I want to persist an entity that has a @OneToMany relationship to a child entity. I'm using Quarkus 1.13.1 with Quarkus Panache.
Example
public class User {
private List<Item> items;
@OneToMany(cascade = CascadeType.ALL)
public List<Item> getItems()...
}
If I want to persist a user (user.persist()) with a few items that already exist in the item table, then I get of course a "duplicate key" exception. So far so good.
But I was wondering if there is a descent way to skip/ignore an insert if an item already exists in the table items.
Of course, I could query the database to check if the child value exists, but this seems somehow tedious and bloats the code with data checks, so I was wondering if there was some annotation or other shortcut to handle this.
A persist operation should be used exclusively to create (store) new objects in the database, and makes the Java objects managed by Hibernate until the Session is closed.
It's really important that you know which objects are managed, and which are not, and distinguish wich ones are newly made persistent rather than just represent an existing object in the database.
To this end, it would indeed be better to load the existing Items first; if you know for sure which ones are already existing in the DB you can use a lazy proxy to represent them and put those in the list before persisting the User.
If you don't know which Items already exist in the database, then you should indeed have to query the database first. There is no shortcut for this operation; I guess we could explore some improvements but generally automating such things is tricky.
I would suggest implement the checks explicitly so you have full control over the strategy. It might be a good idea to make Item a cached entity so you can implement safe validations without performance drawbacks.