java - 无法延迟初始化角色集合无法初始化代理 - 无 session

Question

我正在使用 spring mvc 并使用 spring data jpa，并且遇到此错误无法延迟初始化角色集合无法初始化代理 - 无 session 我知道发生这种情况是因为我没有打开的 session ，但我不知道如何在连接到数据库后保持 session 打开这是到目前为止我的代码:

我的jpa配置类

@Configuration
@EnableSpringConfigured
@ComponentScan( basePackages = {"com.abc.domain", "com.abc.repository", "com.abc.service","com.abc.authenticate"})
@EnableJpaRepositories(basePackages="com.abc.repository")
public class ConfigJPA 
{

       @Bean
       public LocalContainerEntityManagerFactoryBean entityManagerFactory() 
               throws ClassNotFoundException 
       {

          LocalContainerEntityManagerFactoryBean  em  = 
                  new LocalContainerEntityManagerFactoryBean();

          em.setDataSource( dataSource() );
          em.setPackagesToScan("com.abc.domain");
          em.setPersistenceProviderClass(HibernatePersistence.class);
          em.setJpaProperties( asignarPropiedades() );

          return em;
       }

       //Propiedades Hibernate
       Properties asignarPropiedades() {

           Properties jpaProperties = new Properties();

           jpaProperties.put("hibernate.dialect", "org.hibernate.dialect.Oracle10gDialect");
           jpaProperties.put("hibernate.format_sql", true);
           jpaProperties.put("hibernate.ejb.naming_strategy", "org.hibernate.cfg.ImprovedNamingStrategy");
           jpaProperties.put("hibernate.show_sql", true);

           return jpaProperties;
       }


       @Bean
       public DataSource dataSource(){

          DriverManagerDataSource dataSource = 
                  new DriverManagerDataSource();

          dataSource.setDriverClassName("oracle.jdbc.driver.OracleDriver");

          //farmatodo22
            dataSource.setUrl("jdbc:oracle:thin:@127.0.0.1:1521:XE");
            dataSource.setUsername("DATBASE");
            dataSource.setPassword("mypassword");

          return dataSource;
       }

       @Bean
       public JpaTransactionManager transactionManager() 
               throws ClassNotFoundException 
       {

            JpaTransactionManager transactionManager = 
                    new JpaTransactionManager();

            transactionManager.setEntityManagerFactory(entityManagerFactory().getObject());

            return transactionManager;
        }
}

这是我的域类

用户类别

@Entity
@Table(name="User")
public class User implements Serializable {

    private static final long serialVersionUID = 1L;


    //**PRIMARY KEY**//

    @Id
    @SequenceGenerator(name="User_id_seq", sequenceName="SEQ_User")
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="User_id_seq")
    @Column(name="ID_USER", unique=true, nullable=false, precision=8)
    private Long idUser;




    @ManyToOne
    @JoinColumn(name="id_ldap_server", nullable = false)
    private ServerLdap serverLdap;




    @ManyToMany
    @JoinTable
    (
      name="Usuario_Rol",
      joinColumns = 
      {
        @JoinColumn (name="ID_USER", referencedColumnName="ID_USER")
      },
      inverseJoinColumns = 
      {
        @JoinColumn (name="id_rol", referencedColumnName="id_rol")
      }
    )
    private List<Rol> roles;

这是我的角色类

    @Entity
    @Table(name="ROL")
    public class Rol implements Serializable 
    {

    private static final long serialVersionUID = 1L;


    //***PRIMARY KEY***///

    @Id
    @SequenceGenerator(name="ROL_ID_GENERATOR", sequenceName="SEQ_ROL")
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="ROL_ID_GENERATOR")
    @Column(name="id_rol", unique=true, nullable=false, precision=8)
    private Long idRol;




    @ManyToMany(mappedBy="roles")
    private List<User> users;

}

我正在尝试在 User 和 Rol 之间建立 ManyToMany 关系，如果我输入 fetchType = EAGER，效果会很好，但我不想使用该获取类型，因为我认为效率不高。

Answer 1

其中一个选项是使用在 View 中打开 session ，但它几乎没有缺点，并且被一些人认为是一种不好的做法。我建议阅读以下 SO 帖子以做出明智的选择。

• Open Session In View Pattern
• Why is Hibernate Open Session in View considered a bad practice?
• http://java.dzone.com/articles/open-session-view-design