Use DateTimeFormatterBuilder for parsing dates of missing day and default to end of month

user_1298461 0
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I need to parse dates in various formats. Some of these are missing the "day". Now I want to default to end of month. I am not sure if there is a direct way where we can default to end of month

DateTimeFormatter f = new DateTimeFormatterBuilder()
        .appendPattern("MM.yyyy")
        .parseDefaulting(DAY_OF_MONTH, 1) // can we default to end of month?
        .toFormatter();

LocalDate.parse("11.2017", f)

But if not, how can I know that the original date was missing the day (or that a default was used) and therefore I need to adjust it manually by using LocalDate.parse("11.2017", f).with(TemporalAdjusters.lastDayOfMonth()).

localdate java
1 Answers
DateTimeFormatter f = new DateTimeFormatterBuilder()
      .appendPattern("MM.yyyy")
      .parseDefaulting(DAY_OF_MONTH, 31) // set 31
      .toFormatter();

Just set DAY_OF_MONTH with 31, DAY_OF_MONTH will base on rangeUnit to choose the last day of parse month.

/**
 * The day-of-month.
 * <p>
 * This represents the concept of the day within the month.
 * In the default ISO calendar system, this has values from 1 to 31 in most months.
 * April, June, September, November have days from 1 to 30, while February has days
 * from 1 to 28, or 29 in a leap year.
 * <p>
 * Non-ISO calendar systems should implement this field using the most recognized
 * day-of-month values for users of the calendar system.
 * Normally, this is a count of days from 1 to the length of the month.
 */
DAY_OF_MONTH("DayOfMonth", DAYS, MONTHS, ValueRange.of(1, 28, 31), "day"),

Example:

LocalDate parse = LocalDate.parse("02.2017", f);
System.out.println(parse.getDayOfMonth());

Output:

28

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