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我正在尝试强制 HQL 急切加载所有 Lazy 属性。使用 FETCH 给了我“多袋”异常，我无法将其更改为 Set，因为它们在其他地方被引用。我尝试了其他映射注释，但没有运气。“FETCH ALL PROPERTIES”被忽略，没有加载任何内容。我一直不明白它的真正作用。显然，当 JSF 尝试读取列表时，我得到“无法初始化惰性列表。连接关闭”

我正在使用休眠 4.0。所以这是我的 HQL 查询：

Select r from Responsavel r FETCH ALL PROPERTIES 
where r.unidadeEmpresarial.id = :idUnidade

这是我的模型课：

public class Responsavel {

@Id
@SequenceGenerator(name = "SQ_RESPONSAVEL_GENERATOR", sequenceName = "SE_RESPONSAVEL", allocationSize = 1)
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "SQ_RESPONSAVEL_GENERATOR")
@Column(name = "SQ_RESPONSAVEL")
private Long id;

@Temporal(TemporalType.DATE)
@Column(name = "DT_FIM")
private Date dataFim;

@NotNull
@Temporal(TemporalType.DATE)
@Column(name = "DT_INICIO")
private Date dataInicio;

@Temporal(TemporalType.DATE)
@Column(name = "DT_NASCIMENTO")
private Date dataNascimento;

@Type(type = "org.hibernate.type.NumericBooleanType")
@Column(name = "IN_PODER_ADM_REPRES")
private Boolean temPoderAdministrativo;

@Column(name = "NO_RESPONSAVEL")
private String nome;

@Column(name = "NU_CPF_CNPJ_RESPONSAVEL")
private String numeroCpfCnpj;

@NotNull
@Enumerated(EnumType.STRING)
@Column(name = "TP_PESSOA")
private String tipoPessoa;

@ManyToOne
@JoinColumn(name = "SQ_PAIS")
private Pais pais;

@Column(name = "VL_CAPITAL_SOCIAL")
private BigDecimal valorCapitalSocial;

@Column(name = "PC_CAPITAL_SOCIAL")
private BigDecimal percentualCapitalSocial;

@ManyToOne
@JoinColumn(name = "SQ_UNIDADE_EMPRESARIAL")
private UnidadeEmpresarial unidadeEmpresarial;

@OneToMany(mappedBy = "responsavel", fetch = FetchType.LAZY)
@Fetch(FetchMode.SUBSELECT)
private Set<ResponsavelQualificacaoResponsavel> listaResponsavelQualificacaoResponsavel;

@OneToMany(mappedBy = "responsavel")
private List<Representante> representantes;

@OneToMany(mappedBy = "responsavel")
private List<ContatoResponsavel> contatoResponsavel;

@OneToMany(mappedBy = "responsavel", cascade = CascadeType.ALL)
private List<EnderecoResponsavel> enderecoResponsavel;

}

谢谢

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1 回答 1

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如果您遇到多包异常，那是因为 Hibernate 无法一次获取所有集合。您应该选择一个您希望使用基本查询选择的集合并使用连接来完成，然后使用 Hibernate.initialize 获取您需要的其余集合。

所以它看起来像这样（我不选择，因为它不是必需的）：

from Responsavel r 
join fetch pais p
join fetch unidadeEmpresarial u
join fetch representantes s
where r.unidadeEmpresarial.id = :idUnidade

您可以尝试包含哪些内容。如果您希望从 Hibernate 会话中分离以使用前端或服务类中的集合，则剩下的需要初始化。

Hibernate.initialize(listaResponsavelQualificacaoResponsavel);
Hibernate.initialize(contatoResponsavel);
Hibernate.initialize(enderecoResponsavel);

如果您将保留会话，则无需初始化，它们将按需初始化（尽管这可能会导致性能问题，具体取决于您的用例）。

于 2013-07-29T18:22:27.177 回答