Using a Hibernate filter with Spring Boot JPA

Question

I have found the need to limit the size of a child collection by a property in the child class.

I have the following after following this guide:

@FilterDef(name="dateFilter", parameters=@ParamDef( name="fromDate", type="date" ) )
public class SystemNode implements Serializable {

    @Getter
    @Setter
    @Builder.Default
    // "startTime" is a property in HealthHistory
    @Filter(name = "dateFilter", condition = "startTime >= :fromDate")
    @OneToMany(mappedBy = "system", targetEntity = HealthHistory.class, fetch = FetchType.LAZY)
    private Set<HealthHistory> healthHistory = new HashSet<HealthHistory>();

    public void addHealthHistory(HealthHistory health) {
        this.healthHistory.add(health);
        health.setSystem(this);
    }
}

However, I don't really understand how to toggle this filter when using Spring Data JPA. I am fetching my parent entity like this:

public SystemNode getSystem(UUID uuid) {
    return systemRepository.findByUuid(uuid)
        .orElseThrow(() -> new EntityNotFoundException("Could not find system with id " + uuid));
}

And this method in turn calls the Spring supported repository interface:

public interface SystemRepository extends CrudRepository<SystemNode, UUID> {

    Optional<SystemNode> findByUuid(UUID uuid);

}

How can I make this filter play nicely together with Spring? I would like to activate it programatically when I need it, not globally. There are scenarios where it would be viable to disregard the filter.

I am using Spring Boot 1.3.5.RELEASE, I cannot update this at the moment.

I think you could enable it for 1 call and immediatly disable it again. An example can be found here — XtremeBaumer
– XtremeBaumer, Commented Apr 10, 2018 at 11:31
@XtremeBaumer I have found that source, but that forces me to circumvent the Spring repository entirely and use HQL to define my own query. I want to avoid that if possible, but I don't know if it is. — jokarl
– jokarl, Commented Apr 10, 2018 at 11:38
No it doesn't. Before calling the repository method get hold of the EntityManager. Use EntityManager.unwrap(Session.class) to get the underlying Hibernate Session. Call enableFilter, call the repository method, call disableFilter. — M. Deinum
– M. Deinum, Commented Apr 10, 2018 at 11:43

jokarl · Accepted Answer · 2018-04-10 14:36:59Z

6

Update and solution

I tried the following as suggested to me in the comments above.

@Autowired
private EntityManager entityManager;

public SystemNode getSystemWithHistoryFrom(UUID uuid) {
    Session session = entityManager.unwrap(Session.class);

    Filter filter = session.enableFilter("dateFilter");
    filter.setParameter("fromDate", new DateTime().minusHours(4).toDate());

    SystemNode systemNode = systemRepository.findByUuid(uuid)
            .orElseThrow(() -> new EntityNotFoundException("Could not find system with id " + uuid));

    session.disableFilter("dateFilter");

    return systemNode;
}

I also had the wrong type in the FilterDef annotation:

@FilterDef(name="dateFilter", parameters=@ParamDef( name="fromDate", type="timestamp" ) )

I changed from date to timestamp.

This returns the correct number of objects, verified against the database.

Thank you!

edited Apr 10, 2018 at 14:36

answered Apr 10, 2018 at 14:28

jokarl

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3 Comments

Matexon Over a year ago

And what when in parent entity I have OneToMany and then I'd like to add condition on third nested entity like A.fieldB.fieldC.fieldD.id = x... How to achieve this with Filter in this case?

Adrien H Over a year ago

I don't understand why you need a hibernate filter here. You could use any filtering method you want here. The only way a hibernate filter could be useful is if you can enable it from several endpoints at once (using an aspect, for example ?).

jokarl Over a year ago

@AdrienH this answer is over four years old, a lot has probably happened since then.

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