How can I save ordered collection in JPA/Hibernate? - spring

Question

I have an main entity:
#Data
#Entity
#Table(name = "MAINS")
public class Main {
...
#OneToMany(mappedBy = "main", cascade = {CascadeType.PERSIST, CascadeType.MERGE}, fetch = LAZY)
private List<Chield> children;
And I have a child entity:
#Data
#Entity
#Table(name = "CHILDS")
public class Child {
#Id
#Column(name = "GUID")
private String guid;
#Column(name = "NAME")
private String name;
#Column(name = "Age")
private Integer age;
And I try to save one Main with many Child:
List<Child> children = new ArrayList<>();
for (int i = 0, i< 10; i++) {
Child child = new Child();
child.setId(i);
child.setName("Name"+i);
child.setAge(10+i);
children.add(child);
}
main.setChildren(children);
MainRepository.save(main);
But I need to save these children ordered. I can add a new column to Child - order and set child.setOrder(order++); and when I select this list I can sort it by order field.
But can I do it differently? Without adding of the new column?

Answer 1

If I get your question correctly it could be helpful to check such options:
OrderColumn
#OrderColumn(name = "index_id")
private List<Child> changes = new ArrayList<>();
OR
#OrderBy("id")
private Set<Child>list = new LinkedHashSet<>();
OR
#SortNatural
private SortedSet<Child> children = new TreeSet<Child>();
DOCS: naturalSort
OR for custom comparator: #SortComparator(SortById.class)
DOCS: customSort
public class SortById implements Comparator<Child> {
Logger log = Logger.getLogger(SortById.class.getSimpleName());
#Override
public int compare(Child o1, Child o2) {
log.info("Child.compare");
return o1.getId().compareTo(o2.getId());
}
}