Java CriteriaBuilder加入 - 无法解决或不是字段

时间:2016-11-09 18:35:49

标签: java jpa java-ee hibernate-criteria

我尝试使用CriteriaBuilder中的联接进行选择,但我在Eclipse中遇到此错误。我该如何解决?

Hibernate version: hibernate-jpa-2.0-api<br />

Java Version: 1.8

fonte cannot be solved or is not a field

NotificacaoDao.java

@Stateless
public class NotificacaoDao {
    @PersistenceContext(unitName = "PostgreSQLDS")
    private EntityManager em;

    @EJB
    private NotificacaoDao NotificacaoDao;

    public List<Notificacao> getResultList(int first, int pageSize, String sortField, SortOrder sortOrder, Map<String, Object> filters)  throws ApplicationException{
        try {

            CriteriaBuilder cb = em.getCriteriaBuilder();
            CriteriaQuery<Notificacao> cq = cb.createQuery(Notificacao.class);

            Metamodel m = em.getMetamodel();
            EntityType<Notificacao> Notificacao_ = m.entity(Notificacao.class);

            Root<Notificacao> myObj = cq.from(Notificacao_);
            Join<Notificacao, Fonte> fontes = myObj.join(Notificacao_.fonte); // HERE I'M GETTING THE ERROR 
            cq.where(NotificacaoDao.getFilterCondition(cb, myObj, filters));
            Predicate filterCondition = NotificacaoDao.getFilterCondition(cb, myObj, filters);
            filterCondition = cb.and(filterCondition, cb.equal(myObj.get("excluido"), "N"));
            cq.where(filterCondition);
            if (sortField != null) {
                if (sortOrder == SortOrder.ASCENDING) {
                    cq.orderBy(cb.asc(myObj.get(sortField)));
                } else if (sortOrder == SortOrder.DESCENDING) {
                    cq.orderBy(cb.desc(myObj.get(sortField)));
                }
            }
            return em.createQuery(cq).setFirstResult(first).setMaxResults(pageSize).getResultList();
        } catch(Exception e) {
            throw new ApplicationException("myException", e);
        }
    }

Notificacao.java

@Entity
@Table(name = "tb_notificacao", schema = "indicadores")
@NamedQuery(name = "Notificacao.findAll", query = "SELECT n FROM Notificacao n")
@FilterDef(name="notificacaoNaoExcluido", defaultCondition="excluido = 'N'")
public class Notificacao implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @SequenceGenerator(name = "tb_notificacao_codnotificacao_seq", sequenceName = "TB_NOTIFICACAO_CODNOTIFICACAO_SEQ", schema = "indicadores", allocationSize = 1)
    @GeneratedValue(generator = "tb_notificacao_codnotificacao_seq")
    @Column(name = "codnotificacao", nullable = false)
    private Integer codnotificacao;

    private String descricao;

    private String excluido;

    private String nome;

    // bi-directional many-to-one association to CargaNotificacao
    @OneToMany(mappedBy = "notificacao")
    private List<CargaNotificacao> cargaNotificacoes;

    // bi-directional many-to-one association to Fonte
    @Inject
    @ManyToOne
    @JoinColumn(name = "codfonte")
    private Fonte fonte;

    // bi-directional many-to-one association to UsuarioNotificacao
    @OneToMany(mappedBy = "notificacao")
    @Filter(name="usuarioNaoExcluido", condition="excluido = 'N'")
    private List<UsuarioNotificacao> usuarioNotificacoes;

    public Notificacao() {
    }
    // getters and setters
}

Fonte.java

@Entity
@Table(name = "tb_fonte", schema = "indicadores")
@NamedQuery(name = "Fonte.findAll", query = "SELECT f FROM Fonte f")
public class Fonte implements Serializable {
    private static final long serialVersionUID = 1L;

    @Id
    @SequenceGenerator(name = "tb_fonte_codfonte_seq", sequenceName = "TB_FONTE_CODFONTE_SEQ", schema = "indicadores", allocationSize = 1)
    @GeneratedValue(generator = "tb_fonte_codfonte_seq")
    @Column(name = "codfonte", nullable = false)
    private Integer codfonte;

    private String nome;

    // bi-directional many-to-one association to Indicador
    @OneToMany(mappedBy = "fonte")
    @Filter(name="indicadorNaoExcluido", condition="excluido = 'N'")
    private List<Indicador> indicadores;

    // bi-directional many-to-one association to Notificacao
    @OneToMany(mappedBy = "fonte")
    @Filter(name="notificacaoNaoExcluido", condition="excluido = 'N'")
    private List<Notificacao> notificacoes;

    public Fonte() {
    }
    // getters and setters
}

1 个答案:

答案 0 :(得分:2)

嗯,在Metamodels上,基本上有三种使用方法:

  1. 使用基于IDE的元模型生成工具
  2. 使用静态规范元模型类
  3. 使用em.getMetamodel()API,即您正在使用的
  4. 我建议您使用的解决方案更接近您所做的是在第3点。

    第3点解决方案: 替换以下代码:

        Metamodel m = em.getMetamodel();
        EntityType<Notificacao> Notificacao_ = m.entity(Notificacao.class);
        Root<Notificacao> myObj = cq.from(Notificacao_);
        Join<Notificacao, Fonte> fontes = myObj.join(Notificacao_.fonte); // HERE I'M GETTING THE ERROR 
    

    使用新代码:

        Metamodel m = em.getMetamodel();
        EntityType<Notificacao> notificacao_ = m.entity(Notificacao.class);
        Root<Notificacao> myObj = cq.from(notificacao_);
        Join<Notificacao, Fonte> fontes = myObj.join(notificacao_.getSingularAttribute("fonte",Fonte.class));
    

    积分1&amp; 2解决方案 请注意Notificacao_必须是静态或生成的类,并且绝不能是em.getMetamodel()的实例。在你的情况下还要注意Notificacao_之前是变量而不是如下所示的类:

    EntityType<Notificacao> Notificacao_ = m.entity(Notificacao.class);
    

    如果您需要更多信息,请告诉我。