Develop Reference

LocalTime.MIDNIGHT vs. LocalTime.MIN - is there any difference?

I recently answered some questions using LocalDate.atStartOfDay() and LocalDate.atTime(LocalTime.MIN).
I was wondering why there is no LocalDate.atEndOfDay() or similar, so one has to use LocalDate.atTime(LocalTime.MAX) in order to get the very last moment (in nanos, I think) of that specific day.
I had a look at the source of LocalDate and LocalTime and got slightly confused by this:
/**
* Combines this date with the time of midnight to create a {#code LocalDateTime}
* at the start of this date.
* <p>
* This returns a {#code LocalDateTime} formed from this date at the time of
* midnight, 00:00, at the start of this date.
*
* #return the local date-time of midnight at the start of this date, not null
*/
public LocalDateTime atStartOfDay() {
return LocalDateTime.of(this, LocalTime.MIDNIGHT);
}
Contrary to my expectation, this method returns a LocalDateTime using LocalTime.MIDNIGHT instead of LocalTime.MIN.
Of course, I opened the OpenJDK source of LocalTime and was sure to find out the difference myself, but I found out there is no difference apart from the name of the constant:
/**
* Constants for the local time of each hour.
*/
private static final LocalTime[] HOURS = new LocalTime[24];
static {
for (int i = 0; i < HOURS.length; i++) {
HOURS[i] = new LocalTime(i, 0, 0, 0);
}
MIDNIGHT = HOURS[0]; // <--- == MIN
NOON = HOURS[12];
MIN = HOURS[0]; // <--- == MIDNIGHT
MAX = new LocalTime(23, 59, 59, 999_999_999);
}
While I totally understand the presence of NOON and MAX, I don't really get why there are MIN and MIDNIGHT when obviously one of them would be enough since they have the very same value.
Can anyone tell me the reason why...
... there are two constants having the very same value and
... why the code uses MIDNIGHT for the start of a day?
Is it just for having something more readable in some situations?
But why isn't MIN used in LocalTime.atStartOfDay() but rather LocalTime.MIDNIGHT?

MIN exists to provide the minimum value, which is consistent with other java.time.* classes.
MIDNIGHT exists to provide semantic meaning to developers, and as a place to indicate to Javadoc readers that midnight is considered to be at the start of the day (not the end).
Summary, the semantic benefits in code reading outweigh the cost of the extra constant.
(Source: I'm the main java.time.* author)

Java 8 calculate months between two dates

NOTE THIS IS NOT A DUPLICATE OF EITHER OF THE FOLLOWING
Calculating the difference between two Java date instances
calculate months between two dates in java [duplicate]
I have two dates:
Start date: "2016-08-31"
End date: "2016-11-30"
Its 91 days duration between the above two dates, I expected my code to return 3 months duration, but the below methods only returned 2 months. Does anyone have a better suggestion? Or do you guys think this is a bug in Java 8? 91 days the duration only return 2 months.
Thank you very much for the help.
Method 1:
Period diff = Period.between(LocalDate.parse("2016-08-31"),
LocalDate.parse("2016-11-30"));
Method 2:
long daysBetween = ChronoUnit.MONTHS.between(LocalDate.parse("2016-08-31"),
LocalDate.parse("2016-11-30"));
Method 3:
I tried to use Joda library instead of Java 8 APIs, it works. it loos will return 3, It looks like Java duration months calculation also used days value. But in my case, i cannot use the Joda at my project. So still looking for other solutions.
LocalDate dateBefore= LocalDate.parse("2016-08-31");
LocalDate dateAfter = LocalDate.parse("2016-11-30");
int months = Months.monthsBetween(dateBefore, dateAfter).getMonths();
System.out.println(months);

Since you don't care about the days in your case. You only want the number of month between two dates, use the documentation of the period to adapt the dates, it used the days as explain by Jacob. Simply set the days of both instance to the same value (the first day of the month)
Period diff = Period.between(
LocalDate.parse("2016-08-31").withDayOfMonth(1),
LocalDate.parse("2016-11-30").withDayOfMonth(1));
System.out.println(diff); //P3M
Same with the other solution :
long monthsBetween = ChronoUnit.MONTHS.between(
LocalDate.parse("2016-08-31").withDayOfMonth(1),
LocalDate.parse("2016-11-30").withDayOfMonth(1));
System.out.println(monthsBetween); //3
Edit from #Olivier Grégoire comment:
Instead of using a LocalDate and set the day to the first of the month, we can use YearMonth that doesn't use the unit of days.
long monthsBetween = ChronoUnit.MONTHS.between(
YearMonth.from(LocalDate.parse("2016-08-31")),
YearMonth.from(LocalDate.parse("2016-11-30"))
)
System.out.println(monthsBetween); //3

Since Java8:
ChronoUnit.MONTHS.between(startDate, endDate);

//Backward compatible with older Java
public static int monthsBetween(Date d1, Date d2){
if(d2==null || d1==null){
return -1;//Error
}
Calendar m_calendar=Calendar.getInstance();
m_calendar.setTime(d1);
int nMonth1=12*m_calendar.get(Calendar.YEAR)+m_calendar.get(Calendar.MONTH);
m_calendar.setTime(d2);
int nMonth2=12*m_calendar.get(Calendar.YEAR)+m_calendar.get(Calendar.MONTH);
return java.lang.Math.abs(nMonth2-nMonth1);
}

The documentation of Period#between states the following:
The start date is included, but the end date is not.
Furthermore:
A month is considered if the end day-of-month is greater than or equal to the start day-of-month.
Your end day-of-month 30 is not greater than or equal to your start day-of-month 31, so a third month is not considered.
Note the parameter names:
public static Period between​(LocalDate startDateInclusive, LocalDate endDateExclusive)
To return 3 months, you can increment the endDateExclusive by a single day.

In case you want stick to java.time.Period API
As per java.time.Period documentation
Period between(LocalDate startDateInclusive, LocalDate endDateExclusive)
where
#param startDateInclusive the start date, inclusive, not null
#param endDateExclusive the end date, exclusive, not null
So it is better to adjust your implementation to make your end date inclusive and get your desired result
Period diff = Period.between(LocalDate.parse("2016-08-31"),
LocalDate.parse("2016-11-30").plusDays(1));
System.out.println("Months : " + diff.getMonths());
//Output -> Months : 3

You have to be careful, never use LocalDateTime to calculate months between two dates the result is weird and incorrect, always use LocalDate !
here's is some code to prove the above:
package stack.time;
import java.time.LocalDate;
import java.time.LocalDateTime;
import java.time.temporal.ChronoUnit;
public class TestMonthsDateTime {
public static void main(String[] args) {
/**------------------Date Time----------------------------*/
LocalDateTime t1 = LocalDateTime.now();
LocalDateTime t2 = LocalDateTime.now().minusMonths(3);
long dateTimeDiff = ChronoUnit.MONTHS.between(t2, t1);
System.out.println("diff dateTime : " + dateTimeDiff); // diff dateTime : 2
/**-------------------------Date----------------------------*/
LocalDate t3 = LocalDate.now();
LocalDate t4 = LocalDate.now().minusMonths(3);
long dateDiff = ChronoUnit.MONTHS.between(t4, t3);
System.out.println("diff date : " + dateDiff); // diff date : 3
}
}
My 2%

This example checks to see if the second date is the end of that month. If it is the end of that month and if the first date of month is greater than the second month date it will know it will need to add 1
LocalDate date1 = LocalDate.parse("2016-08-31");
LocalDate date2 = LocalDate.parse("2016-11-30");
long monthsBetween = ChronoUnit.MONTHS.between(
date1,
date2);
if (date1.isBefore(date2)
&& date2.getDayOfMonth() == date2.lengthOfMonth()
&& date1.getDayOfMonth() > date2.getDayOfMonth()) {
monthsBetween += 1;
}

After the short investigation, still not totally fix my question, But I used a dirty solution to avoid return the incorrect duration. At least, we can get the reasonable duration months.
private static long durationMonths(LocalDate dateBefore, LocalDate dateAfter) {
System.out.println(dateBefore+" "+dateAfter);
if (dateBefore.getDayOfMonth() > 28) {
dateBefore = dateBefore.minusDays(5);
} else if (dateAfter.getDayOfMonth() > 28) {
dateAfter = dateAfter.minusDays(5);
}
return ChronoUnit.MONTHS.between(dateBefore, dateAfter);
}

The Java API response is mathematically accurate according to the calendar. But you need a similar mechanism, such as rounding decimals, to get the number of months between dates that matches the human perception of the approximate number of months between two dates.
Period period = Period.between(LocalDate.parse("2016-08-31"), LocalDate.parse("2016-11-30"));
long months = period.toTotalMonths();
if (period.getDays() >= 15) {
months++;
}

Strip seconds and milliseconds from System.currentTimeMillis value

Suppose I have a System.currentTimeMillis() value as a long number.
How do I modify it to match the instant when last minute started? I.e., zero out seconds and milliseconds.
I would prefer to not use magic constants. Using java.time is fine.

I agree with the answers recommending java.time, but it can be done yet simpler as in those answers:
long lastWholeMinute = Instant.now().truncatedTo(ChronoUnit.MINUTES).toEpochMilli();
This just gave 1517940060000. Of course, if it makes sense for you to keep the Instant object, by all means do that rather than converting to a naked primitive long.
If your long value was one you had stored rather than the time now, it’s quite similar:
long someEpochMilliValue = 1_517_941_234_567L;
long lastWholeMinute = Instant.ofEpochMilli(someEpochMilliValue)
.truncatedTo(ChronoUnit.MINUTES)
.toEpochMilli();

Using java.time is probably the easiest way. You could use withNano and withSecond, like
java.time.ZonedDateTime zdt = java.time.ZonedDateTime.now().withNano(0).withSecond(0);
long millis = zdt.toInstant().toEpochMilli();

Since the value is in milliseconds, if we assume an idealized day (no leap seconds, etc.), then given l you could do it by simply removing the value of l % 60000L from it. I realize that's a magic constant, but it's truly a constant, there are always going to be 60,000 milliseconds in a minute. I'd give it symbolic name:
private static long SIXTY_SECONDS_IN_MS = 60000L;
and not worry about it. Then it's:
long l = /*...your number...*/;
l = l - (l % SIXTY_SECONDS_IN_MS);
Why this works: The Epoch value is from midnight Jan 1st 1970, and so at 0L, 60000L, 120000L, etc., the seconds and milliseconds of an idealized day based on that value are 0. So we use the remainder operator (%) to isolate the part of the value that would remain if we divided by 60000L and remove it. Thus the resulting value, again assuming idealized days, has 0 for seconds and milliseconds. It also works across timezones if we assume all timezones are going to be at whole-minute offsets to UTC. I've only ever heard of timezones that were multiples of hours or half-hours offset from UTC ("GMT plus five hours", "GMT plus 5.5 hours"), never (say) "GMT plus five hours seven minutes and 20 seconds". (And indeed, the standard notation for timezome offsets, +0600 or similar, only includes hours and minutes, not fractional minutes.)
Live Example:
import java.time.*;
public class Example
{
private static long SIXTY_SECONDS_IN_MS = 60000L;
public static void main (String[] args) throws java.lang.Exception
{
long l = System.currentTimeMillis();
l = l - (l % SIXTY_SECONDS_IN_MS);
System.out.println("l = " + l);
// Checking the result
LocalDateTime dt = Instant.ofEpochMilli(l).atZone(ZoneId.systemDefault()).toLocalDateTime();
System.out.println(dt);
System.out.println(dt.getSecond()); // 0
System.out.println(dt.getNano()); // 0
}
}
Still, though, if that constant violates the terms of the question such that you think I shouldn't have answered, let me know and I'll delete the answer. :-)

How many instances of a partial date (month & day) appear in a range, and its day of the week

In Java, how would I go about constructing a utility that would take a range of dates (start and end date) and then would see how many times a given partial date ( the month and day-of-month) appears in that range, and will add an entry to a list for each match.
In my instance, I want to give it a range of say 5 years - starting Jan 1st 2014 and going to Dec 31st 2019. My check date is the 2nd August. I want the method to return the full information about each match of any August 2 of any year in the range. So for 2014 is will return Saturday 2nd August 2014, then Sunday 2nd August 2015 etc and so on.
I've been trying to get something working so far with Joda Time and the default date/calendar classes in Java and I'm just getting myself in a mess.
Thanks,
S
Edit: How silly of me, apologies for not adding my code :(
public static List<Date> getDaysInRange(Date startdate,
Date enddate,
Date checkDate) {
SimpleDateFormat sdf = new SimpleDateFormat("MMdd");
List<Date> dates = new ArrayList<>();
Calendar cal = new GregorianCalendar();
cal.setTime(startdate);
while (cal.getTime().before(enddate)) {
if (sdf.format(cal.getTime()).equals(sdf.format(checkDate))) {
Date result = cal.getTime();
dates.add(result);
}
cal.add(Calendar.DATE, 1);
}
return dates;
}

Date-Only
Since you want only a date without time-of-day and without time zone, use a date-only class. The old java.util.Date/.Calendar classes lack such a class. And those old classes are notoriously troublesome and flawed.
Instead use either:
Joda-Time
java.time, built into Java 8, inspired by Joda-Time.
Joda-Time
Here is some untested code using Joda-Time 2.6.
The main idea is to focus on the small set of possible year numbers rather than test every day of year. In the example below, that means six date-time values to compare rather than thousands. Besides efficiency, the purpose of the code becomes more apparent.
The arguments to your routine should be a month number and a day-of-month number, a pair of ints or Integers, rather than a Date. As seen in this examples two int variables, month and day.
LocalDate start = new LocalDate( 2011, 2, 3 );
LocalDate stop = new LocalDate( 2016, 4, 5 );
int yearStart = start.getYear();
int yearStop = stop.getYear();
int month = 11;
int day = 22;
for ( i = yearStart, i <= yearStop, i++ )
{
LocalDate x = new LocalDate( i, month, day );
boolean matchStart = ( x.isEqual( start ) || x.isAfter( start ) );
boolean matchStop = x.isBefore( stop ); // Half-Open approach where beginning of range is inclusive while ending is exclusive.
if ( matchStart && matchStop )
{
// Add to collection of LocalDate objects.
// Later you can ask each LocalDate object for its day-of-week.
{
}
java.time
The java.time package also offers a LocalDate class. The code would be similar to the above Joda-Time example.

I think using SimpleDateFormat is a bad idea. Use Calendar for comparison directly, like this
cal1.get(Calendar.MONTH) == cal2.get(Calendar.MONTH) && cal1.get(Calendar.DAY_OF_MONTH) == cal2.get(Calendar.DAY_OF_MONTH)

Check if two date periods overlap [duplicate]

This question already has answers here:
Determine Whether Two Date Ranges Overlap
(39 answers)
Closed 8 years ago.
I have two date ranges, (start1,end1):::>>date1 && (start2,end2):::>>date2 .
I want to check if the two dates isOverLaped.
My flow chart I assume "<>=" operators is valid for comparing.
boolean isOverLaped(Date start1,Date end1,Date start2,Date end2) {
if (start1>=end2 && end2>=start2 && start2>=end2) {
return false;
} else {
return true;
}
}
Any Suggestion will be appreciated.

You can use Joda-Time for this.
It provides the class Interval which specifies a start and end instants and can check for overlaps with overlaps(Interval).
Something like
DateTime now = DateTime.now();
DateTime start1 = now;
DateTime end1 = now.plusMinutes(1);
DateTime start2 = now.plusSeconds(50);
DateTime end2 = now.plusMinutes(2);
Interval interval = new Interval( start1, end1 );
Interval interval2 = new Interval( start2, end2 );
System.out.println( interval.overlaps( interval2 ) );
prints
true
since the end of the first interval falls between the start and end of the second interval.

boolean overlap(Date start1, Date end1, Date start2, Date end2){
return start1.getTime() <= end2.getTime() && start2.getTime() <= end1.getTime();
}

//the inserted interval date is start with fromDate1 and end with toDate1
//the date you want to compare with start with fromDate2 and end with toDate2
if ((int)(toDate1 - fromDate2).TotalDays < 0 )
{ return true;}
else
{
Response.Write("<script>alert('there is an intersection between the inserted date interval and the one you want to compare with')</script>");
return false;
}
if ((int)(fromDate1 - toDate2).TotalDays > 0 )
{ return true;}
else
{
Response.Write("<script>alert('there is an intersection between the inserted date interval and the one you want to compare with')</script>");
return false;
}

You have two intervals, i1 and i2. There are six cases for how the intervals can be temporally related (at least in a Newtonian world view) but only two are important: if i1 is entirely before i2 or i1 is entirely after i2; otherwise the two intervals are overlapping (the other four cases are i1 contains i2, i2 contains i1, i1 contains the start of i2 and i1 contains the end of i2). Assume i1 and i2 are of type Interval that have Date fields beginTime and endTime. The function then is (note, the assumption here is that if i1 starts at the same time i2 ends, or vice versa, we don't consider that an overlap and we assme for a given interval endTime.before(beginTime) is false):
boolean isOverlapped(Interval i1, Interval i2) {
return i1.endTime.before(i2.beginTime) || i1.beginTime.after(i2.endTime);
}
In the original question, you specify DateTime instead of Date. In java, Date has both date and time. This is in contrast to sql where Date does not have a time element while DateTime does. That is a point of confusion that I stumbled across when I first started using sql after having done only java for many years. Anyway, I hope this explanation is helpful.