Calculate number of weekdays between two dates in Java

Java

Java Problem Overview


Can anyone point me to some Java snippet wherein i can get business (except Sat and Sun) days between two dates.

Java Solutions


Solution 1 - Java

public static int getWorkingDaysBetweenTwoDates(Date startDate, Date endDate) {
    Calendar startCal = Calendar.getInstance();
	startCal.setTime(startDate);		
	
	Calendar endCal = Calendar.getInstance();
	endCal.setTime(endDate);

    int workDays = 0;

    //Return 0 if start and end are the same
    if (startCal.getTimeInMillis() == endCal.getTimeInMillis()) {
        return 0;
    }
    
    if (startCal.getTimeInMillis() > endCal.getTimeInMillis()) {
        startCal.setTime(endDate);
        endCal.setTime(startDate);
    }

    do {
       //excluding start date
        startCal.add(Calendar.DAY_OF_MONTH, 1);
        if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY) {
            ++workDays;
        }
    } while (startCal.getTimeInMillis() < endCal.getTimeInMillis()); //excluding end date

    return workDays;
}

> Start date and end date are exclusive, Only the days between given > dates will be counted. Start date and end date will not be included.

Solution 2 - Java

Solution without loop:

static long days(Date start, Date end){
	//Ignore argument check
	
	Calendar c1 = Calendar.getInstance();
	c1.setTime(start);
	int w1 = c1.get(Calendar.DAY_OF_WEEK);
	c1.add(Calendar.DAY_OF_WEEK, -w1);
  
	Calendar c2 = Calendar.getInstance();
	c2.setTime(end);
	int w2 = c2.get(Calendar.DAY_OF_WEEK);
	c2.add(Calendar.DAY_OF_WEEK, -w2);

	//end Saturday to start Saturday 
	long days = (c2.getTimeInMillis()-c1.getTimeInMillis())/(1000*60*60*24);
	long daysWithoutWeekendDays = days-(days*2/7);

    // Adjust days to add on (w2) and days to subtract (w1) so that Saturday
    // and Sunday are not included
    if (w1 == Calendar.SUNDAY && w2 != Calendar.SATURDAY) {
        w1 = Calendar.MONDAY;
    } else if (w1 == Calendar.SATURDAY && w2 != Calendar.SUNDAY) {
        w1 = Calendar.FRIDAY;
    } 
    
    if (w2 == Calendar.SUNDAY) {
        w2 = Calendar.MONDAY;
    } else if (w2 == Calendar.SATURDAY) {
        w2 = Calendar.FRIDAY;
    }
	
	return daysWithoutWeekendDays-w1+w2;
}

Solution 3 - Java

Solution without loop in 5 lines of code

Days between are defined in the same way as ChronoUnit.DAYS.between(start, end) which means there are 4 days between Monday and Friday. Since we are only interested in weekdays we have to subtract weekends, therefore from Friday until Tuesday there will be 2 weekdays(just compute endDay - startDay and subtract 2 for the weekend). Add 1 to the result if you want an inclusive result, i.e. not days between.

I present two solutions.

First solution (5-liner, short and cryptic):

import java.time.*;
import java.time.temporal.*;

public static long calcWeekDays1(final LocalDate start, final LocalDate end) {
    final DayOfWeek startW = start.getDayOfWeek();
    final DayOfWeek endW = end.getDayOfWeek();

    final long days = ChronoUnit.DAYS.between(start, end);
    final long daysWithoutWeekends = days - 2 * ((days + startW.getValue())/7);

    //adjust for starting and ending on a Sunday:
    return daysWithoutWeekends + (startW == DayOfWeek.SUNDAY ? 1 : 0) + (endW == DayOfWeek.SUNDAY ? 1 : 0);
}

Second solution:

public static long calcWeekDays2(final LocalDate start, final LocalDate end) {
    final int startW = start.getDayOfWeek().getValue();
    final int endW = end.getDayOfWeek().getValue();

    final long days = ChronoUnit.DAYS.between(start, end);
    long result = days - 2*(days/7); //remove weekends

    if (days % 7 != 0) { //deal with the rest days
        if (startW == 7) {
            result -= 1;
        } else if (endW == 7) {  //they can't both be Sunday, otherwise rest would be zero
            result -= 1;
        } else if (endW < startW) { //another weekend is included
            result -= 2;
        }
    }

    return result;
}

Solution 4 - Java

java.time

The modern way is with the java.time classes.

LocalDate

The LocalDate class represents a date-only value without time-of-day and without time zone.

LocalDate start = LocalDate.of( 2016 , 1 , 23 );
LocalDate stop = start.plusMonths( 1 );

DayOfWeek enum

The DayOfWeek enum provides a singleton instance for each of the sever days of the week.

DayOfWeek dow = start.getDayOfWeek();
if( dow.equals( DayOfWeek.SATURDAY ) || dow.equals( DayOfWeek.SUNDAY ) ) …

We can collect the desired dates in a List.

int initialCapacity = Duration.between( start , stop ).toDays() ;
List<LocalDate> dates = new ArrayList<>( initialCapacity );
…
if( dow.equals( DayOfWeek.SATURDAY ) || dow.equals( DayOfWeek.SUNDAY ) ) {
    dates.add( date );

An EnumSet is an extremely efficient, fast and low-memory, implementation of Set. We can use an EnumSet instead of the if statement seen above.

Set<DayOfWeek> weekend = EnumSet.of( DayOfWeek.SATURDAY , DayOfWeek.SUNDAY ) ;
…
if( weekend.contains( dayOfWeek ) ) …

Put that all together.

LocalDate date = start ;
while( date.isBefore( stop ) ) {
    if( ! weekend.contains( date.getDayOfWeek() ) ) { // If not weekend, collect this LocalDate.
        dates.add( date ) ;
    }
    // Prepare for next loop.
    date = date.plusDays( 1 ); // Increment to next day.
}

nextWorkingDay TemporalAdjuster

Another approach uses the ThreeTen-Extra project to add classes that work with java.time.

The Temporals class adds additional implementations of TemporalAdjuster for manipulating date-time values. We want the nextWorkingDay adjuster to increment the date while skipping over Saturday & Sunday.

LocalDate start = LocalDate.of( 2016 , 1 , 23 );
LocalDate stop = start.plusMonths( 1 );

int initialCapacity = Duration.between( start , stop ).toDays() ;
List<LocalDate> dates = new ArrayList<>( initialCapacity );

LocalDate date = start.minusDays( 1 );  // Start a day ahead.
while( date.isBefore( stop ) ) {
    date = date.with( org.threeten.extra.Temporals.nextWorkingDay() );
    // Double-check ending date as the `nextWorkingDay` adjuster could move us past the stop date.
    if( date.isBefore( stop ) ) { 
        dates.add( date ) ;
    }
}

Performance

I am curious about the performance of the various approach in various Answers on this page. I am considering only the modern java.time code, not the code using troublesome legacy Date/Calendar classes.

Here are four methods that each return the number of days elapsed.

One uses the clever math-based approach seen in the Answer by Roland.

private long countWeekDaysMath ( LocalDate start , LocalDate stop ) {
    // Code taken from Answer by Roland.
    // https://stackoverflow.com/a/44942039/642706
    long count = 0;
    final DayOfWeek startW = start.getDayOfWeek();
    final DayOfWeek stopW = stop.getDayOfWeek();

    final long days = ChronoUnit.DAYS.between( start , stop );
    final long daysWithoutWeekends = days - 2 * ( ( days + startW.getValue() ) / 7 );

    //adjust for starting and ending on a Sunday:
    count = daysWithoutWeekends + ( startW == DayOfWeek.SUNDAY ? 1 : 0 ) + ( stopW == DayOfWeek.SUNDAY ? 1 : 0 );

    return count;
}

Two use approaches seen in this Answer of mine: (a) Visit each date, incrementing one-by-one in a conventional loop.

private long countWeekDaysVisit ( LocalDate start , LocalDate stop ) {
    // Code taken from Answer by Basil Bourque.
    // https://stackoverflow.com/a/40369140/642706
    long count = 0;
    Set < DayOfWeek > weekend = EnumSet.of( DayOfWeek.SATURDAY , DayOfWeek.SUNDAY );
    LocalDate ld = start;
    while ( ld.isBefore( stop ) ) {
        if ( ! weekend.contains( ld.getDayOfWeek() ) ) { // If not weekend, collect this LocalDate.
            count++;
        }
        // Prepare for next loop.
        ld = ld.plusDays( 1 ); // Increment to next day.
    }
    return count;
}

…and, (b) Using the TemporalAdjuster implementation org.threeten.extra.Temporals.nextWorkingDay().

private long countWeekDaysAdjuster ( LocalDate start , LocalDate stop ) {
    // Code taken from Answer by Basil Bourque.
    // https://stackoverflow.com/a/40369140/642706
    long count = 0;
    Set < DayOfWeek > weekend = EnumSet.of( DayOfWeek.SATURDAY , DayOfWeek.SUNDAY );
    TemporalAdjuster nextWorkingDayTA = org.threeten.extra.Temporals.nextWorkingDay();
    LocalDate ld = start;
    if ( weekend.contains( ld.getDayOfWeek() ) ) {
        ld = ld.with( nextWorkingDayTA );
    }
    while ( ld.isBefore( stop ) ) {
        count++;
        // Prepare for next loop.
        ld = ld.with( nextWorkingDayTA ); // Increment to next working day (non-weekend day).
    }
    return count;
}

The last uses Java Streams approach seen in the Answer by Ravindra Ranwala.

private long countWeekDaysStream ( LocalDate start , LocalDate stop ) {
    // Code taken from the Answer by Ravindra Ranwala.
    // https://stackoverflow.com/a/51010738/642706
    long count = 0;
    Set < DayOfWeek > weekend = EnumSet.of( DayOfWeek.SATURDAY , DayOfWeek.SUNDAY );
    final long weekDaysBetween = start.datesUntil( stop )
                                     .filter( d -> ! weekend.contains( d.getDayOfWeek() ) )
                                     .count();
    return count;
}

And the test harness.

Caveats:

  • Well, the usual caveats about micro-benchmarking being untrustworthy, prone to unjustified or unrealistic conclusions.
  • I wish I'd learned to use the JMH micro-benchmarking framework.
  • I have not bothered to try optimizing any of this code. For example, in real work, the TemporalAdjuster could be cached outside our method.

Test harness.

LocalDate start = LocalDate.of( 2018 , Month.JANUARY , 1 );
LocalDate stop = start.plusYears( 1 );

int runs = 100_000;

long go = System.nanoTime();
for ( int i = 1 ; i <= runs ; i++ ) {
    long count = this.countWeekDaysMath( start , stop );
}
long elapsedMath = ( System.nanoTime() - go );

go = System.nanoTime();
for ( int i = 1 ; i <= runs ; i++ ) {
    long count = this.countWeekDaysVisit( start , stop );
}
long elapsedVisit = ( System.nanoTime() - go );

go = System.nanoTime();
for ( int i = 1 ; i <= runs ; i++ ) {
    long count = this.countWeekDaysStream( start , stop );
}
long elapsedAdjuster = ( System.nanoTime() - go );

go = System.nanoTime();
for ( int i = 1 ; i <= runs ; i++ ) {
    long count = this.countWeekDaysStream( start , stop );
}
long elapsedStream = ( System.nanoTime() - go );

System.out.println( "math: " + elapsedMath + " each: " + ( elapsedMath / runs ) );
System.out.println( "visit: " + elapsedVisit + " each: " + ( elapsedVisit / runs ) );
System.out.println( "adjuster: " + elapsedAdjuster + " each: " + ( elapsedAdjuster / runs ) );
System.out.println( "stream: " + elapsedStream + " each: " + ( elapsedStream / runs ) );

When run on my MacBook Pro (Sierra) with Oracle JDK 10.0.1 and ThreeTen-Extra version 1.3.2, I get results consistently close to the following. The math solution is a tiny fraction of the others at a couple hundred nanos versus several thousand, as we would expect obviously. Of the other three, the TemporalAdjuster is the longest, always over 10,000 nanos each. The visit and stream both come in well under that 10,000 nanos each, with visit being noticeably faster than streams. As seen in other examples around the internets, Java Streams usually make for nifty short code while often running significantly longer, about 20% longer in this case.

>math: 18313309 each: 183

>visit: 708420626 each: 7084

>adjuster: 1002157240 each: 10021

>stream: 924724750 each: 9247


About java.time

The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.

The Joda-Time project, now in maintenance mode, advises migration to java.time.

To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.

Where to obtain the java.time classes?

  • Java SE 8 and SE 9 and later
  • Built-in.
  • Part of the standard Java API with a bundled implementation.
  • Java 9 adds some minor features and fixes.
  • Java SE 6 and SE 7
  • Much of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
  • Android
  • The ThreeTenABP project adapts ThreeTen-Backport (mentioned above) for Android specifically.
  • See How to use….

The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.

Solution 5 - Java

I used Shengyuan Lu's solution, but I needed to make a fix for the case where the method is called when one of the dates is on a Saturday and the other a Sunday - otherwise the answer is off by a day:

static long days(Date start, Date end){
    //Ignore argument check

    Calendar c1 = GregorianCalendar.getInstance();
    c1.setTime(start);
    int w1 = c1.get(Calendar.DAY_OF_WEEK);
    c1.add(Calendar.DAY_OF_WEEK, -w1 + 1);

    Calendar c2 = GregorianCalendar.getInstance();
    c2.setTime(end);
    int w2 = c2.get(Calendar.DAY_OF_WEEK);
    c2.add(Calendar.DAY_OF_WEEK, -w2 + 1);

    //end Saturday to start Saturday 
    long days = (c2.getTimeInMillis()-c1.getTimeInMillis())/(1000*60*60*24);
    long daysWithoutSunday = days-(days*2/7);

    if (w1 == Calendar.SUNDAY) {
        w1 = Calendar.MONDAY;
    }
    if (w2 == Calendar.SUNDAY) {
        w2 = Calendar.MONDAY;
    }
    return daysWithoutSunday-w1+w2;
}

Solution 6 - Java

Almost all the solutions are pretty much obsoleted and narrative. However here's a much condensed and readable solution.

This approach uses a Java Stream provided by the LocalDate::datesUntil method built into in Java 9 and later.

LocalDate startDate = LocalDate.of(2018, 5, 2);
LocalDate endDate = LocalDate.now();
Set<DayOfWeek> weekend = EnumSet.of(DayOfWeek.SATURDAY, DayOfWeek.SUNDAY);
final long weekDaysBetween = startDate.datesUntil(endDate)
		.filter(d -> !weekend.contains(d.getDayOfWeek()))
		.count();

> .datesUntil returns a sequential ordered stream of dates. The > returned stream starts from this date (inclusive) and goes to > endExclusive (exclusive) by an incremental step of 1 day.

Then all the Saturdays and Sundays are filtered out. Final step is to get the count of the remaining week days.

Java-9 has been released one year ago, since using it now seems reasonable to me.

Solution 7 - Java

I don't have a Java based solution, but have a PHP one, hope it helps:

function getDate($days) {	
    for ($i = 0; $i < $days; $i ++) {                                      
	    if (date('N' , strtotime('+' . ($i + 1) . ' days')) > 5) {  
		    $days++;                                                        
	    }                                                                   
    }                                                                       
                                                                            
    return date('l, F jS', strtotime('+' . $days . ' days', time()));
}

Solution 8 - Java

This is my example without looping. Algorithm is same as 卢声远 Shengyuan Lus one but I used some features of JodaTime.

public static int getNumberOfBusinessDays(@Nonnull LocalDate from, @Nonnull LocalDate to) {
	int fromDateDayOfWeek = from.getDayOfWeek();
	int toDateDayOfWeek = to.getDayOfWeek();

	int daysWithoutWeekends = 5 * Weeks.weeksBetween(
			from.withDayOfWeek(DateTimeConstants.MONDAY), to).getWeeks();

	if (fromDateDayOfWeek == DateTimeConstants.SUNDAY) {
		fromDateDayOfWeek = DateTimeConstants.SATURDAY;
	}
	if (toDateDayOfWeek == DateTimeConstants.SUNDAY) {
		toDateDayOfWeek = DateTimeConstants.SATURDAY;
	}

	return daysWithoutWeekends - (fromDateDayOfWeek - toDateDayOfWeek);
}

Solution 9 - Java

This thread is filled with failing solutions... I started by writing a little test file which met my needs, and saw that Roland's both solutions fails, Amir's too. I wanted a solution that uses java 8 and that does not uses loops because, do I have to say why ?

So here's the test file :

@Test
public void test() {
	LocalDate d1 = LocalDate.of(2018, 8, 1);
	LocalDate d2 = LocalDate.of(2018, 8, 2);
	LocalDate d3 = LocalDate.of(2018, 8, 3);
	LocalDate d4 = LocalDate.of(2018, 8, 4);
	LocalDate d5 = LocalDate.of(2018, 8, 5);
	LocalDate d6 = LocalDate.of(2018, 8, 6);
	LocalDate d7 = LocalDate.of(2018, 8, 7);
	LocalDate d8 = LocalDate.of(2018, 8, 8);
	LocalDate d9 = LocalDate.of(2018, 8, 9);
	LocalDate d10 = LocalDate.of(2018, 8, 10);
	LocalDate d15 = LocalDate.of(2018, 8, 15);
	LocalDate dsep = LocalDate.of(2018, 9, 5);

	// same day : 0 days between
	Assert.assertEquals(0, DateUtils.calcWeekDays1(d1, d1));
	Assert.assertEquals(1, DateUtils.calcWeekDays1(d1, d2));
	Assert.assertEquals(2, DateUtils.calcWeekDays1(d1, d3));
	// end on week-end 
	Assert.assertEquals(2, DateUtils.calcWeekDays1(d1, d4));
	Assert.assertEquals(2, DateUtils.calcWeekDays1(d1, d5));
	// next week
	Assert.assertEquals(3, DateUtils.calcWeekDays1(d1, d6));
	Assert.assertEquals(4, DateUtils.calcWeekDays1(d1, d7));
	Assert.assertEquals(5, DateUtils.calcWeekDays1(d1, d8));
	Assert.assertEquals(6, DateUtils.calcWeekDays1(d1, d9));
	Assert.assertEquals(7, DateUtils.calcWeekDays1(d1, d10));
	// start on saturday
	Assert.assertEquals(0, DateUtils.calcWeekDays1(d4, d5));
	Assert.assertEquals(0, DateUtils.calcWeekDays1(d4, d6));
	Assert.assertEquals(1, DateUtils.calcWeekDays1(d4, d7));
	// start on sunday
	Assert.assertEquals(0, DateUtils.calcWeekDays1(d5, d5));
	Assert.assertEquals(0, DateUtils.calcWeekDays1(d5, d6));
	Assert.assertEquals(1, DateUtils.calcWeekDays1(d5, d7));
	// go to next week
	Assert.assertEquals(10, DateUtils.calcWeekDays1(d1, d15));
	// next month
	Assert.assertEquals(25, DateUtils.calcWeekDays1(d1, dsep));
	// start sat, go to next month
	Assert.assertEquals(22, DateUtils.calcWeekDays1(d4, dsep));

}

And here is my proposed solution, quite simple. Just let java count the number of weeks, multiply by five, and add the number of days needed to compensate the difference ; the only trick is adjusting the start and end to avoid week-ends :

public static long calcWeekDays1(LocalDate start, LocalDate end) {
	if (start.getDayOfWeek().getValue() > 5) {
		start = start.with(TemporalAdjusters.next(DayOfWeek.MONDAY));
	}
	if (end.getDayOfWeek().getValue() > 5) {
		end = end.with(TemporalAdjusters.previous(DayOfWeek.FRIDAY));
	}
	if (start.isAfter(end)) { // may happen if you start sat. and end sunday
		return 0;
	}
	long weeks = ChronoUnit.WEEKS.between(start, end);
	if (start.getDayOfWeek().getValue() > end.getDayOfWeek().getValue()) {
		weeks += 1;
	}
	return 5 * weeks + end.getDayOfWeek().getValue() - start.getDayOfWeek().getValue();
}

And now I will look stupid if my code fails too :)

Solution 10 - Java

The do while in the solution of Piyush is wrong, it should be :

do {
    if (startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY && startCal.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY) {
        ++workDays;
    }
    startCal.add(Calendar.DAY_OF_MONTH, 1);
} while (startCal.getTimeInMillis() < endCal.getTimeInMillis());

Solution 11 - Java

The startCal.add should add onto the Calendar.DATE field, not the Calendar.DAY_OF_MONTH, I was getting weird results with over Decemeber / January period.

Solution 12 - Java

This is my example without looping. It is a class in this example because I serialize it in some JSON output. Basically I work out the number of days between the two dates, divide by 7 and assign to a long to have a integer value for the number of weeks. Take the original number of days and subtract the number of weekends*2. This isn't quite perfect - you need to work out if there is a 'hangover' where the start is close to the end of the week and goes over the weekend. To correct for this I find the day of the week at the start and find the remainder of the number of days, and add those together to find the 'hangover' - and if it is more than 5 it is a weekend. It isn't quite perfect, and does not account for holidays at all. And no Joda in sight. That said there is also a issue with timezones.

import java.io.Serializable;
import java.util.Date;

public class BusinessDayCalculator implements Serializable {

	private static long DAY = 86400000l;

	private Date startTime;
	private Date endTime;
	
	public void setStartTime(Date startTime) {
		this.startTime = startTime;
	}

	public Date getStartTime() {
		return startTime;
	}

	public void setEndTime(Date endTime) {
		this.endTime = endTime;
	}

	public Date getEndTime() {
		return endTime;
	}
	
	public long getHours() {
		return (this.endTime.getTime() - this.startTime.getTime())/(1000*60*60);
	}

	public long getBusinessDays(){
	
		long startDay = getDayFromDate(this.startTime);
		long endDay = getDayFromDate(this.endTime);
	
		long totalDays = endDay-startDay;
		long totalWeekends = totalDays/7;
	
		long day = getDay(this.startTime);
	
		long hangover = totalDays % 7;
	
		long intoWeekend = day + hangover;
		if(intoWeekend>5){
			totalWeekends++;
		}
	
		long totalBusinessDays = totalDays - (totalWeekends *2);
	
		/*
		System.out.println("Days = " + day );
		System.out.println("Hangover = " + hangover );
		System.out.println("Total Days = " + totalDays);
		System.out.println("Total Weekends = " + totalWeekends);
		System.out.println("Total Business Days = " + totalBusinessDays);
		*/
	
		return totalBusinessDays;
	}

	private long getDayFromDate( Date date ){
		long d = date.getTime() / DAY;
		return d;
	}

	private long getDay( Date date ){
		long daysSinceEpoc = getDayFromDate(date);
		long day = daysSinceEpoc % 7;
		day = day + 4;
		if(day>6) day = day - 7;
		return day;
	}

}

Solution 13 - Java

Solution for Java 8 without loop and INCLUSIVE intervals:

public long getDaysWithoutSundays(LocalDate startDate, LocalDate endDate) {
	long numberOfDays = ChronoUnit.DAYS.between(startDate, endDate) + 1;
	long numberOfSundays = numberOfDays / 7;
	long rest = numberOfDays % 7;
	if (rest > 0) {
		int startToEnd = startDate.getDayOfWeek().getValue() - endDate.getDayOfWeek().getValue();
		if (startToEnd > 0) {
			numberOfSundays++;
		}
		else {
			if (endDate.getDayOfWeek().equals(DayOfWeek.SUNDAY)) {
				numberOfSundays++;
			}
		}
	}
	return numberOfDays - numberOfSundays;
}

Solution 14 - Java

import java.util.ArrayList;
import java.util.Calendar;
import java.util.Date;
import java.util.HashMap;
import java.util.List;

/**
 * 
 * @author varun.vishwakarma
 *
 */
public class FindWeekendsInDateRange {
	static HashMap<Integer, String> daysOfWeek=null;
	
	
	static {
		daysOfWeek = new HashMap<Integer, String>();
		daysOfWeek.put(new Integer(1), "Sun");
		daysOfWeek.put(new Integer(2), "Mon");
		daysOfWeek.put(new Integer(3), "Tue");
		daysOfWeek.put(new Integer(4), "Wed");
		daysOfWeek.put(new Integer(5), "Thu");
		daysOfWeek.put(new Integer(6), "Fri");
		daysOfWeek.put(new Integer(7), "Sat");
	}

	/**
	 * 
	 * @param from_date 
	 * @param to_date
	 * @return 
	 */
	public static List<Date>  calculateWeekendsInDateReange(Date fromDate, Date toDate) {
		List<Date> listOfWeekends = new ArrayList<Date>();
			Calendar from = Calendar.getInstance();
			Calendar to = Calendar.getInstance();
			from.setTime(fromDate);
			to.setTime(toDate);
			while (from.getTimeInMillis() < to.getTimeInMillis()) {
				if (daysOfWeek.get(from.get(Calendar.DAY_OF_WEEK)) == "Sat") {
					Date sat = from.getTime();
					listOfWeekends.add(sat);
				} else if (daysOfWeek.get(from.get(Calendar.DAY_OF_WEEK)) == "Sun") {
					Date sun = from.getTime();
					listOfWeekends.add(sun);
				}
				from.add(Calendar.DAY_OF_MONTH, 1);
			}
		return listOfWeekends;
	}
	public static void main(String[] args) {
		String fromDate = "7-Oct-2019";
		String toDate = "25-Oct-2019";
		System.out.println(FindWeekendsInDateRange.calculateWeekendsInDateReange(new Date(fromDate), new Date(toDate)));
	

	}

}

Solution 15 - Java

This program considers loop approach but consider activities happened on after work hours to next working day office start hour

public class BusinessDayCalculator {

private final String DATE_FORMAT = "dd/MM/yyyy HH:mm:ss";
private final int OFFICE_START_HOUR = 9;
private final int OFFICE_CLOSE_HOUR = 17;
private final int TOTAL_MINS_IN_BUSINESS_DAY = (OFFICE_CLOSE_HOUR - OFFICE_START_HOUR)*60;

public void dateDifference(String start, String end){
	Date startDate = validateStringToDate(start);
	Date endDate = validateStringToDate(end);
	System.out.println(startDate);
	System.out.println(endDate);
	Calendar startDay = convertDateToCalendar(startDate);
	Calendar tempDay = (Calendar) startDay.clone();
	Calendar endDay = convertDateToCalendar(endDate);

	System.out.println(startDay.getTime());
	System.out.println(endDay.getTime());
	int workDays = -1;

	int startDayDifference = 0;
	int endDayDifference = 0;
    int hours = 0;
	int minsRemainder = 0;

	if(!(startDay.get(Calendar.DAY_OF_YEAR) == endDay.get(Calendar.DAY_OF_YEAR)
			&& startDay.get(Calendar.YEAR) == endDay.get(Calendar.YEAR))){

		do{
			tempDay.add(Calendar.DAY_OF_MONTH, 1);
			if(tempDay.get(Calendar.DAY_OF_WEEK) != Calendar.SATURDAY 
					&& tempDay.get(Calendar.DAY_OF_WEEK) != Calendar.SUNDAY){
				workDays++;
			}
		}while(tempDay.getTimeInMillis() <= endDay.getTimeInMillis());

		if(workDays > 0){
			workDays = workDays - 1;
		}

	}

	startDayDifference = hourDifferenceInMinutesOfStartDay(startDay);
	endDayDifference = hourDifferenceInMinutesOfEndDay(endDay);

	minsRemainder = (startDayDifference + endDayDifference) % TOTAL_MINS_IN_BUSINESS_DAY;
	workDays = workDays + ((startDayDifference + endDayDifference) / TOTAL_MINS_IN_BUSINESS_DAY);

	hours = minsRemainder/60;
	minsRemainder = minsRemainder % 60;

	System.out.println(workDays + "d "+ hours + "hrs " + minsRemainder + " mins");

}


private int hourDifferenceInMinutesOfEndDay(Calendar endDay) {
	long endTimestamp = endDay.getTimeInMillis();
	System.out.println(endTimestamp);
	endDay.set(Calendar.HOUR_OF_DAY, OFFICE_START_HOUR);
	endDay.set(Calendar.MINUTE,0);
	long endDayOfficeStartTimestamp = endDay.getTimeInMillis();
	System.out.println(endDayOfficeStartTimestamp);
	int difference = (int)((endTimestamp - endDayOfficeStartTimestamp) / 1000) / 60;
	System.out.println(difference);
	return difference;
}


private int hourDifferenceInMinutesOfStartDay(Calendar startDay) {
	long starttimestamp = startDay.getTimeInMillis();
	System.out.println(starttimestamp);
	startDay.set(Calendar.HOUR_OF_DAY, OFFICE_CLOSE_HOUR);
	startDay.set(Calendar.MINUTE,0);
	long startDayOfficeCloseTimestamp = startDay.getTimeInMillis();
	System.out.println(startDayOfficeCloseTimestamp);
	int difference = (int)((startDayOfficeCloseTimestamp - starttimestamp) / 1000) / 60;
	System.out.println(difference);
	return difference;
}

public Calendar convertDateToCalendar(Date date){
	Calendar calendar = Calendar.getInstance();
	calendar.setTime(date);

	if(calendar.get(Calendar.DAY_OF_WEEK) == Calendar.SATURDAY 
			|| calendar.get(Calendar.DAY_OF_WEEK) == Calendar.SUNDAY){
		calendar = handleActivityOnAfterWorkHoursOrWeekendOrHolidays(calendar);
	}

	if(calendar.get(Calendar.HOUR_OF_DAY) >= OFFICE_CLOSE_HOUR 
			&& calendar.get(Calendar.MINUTE) > 0){
		calendar = handleActivityOnAfterWorkHoursOrWeekendOrHolidays(calendar);
	}

	if(calendar.get(Calendar.HOUR_OF_DAY) < OFFICE_START_HOUR){
		calendar.set(Calendar.HOUR_OF_DAY, OFFICE_START_HOUR);
		calendar.set(Calendar.MINUTE,0);
	}

	return calendar;
}

private Calendar handleActivityOnAfterWorkHoursOrWeekendOrHolidays(Calendar calendar) {
	do{
		calendar.add(Calendar.DAY_OF_MONTH, 1);
	}while(isHoliday(calendar));
	calendar.set(Calendar.HOUR_OF_DAY, OFFICE_START_HOUR);
	calendar.set(Calendar.MINUTE,0);
	return calendar;
}

private boolean isHoliday(Calendar calendar) {
	if(calendar.get(Calendar.DAY_OF_WEEK) == Calendar.SATURDAY 
			|| calendar.get(Calendar.DAY_OF_WEEK) == Calendar.SUNDAY){
		return true;
	}
	return false;
}

public Date validateStringToDate(String input){
	SimpleDateFormat dateFormat = new SimpleDateFormat(DATE_FORMAT);
	Date date = null;
	try{
		date = dateFormat.parse(input);
	}catch(ParseException exception){
		System.out.println("invalid date format");
		throw new RuntimeException("invalid date format");
	}
	return date;
}

public static void main(String[] args){
	BusinessDayCalculator calc = new BusinessDayCalculator();
	String startDate = "27/12/2016 11:38:00";
	String endDate = "04/01/2017 12:38:00";
	calc.dateDifference(startDate, endDate);
}

}

Solution 16 - Java

Here is a set-based solution that completes in constant time for any given subset of weekdays, not just Monday-Friday. It splits the problem into counting full weeks and counting the days in the residual week. If you are interested, here's a detailed explanation and a formal proof that the algorithm is correct. Note that the intervals are inclusive, i.e. startDate and endDate are counted in. If startDate is after endDate, the result is zero rather than negative.

long countWeekDays(LocalDate startDate, LocalDate endDate, Set<DayOfWeek> daysOfWeek) {

    long periodLength = Math.max(0, ChronoUnit.DAYS.between(startDate, endDate) + 1);
    long fullWeeks = periodLength / 7;
    long residualWeekLength = periodLength % 7;

    Set<DayOfWeek> residualWeekDays = LongStream.range(0, residualWeekLength)
        .mapToObj(offset -> startDate.plusDays(offset).getDayOfWeek())
        .collect(Collectors.toSet());
    residualWeekDays.retainAll(daysOfWeek);

    return fullWeeks * daysOfWeek.size() + residualWeekDays.size();
}

For the original problem (Monday-Friday) it is called e.g. with:

countWeekDays(
    LocalDate.of(2016, 2, 8),
    LocalDate.of(2016, 2, 26),
    new HashSet(Arrays.asList(
        DayOfWeek.MONDAY,
        DayOfWeek.TUESDAY,
        DayOfWeek.WEDNESDAY,
        DayOfWeek.THURSDAY,
        DayOfWeek.FRIDAY
        )
    )
)

This assumes you are working with inclusive intervals. If you want to skip the first day of the interval, just add one day to the first parameter:

countWeekDays(
    LocalDate.of(2016, 2, 8).plusDays(1),
    LocalDate.of(2016, 2, 26),
    new HashSet(Arrays.asList(
        DayOfWeek.MONDAY,
        DayOfWeek.TUESDAY,
        DayOfWeek.WEDNESDAY,
        DayOfWeek.THURSDAY,
        DayOfWeek.FRIDAY
        )
    )
)

Likewise, if you want to skip the last day of the inclusive interval, subtract one day from the second parameter:

countWeekDays(
    LocalDate.of(2016, 2, 8),
    LocalDate.of(2016, 2, 26).minusDays(1),
    new HashSet(Arrays.asList(
        DayOfWeek.MONDAY,
        DayOfWeek.TUESDAY,
        DayOfWeek.WEDNESDAY,
        DayOfWeek.THURSDAY,
        DayOfWeek.FRIDAY
        )
    )
)

Finally, if you need to skip both interval-delimiting days, combine the two previous modifications:

countWeekDays(
    LocalDate.of(2016, 2, 8).plusDays(1),
    LocalDate.of(2016, 2, 26).minusDays(1),
    new HashSet(Arrays.asList(
        DayOfWeek.MONDAY,
        DayOfWeek.TUESDAY,
        DayOfWeek.WEDNESDAY,
        DayOfWeek.THURSDAY,
        DayOfWeek.FRIDAY
        )
    )
)

Solution 17 - Java

In groovy:

  public static int getWorkingDaysBetweenDates (Date start, Date end) {
    def totalDays = (Integer) (end.getTime() - start.getTime()) / (1000 * 60 * 60 * 24)
    def int workingDays = 0

    (0..totalDays).each { def dow = (start + it)[Calendar.DAY_OF_WEEK]; if(dow != Calendar.SATURDAY && dow != Calendar.SUNDAY){workingDays++} }
 
    workingDays
  }

Solution 18 - Java

Using java 8 it can be easily done, example function:

long getBusinessDaysDifference(LocalDate startDate, LocalDate endDate) { 

    EnumSet<DayOfWeek> weekend = EnumSet.of(DayOfWeek.SATURDAY, DayOfWeek.SUNDAY);
    List<LocalDate> list = Lists.newArrayList();

    LocalDate start = startDate; 		
    while (start.isBefore(endDate)) {
		list.add(start);
		start = start.plus(1, ChronoUnit.DAYS);
    }

    long numberOfDays = list.stream().filter(d -> !weekend.contains(d.getDayOfWeek())).count();

    return numberOfDays;
}

Description:

  1. Define your off-days in an EnumSet (weekends in this case).
  2. Create a list holding all the days between the startDate and endDate.
  3. Reduce the outcome list by removing any occurrence of a day from the EnumSet.
  4. Then finally count the size of this reduced list.

Note: this function can be optimized, but might be helpful as a starting point.

Solution 19 - Java

public long getNumberOfWeekDayBetweenDates(LocalDate startDate, LocalDate endDate, String dayOfWeek) {
        long result = -1;
        if (startDate != null && endDate != null && dayOfWeek != null && (startDate.isBefore(endDate) || startDate.isEqual(endDate))) {
            java.time.DayOfWeek givenDayOfWeek = java.time.DayOfWeek.valueOf(dayOfWeek);
            // find the first given day of week in the interval
            LocalDate firstOccurrence = startDate.with(TemporalAdjusters.nextOrSame(givenDayOfWeek));
            // similarly find last Monday
            LocalDate lastOccurrence = endDate.with(TemporalAdjusters.previousOrSame(givenDayOfWeek));
            if (firstOccurrence != null && lastOccurrence != null) {
                // count the number of weeks between the first and last occurrence, then add 1 as end day is exclusive
                result = ChronoUnit.WEEKS.between(firstOccurrence, lastOccurrence) + 1;
            } else if (firstOccurrence == null && lastOccurrence == null) {
                // no occurrence
                result = 0;
            } else {
                result = 1;
            }
        }
        return result;
    }

Solution 20 - Java

For LocalDate supported by latest java version you can try below function.

It provides support of functiongetDayOfWeek().

The getDayOfWeek() method of LocalDate class in Java gets the day-of-week field, which is an enum DayOfWeek.

public static int getWeekEndCount(LocalDate fromDate, LocalDate toDate) {
        int saturday = 0;
        int sunday = 0;
        while (!fromDate.isAfter(toDate)) {
            if (fromDate.getDayOfWeek().equals(DayOfWeek.SATURDAY))
                saturday++;
            else if (fromDate.getDayOfWeek().equals(DayOfWeek.SUNDAY))
                sunday++;
            fromDate = fromDate.plusDays(1);
        }
        System.out.println("Saturday count=="+saturday);
        System.out.println("Sunday count=="+sunday);
        return saturday+sunday;
    }

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