如何在列表中对日期进行排序？

Question

I have a list of Objects . The list is already sorted in order by date of the booking. I would like to create a new list which takes todays date as an argument and stores the last 10 booking and next 10 bookings as of todays date.

How can this be done?

transaction = session.beginTransaction();

        String hql = "FROM Booking ORDER BY scheduledDate";
        Query query = session.createQuery(hql);
List<Booking> bookingList = query.list();
Date now = new Date();
        List<Booking> lessThan = null;
        List<Booking> moreThan = null;

        for(int i =0; i<bookingList.size();i++){

            if(bookingList.get(i).getScheduledDate().compareTo(now)<0)
                lessThan.add(bookingList.get(i));

            if (bookingList.get(i).getScheduledDate().compareTo(now)>0)
                moreThan.add(bookingList.get(i));
        }

These are my getter methods.

public java.util.Date getScheduledDate() {
    return scheduledDate;
}

Answer 1

Make a new list by taking the absolute values of the result of the difference of today's date and each of the dates in the date-list you have.

Find the position of the minimum number in the new list. You can then get the positions of the dates you require from here. min-pos to (min-pos + 9) will give you 10 records of next 10 bookings and (min-pos - 1) to (min-pos - 10) will give you the other set.

The result will differ a little based on whether the (today's date - date-at-min-pos) is greater than or less than zero. Then (min-pos+1) to (min-pos + 10) will give you 10 records of next 10 bookings and (min-pos) to (min-pos - 9) will give you the other set

Answer 2

You can easily do it with Java 8's stream , eg:

List<Booking> bookingList = query.list();
final Date today = new Date();
List<Booking> greaterThan = bookingList.stream()
        .filter(b -> b.getScheduledDate().after(today))
        .limit(10)
        .collect(Collectors.toList());

List<Booking> lessThan = bookingList.stream()
        .filter(b -> b.getScheduledDate().before(today))
        .limit(10)
        .collect(Collectors.toList());