Check if current time is within the timeframe using Joda-Time

Question

Novice Java developer here. I've never really used a time/date library before and I'm curious how an experienced developer would solve this. You're given 4 ints: startHour, startMinute, endHour and endMinute. Now, check if current time is within the given timeframe. Is there a more clean way of doing this than what I've done here:

private void checkTimeframe(int startHour, int startMinute, int endHour, int endMinute) {
    LocalDateTime now = LocalDateTime.now();
    LocalTime localTimeStart = new LocalTime(startHour, startMinute);
    LocalTime localTimeEnd = new LocalTime(endHour, endMinute);

    LocalDateTime startTime = new LocalDateTime(now.getYear(), now.getMonthOfYear(),
            now.getDayOfMonth(), startHour, startMinute);

    LocalDateTime endTime = new LocalDateTime(now.getYear(), now.getMonthOfYear(),
            now.getDayOfMonth(), endHour, endMinute);

    //Check if start/end is, for instance, 23:00 - 03:00
    if (localTimeStart.isAfter(localTimeEnd) || localTimeStart.equals(localTimeEnd)) {
        endTime = endTime.plusDays(1);
    }

    if ( (now.equals(startTime) || now.isAfter(startTime) ) && now.isBefore(endTime)) {
        System.out.println("Ok, we're within start/end");
    } else {
        System.out.println("Outside start/end");
    }
}

Answer 1

to me it looks like a simple check of number between range of numbers. do I would do like this

1. turn hour and minute into one number that is hhmm to simplify the comparison

2. now its a simple check between range of numbers, taking into account the case of cross-date boundary

complete code:

private static void checkTimeframe(int startHour, int startMinute, int endHour, int endMinute) {
    // "concatanate" hour and minute into one number
    int startHourMinute = startHour * 100 + startMinute;
    int endHourMinute = endHour * 100 + endMinute;
    LocalDateTime now = LocalDateTime.now();
    int nowHourMinute = now.getHour() * 100 + now.getMinute();

    // if range within date - simple between boundaries check
    if (startHourMinute <= endHourMinute) {
        if (nowHourMinute >= startHourMinute && nowHourMinute <= endHourMinute) {
            System.out.println("Ok, we're within start/end");
        } else {
            System.out.println("Outside start/end");
        }
    // else (cross date boundary range) - check if now date is either within range of yesterday or within range tomorrow  
    } else {
        if (nowHourMinute >= startHourMinute || nowHourMinute <= endHourMinute) {
            System.out.println("Ok, we're within start/end");
        } else {
            System.out.println("Outside start/end");
        }
    }
}

Answer 2

Kotlin version of @sharon-ben-asher answer:

fun LocalDateTime.isInTimeFrame(startHour: Int, startMinute: Int, endHour: Int, endMinute: Int): Boolean {
        val startHourMinute = startHour * 100 + startMinute
        val endHourMinute = endHour * 100 + endMinute

        val nowHourMinute: Int = hour * 100 + minute

        return if (startHourMinute <= endHourMinute) {
            // if range within date - simple between boundaries check
            nowHourMinute in startHourMinute..endHourMinute
        } else {
            // else (cross date boundary range) - check if now date is either within range of yesterday or within range tomorrow
            nowHourMinute >= startHourMinute || nowHourMinute <= endHourMinute
        }
    }

And use it like this:

LocalDateTime.parse(dateString, DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")).isInTimeFrame(18,0,6,0)

Is DateTime between 18:00 - 06:00 ?