Joda-Time: Get first/second/last sunday of month

Question

In plain Java, I have this code to get the last Sunday of the month.

Calendar getNthOfMonth(int n, int day_of_week, int month, int year) {
    Calendar compareDate = Date(1, month, year);
    compareDate.set(DAY_OF_WEEK, day_of_week);
    compareDate.set(DAY_OF_WEEK_IN_MONTH, n);
    return compareDate;
}
// Usage
Calendar lastSundayOfNovember = getNthOfMonth(-1, SUNDAY, NOVEMBER, 2012)

What is a clean and elegant way to achieve the same result using Joda-Time?

Solution 1

public class Time {
public static void main(String[] args) {
    System.out.println(getNthOfMonth(DateTimeConstants.SUNDAY, DateTimeConstants.SEP, 2012));
}


public static LocalDate getNthOfMonth(int day_of_week, int month, int year) {
    LocalDate date = new LocalDate(year, month, 1).dayOfMonth()  
             .withMaximumValue()
             .dayOfWeek()
             .setCopy(day_of_week);
    if(date.getMonthOfYear() != month) {
        return date.dayOfWeek().addToCopy(-7);
    }
    return date;
}
}

Solution 2

This is quite an old post but possibly this answer will help someone. Use the java.time classes that replace Joda-Time.

private static LocalDate getNthOfMonth(int type, DayOfWeek dayOfWeek, int month, int year){
    return LocalDate.now().withMonth(month).withYear(year).with(TemporalAdjusters.dayOfWeekInMonth(type, dayOfWeek));
}

Solution 3

You can try something like that:

public class Foo {

  public static LocalDate getNthSundayOfMonth(final int n, final int month, final int year) {
    final LocalDate firstSunday = new LocalDate(year, month, 1).withDayOfWeek(DateTimeConstants.SUNDAY);
    if (n > 1) {
      final LocalDate nThSunday = firstSunday.plusWeeks(n - 1);
      final LocalDate lastDayInMonth = firstSunday.dayOfMonth().withMaximumValue();
      if (nThSunday.isAfter(lastDayInMonth)) {
        throw new IllegalArgumentException("There is no " + n + "th Sunday in this month!");
      }
      return nThSunday;
    }
    return firstSunday;
  }


  public static void main(final String[] args) {
    System.out.println(getNthSundayOfMonth(1, DateTimeConstants.SEPTEMBER, 2012));
    System.out.println(getNthSundayOfMonth(2, DateTimeConstants.SEPTEMBER, 2012));
    System.out.println(getNthSundayOfMonth(3, DateTimeConstants.SEPTEMBER, 2012));
    System.out.println(getNthSundayOfMonth(4, DateTimeConstants.SEPTEMBER, 2012));
    System.out.println(getNthSundayOfMonth(5, DateTimeConstants.SEPTEMBER, 2012));
  }
}

Output:

2012-09-02
2012-09-09
2012-09-16
2012-09-23
2012-09-30

Solution 4

With Joda

Let's say we have a date:

DateTime date = new DateTime(2017, 6, 1, 0, 0, 0);

`First we need to know if the month has 31, 30, 28 or 29 days :

int lastofMonth = date.dayOfMonth().getMaximumValue();

From this we create a new date:

DateTime endOfMonth = new DateTime(2017, 6, lastOfMonth, 0, 0, 0);

The we find out what day of week the last day is:

int whatDayIsLast = endOfMonth.getDayOfWeek();

Now we can create a date, which will be the date of the last Sunday of the month:

DateTime lastSunday = new DateTime(2017, 6, lastOfMonth - whatDayIsLast, 0, 0, 0);

For my example (June 2017), the result will be:

DateTimeFormatter dtf = DateTimeFormat.forPattern("yyyy-MM-dd");
System.out.println(dtf.print(lastSunday);

2017-06-25

Solution 5

   static LocalDate getNthOfMonth(int n, int day_of_week, int month, int year)
   {
      if (n == -1)
      {
         return getNthOfMonth(0, day_of_week, month + 1, year);
      }
      final LocalDate compareDate = new LocalDate(year, month, 1);
      if (compareDate.getDayOfWeek() > day_of_week)
      {
         return compareDate.withDayOfWeek(day_of_week).plusDays(7 * n);   
      }
      else
      {
         return compareDate.withDayOfWeek(day_of_week).plusDays(7 * (n - 1));    
      }
   }  

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