休眠条件列表必须包含另一个列表

Question

Given the simplified model:

public class Access {
  private Set<Tag> tags;
}

public class Item {
  private Set<Tag> tags;
}

An access grant the access on every items that fully contains his tags, no problem if the item has more tags or not.

But I have no idea how to create an hibernate criteria query for this. Could you help me?

Answer 1

You actually don't need to do anything other then tell Hibernate to load it always with your entity.

So option one, load always:

public class Access {

        @OneToMany(cascade = {CascadeType.ALL}, orphanRemoval = true, targetEntity = Tag.class, fetch = FetchType.EAGER)
        @JoinColumn(nullable = true)
        protected Set<Tag> tags;
    }

So the OneToMany annotation is telling Hibernate to look for tags in the Tag table linked to your table and to do this always: FetchType.EAGER .

Actually Hibernate will create two tables with a foreign key in the Tag table.

In case you wan't to lazy load this list you can set FetchType.LAZY , then you need to do some additional work in order to fill this list.

A simple load will fill up all values as stored:

session.get(Access.class, key);

Filtering Access list by Tags see this answer: How to join two different criterias under Hibernate?