If they're paid once or twice a month, you do the easy calculation on the whole months -- starting on the first and ending on the last day of the month, which varies -- and then you have to consider partial months at the beginning and/or end. (Don't forget what happens if the 15th or last day of the month falls on a weekend.) If they're paid every one or two weeks, you can sync on a known payday, and then do the simpler math to figure the whole weeks before and/or since. (Don't forget holidays that fall on the payday.)

There are two tricks here: One is that the rules are different depending on the time frame. I mean, if a person is paid once a week, then in 7 days he gets paid once, in 14 days he gets paid twice, etc. But if a person is paid on the 1st and 16th of every month, I can't tell you how many times he was paid in 60 days without knowing what months were included: where they short months or long months?
The second is that you have to worry about the start and end of the time period. If a person is paid every Monday, then the number of times he gets paid in 8 days depends on whether the first day of the 8 is Monday.
Thus, I think you need to have different logic for schedules that are a fixed number of days and those that are tied to months or something else where the intervals can vary.
For the fixed number of days, the problem is fairly simple. The only complexity is if the time frame is not an exact multiple of the interval. So I'd say, find the first date in the interval on which a payday occurs. Then find the number of days between there and the end of the time period, divide by the interval and drop any fractions.
For example: A person is paid every Monday. How many pay days between March 1 and April 12? Find the first Monday in that range. Say it falls on March 4. Then calculate the number of days from March 4 to April 12. That would be 39. 39/7=5 and a fraction. Therefore he gets paid 5 more paychecks, for a total of 6.
For monthly pay, I think you'd have to separate out the first and last month. You could then count the number of months in the middle and multiply by the number of pays per month. Then for the first and last count how many are in them the hard way.

Just got solutions please check if I did something wrong
import org.joda.time.* ;
def start = new Date().parse("dd/MM/yy","14/08/2009");
def end = new Date().parse("dd/MM/yy","14/09/2009");
println("date range ${start} - ${end}");
def diff = end - start ;
println("diff : ${diff} days ");
println("how many weeks : ${diff/7}");
def payDay = 2 ; // Monday = 1 Sunday = 0
def startDay = new DateTime(start).dayOfWeek ; // 5 = Thursday
def startDayDiff = payDay - startDay ;
if(startDay > payDay){
startDayDiff = 7 + payDay - startDay ;
}
// for example if end on Friday (5) while Pay day is day 1 (Monday) then
// make sure end date is on Monday (same week )
// end date = end - ( endDay - payDay)
def endDay = new DateTime(end).dayOfWeek;
println("original end day: ${endDay}");
def endDayDiff = endDay - payDay ;
// otherwise ... if endDay < payDay (for example PayDay = Friday but End day is on Monday)
// end date = end - 7 + payDay
if(endDay < payDay){
endDayDiff = 7 - endDay - payDay ;
}
println("endDayDiff : ${endDayDiff}");
println("startDayDiff: ${startDayDiff}");
def startedOn = new DateTime(start).plusDays(startDayDiff);
println("started on : ${startedOn.toDate()}");
def endOn = new DateTime(end).minusDays(endDayDiff);
println("End on : ${endOn.toDate()}");
println("occurences : ${Weeks.weeksBetween(startedOn,endOn).getWeeks()+1}");
Tested using groovyConsole with Joda Time help .. :)