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Java CriteriaBuilder Join - 无法解决或不是字段

转载 作者:行者123 更新时间:2023-12-02 03:14:26 25 4
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我尝试使用 CriteriaBuilder 中的联接进行选择,但在 Eclipse 中收到此错误。我该如何修复它?

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
}

最佳答案

嗯,在元模型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 API

我建议您使用的解决方案更接近您正在做的事情,即第 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));

第 1 点和第 2 点解决方案请注意,Notificacao_ 必须是静态或生成的类,并且绝不能是 em.getMetamodel() 的实例。另请注意,在您的情况中,Notificacao_ 是变量而不是类,如下所示:

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

如果您需要更多信息,请告诉我。

关于Java CriteriaBuilder Join - 无法解决或不是字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40513761/

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