java - 在 CriteriaQuery 中使用 @ElementCollection(或查询 @ElementCollection 的内容)

Question

public enum ReportStatus { 
    SUCCCEED, FAILED;
}

public class Work {
    @ElementCollection
    @Enumerated(EnumType.STRING)
    List<ReportStatus> reportStatuses;
}

给定以下结构，我想执行查询以查找由 reportStatuses 过滤的所有工作。它适用于以下 hql 语法:

public List<Long> queryHQL() {
    final String query = "SELECT w.id FROM Work w JOIN w.reportStatuses s WHERE s in (:rs)";

    final List<ReportStatus> reportStatuses = new ArrayList<ReportStatus>();
    reportStatuses.add(ReportStatus.FAILED);

    return this.entityManager.createQuery(query).setParameter("rs", reportStatuses).getResultList();
}

但我想使用标准 API (jpa2)，但不知道该怎么做。这是我认为最接近的尝试:

public List<Long> query() {
    final List<ReportStatus> reportStatuses = new ArrayList<ReportStatus>();
    reportStatuses.add(ReportStatus.FAILED);

    final CriteriaBuilder builder = this.entityManager.getCriteriaBuilder();

    final CriteriaQuery<Long> criteriaQuery = builder.createQuery(Long.class);
    final Root<Work> workModel = criteriaQuery.from(Work.class);

    final ListJoin<Work, ReportStatus> status = workModel.joinList("reportStatuses");

    final Predicate predicate = status.in(reportStatuses);

    criteriaQuery.where(predicate);
    criteriaQuery.select(workModel.<Long> get("id"));

    return this.entityManager.createQuery(criteriaQuery).getResultList();
}

我也尝试过使用 hibernate criteria API，但是作为 jpa2，我没能找到正确的语法。

Answer 1

我会使用以下 CriteriaQuery 语法来做同样的事情。

EntityManager em = entityManagerFactory.createEntityManager();

final List<ReportStatus> reportStatuses = new ArrayList<ReportStatus>();
reportStatuses.add(ReportStatus.FAILED);

final CriteriaBuilder builder = em.getCriteriaBuilder();

final CriteriaQuery<Long> criteriaQuery = builder.createQuery(Long.class);
final Root<Work> _work = criteriaQuery.from(Work.class);

/*final ListJoin<Work, ReportStatus> status = _work.joinList("reportStatuses");
final Predicate predicate = status.in(reportStatuses);
criteriaQuery.where(predicate);*/

final Expression<List<ReportStatus>> _status = _work.get(Work_.reportStatuses);
_status.in(reportStatuses);

criteriaQuery.select(_work.get(Work_.id));

List<Long> list =  em.createQuery(criteriaQuery).getResultList();

更新

Thanks, but we do not use the generated metamodel. So unfortunately I can't try with your answer. :(

如果您不使用 Metamodel，只需将 _work.get(Work_.reportStatuses) 替换为 _work.get("reportStatuses")。它会起作用的。 :)