JPA / Hibernate: CriteriaBuilder - How to create query using relationship object?

Question

I have the following four tables:

SCHEDULE_REQUEST TABLE: ID, APPLICATION_ID (FK)

APPLICATION TABLE: ID, CODE

USER_APPLICATION TABLE: APPLICATION_ID (FK), USER_ID (FK)

USER TABLE: ID, NAME

Now I wanted to create a CriteriaBuilder where condition is to select ScheduleRequests for specified user Ids.

I have the following codes:

List<User> usersList = getSelectedUsers(); // userList contains users I wanted to select

CriteriaBuilder builder = getJpaTemplate().getEntityManagerFactory().getCriteriaBuilder();
CriteriaQuery<ScheduleRequest> criteria = builder.createQuery(ScheduleRequest.class);
Root<ScheduleRequest> scheduleRequest = criteria.from(ScheduleRequest.class);
criteria = criteria.select(scheduleRequest);

ParameterExpression<User> usersIdsParam = null;
if (usersList != null) {
    usersIdsParam = builder.parameter(User.class);
    params.add(builder.equal(scheduleRequest.get("application.userApplications.user"), usersIdsParam));
}

criteria = criteria.where(params.toArray(new Predicate[0]));

TypedQuery<ScheduleRequest> query = getJpaTemplate().getEntityManagerFactory().createEntityManager().createQuery(criteria);

// Compile Time Error here:
// The method setParameter(Parameter<T>, T) in the type TypedQuery<ScheduleRequest> is not 
// applicable for the arguments (ParameterExpression<User>, List<User>)
query.setParameter(usersIdsParam, usersList);

return query.getResultList();

Can you please help me how to pass query filter to a relationship object? I think what I did in "application.userApplications.user" is wrong? Please really need help.

Thank you in advance!

Defect

Answer 1

Using the canonical Metamodel and a couple of joins, it should work. Try if you get some hints from the following pseudo-code (not tested):

...
Predicate predicate = cb.disjunction();
if (usersList != null) {
    ListJoin<ScheduleRequest, Application> applications = scheduleRequest.join(ScheduleRequest_.applications);
    ListJoin<Application, UserApplication> userApplications = applications.join(Application_.userApplications);
    Join<UserApplication, User> user = userApplications.join(UserApplication_.userId);
    for (String userName : usersList) {
        predicate = builder.or(predicate, builder.equal(user.get(User_.name), userName));
    }
}

criteria.where(predicate); 
...

In order to understand Criteria Queries, have a look at these tutorials: http://www.ibm.com/developerworks/java/library/j-typesafejpa/ http://docs.oracle.com/javaee/6/tutorial/doc/gjitv.html

The second link should also guide you on how to use Metamodel classes, that should be built automatically by the compiler / IDE.

Cavie

Hi perissf. Thank for the helping. But I don't have classes "ScheduleRequest_", "UserApplication_", "User_". How do i create them? And also in your pseudo-code, you didn't include "Application" table? The relationship table is USER_APPLICATION table, relating USER and APPLICATION tables. Please kindly advise. Really appreciate it. Thanks. – Defect

I saw some tutorial here but I'm not sure how to do apply it in my case. altuure.com/2010/09/23/… – Defect

I have updated my answer by adding some references and by adding the third join as per your information – Cavie