Hibernate-一个域的双向插入导致空

Question

所以，我正在使用Hibernate 5和SpringMVC 4制作这个网络应用程序。出于某种原因，我无法插入具有OneTo很多关系的实体。在我首先解释任何事情之前，我想说我尝试了这里和其他论坛上发布的许多解决方案，但都不适合我...我不知道我在尝试解决问题时是否做错了什么。

现在，我最近添加了phone表。在此之前，我可以添加一个配置文件对象，它会级联同一事务中地址和用户实体的插入。现在，随着将电话实体添加到代码中，持久化操作不断抛给我

列“ProfileId”不能为null

所有类都有各自的setter和getter。。。但为了保持代码简短，我删除了它们。

@Entity
@Table(name = "profile")
public class Profile {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    @NotEmpty
    @Size(max = 100)
    @Column(name = "FirstName", nullable = false, unique = false, length = 100)
    private String firstName;

    @NotEmpty
    @Size(max = 100)
    @Column(name = "LastName", nullable = false, unique = false, length = 100)
    private String lastName;

    @NotNull
    @DateTimeFormat(pattern = "MM/dd/yyyy")
    @Column(name = "BirthDate", nullable = false, unique = false)
    private Timestamp birthDate;

    @Valid
    @OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "UserId")
    private User user;

    @Valid
    @OneToOne(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    @JoinColumn(name = "AddressId")
    private Address address;

    @Valid
    @OneToMany(fetch = FetchType.EAGER, mappedBy = "profile", cascade = CascadeType.ALL)
    private List<Phone> phones;
}

_

@Entity
@Table(name = "Phone")
public class Phone {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    @NotEmpty
    @Size(max = 50)
    @Column(name = "PhoneNumber", nullable = false, unique = false, length = 50)
    private String phoneNumber;

    @Valid
    @ManyToOne(cascade = CascadeType.ALL, optional = false)
    @JoinColumn(name = "ProfileId", nullable = false)
    private Profile profile;
}

_

@Entity
@Table(name = "\"User\"", uniqueConstraints = { @UniqueConstraint(columnNames = "Username") })
public class User {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "Id", unique = true, nullable = false)
    private long id;

    @Column(name = "Username", nullable = false, unique = true, length = 80)
    private String username;

    @Column(name = "Password", nullable = false, length = 128)
    private String password;

    @Valid
    @OneToOne(fetch = FetchType.LAZY, mappedBy = "user", cascade = CascadeType.ALL)
    private Profile profile;
}

_

@Entity
@Table(name = "Address")
public class Address {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    @Column(name = "FullAddress", nullable = false, unique = false, length = 100)
    private String fullAddress;

    @Column(name = "City", nullable = false, unique = false, length = 100)
    private String city;

    @Column(name = "PostalCode", nullable = false, unique = false, length = 100)
    private String postalCode;

    @OneToOne(fetch = FetchType.LAZY, mappedBy = "address", cascade = CascadeType.ALL)
    private Profile profile;
}

在我为类似案例找到的解决方案中，我发现：

• 更改Phone表的ProfileId列以使其允许null值，但在JoinColumn注释上保持null=false。这样做的结果是，它实际上插入了所有实体，但ProfileId也保存为null

像我这样的新人。。据我所知，ProfileId接收到null是因为出于某种原因，在插入Profile表之后，尚未在Profile对象上设置生成的id，这导致电话插入失败。但我不知道怎么解决。

如果您需要知道我是如何持久化对象的。。。

sessionFactory.getCurrentSession().persist(profile);

sessionFactory是自动连接的。profile对象具有以下特性：

{"phones":[{"phoneNumber":"123456789"}],"user":{"username":"[e-mail]","password":"123123123"},"firstName":"Nameeeeee","lastName":"Nameeeeee","birthDate":"02/05/2016", "address":{"fullAddress":"laksjdlkas","city":"alksjdlkasjd","postalCode":"101010"}}

最后是完整的错误：

Hibernate: 
    insert 
    into
        Address
        (FullAddress, City, PostalCode) 
    values
        (?, ?, ?)
Hibernate: 
    select
        last_insert_id()
Hibernate: 
    insert 
    into
        `User` (
            Password, Username
        ) 
    values
        (?, ?)
Hibernate: 
    select
        last_insert_id()
Hibernate: 
    insert 
    into
        Profile
        (AddressId, BirthDate, FirstName, LastName, UserId) 
    values
        (?, ?, ?, ?, ?)
Hibernate: 
    select
        last_insert_id()
Hibernate: 
    insert 
    into
        Phone
        (PhoneNumber, ProfileId) 
    values
        (?, ?)
WARN  SqlExceptionHelper::logExceptions:127 - SQL Error: 1048, SQLState: 23000
ERROR SqlExceptionHelper::logExceptions:129 - Column 'ProfileId' cannot be null

正在添加请求的代码。

@Configuration
@EnableTransactionManagement
@ComponentScan(basePackages = { "com.configuration" })
@PropertySource(value = { "classpath:application.properties" })
public class HibernateConfiguration {

    final static Logger logger = Logger.getLogger(HibernateConfiguration.class);
    @Autowired
    private Environment environment;

    @Bean
    public LocalSessionFactoryBean sessionFactoryBean() {
        logger.info("Creating LocalSessionFactoryBean...");
        LocalSessionFactoryBean sessionFactoryBean = new LocalSessionFactoryBean();
        sessionFactoryBean.setDataSource(dataSource());
        sessionFactoryBean.setPackagesToScan(new String[] { "com.model" });
        sessionFactoryBean.setHibernateProperties(hibernateProperties());
        return sessionFactoryBean;
    }

    @Bean
    public DataSource dataSource() {
        DriverManagerDataSource dataSource = new DriverManagerDataSource();
        dataSource.setDriverClassName(environment.getRequiredProperty("jdbc.driverClassName"));
        dataSource.setUrl(environment.getRequiredProperty("jdbc.url"));
        dataSource.setUsername(environment.getRequiredProperty("jdbc.username"));
        dataSource.setPassword(environment.getRequiredProperty("jdbc.password"));
        return dataSource;
    }

    @Bean
    @Autowired
    public HibernateTransactionManager transactionManager(SessionFactory sessionFactory) {
        HibernateTransactionManager transactionManager = new HibernateTransactionManager();
        transactionManager.setSessionFactory(sessionFactory);
        return transactionManager;
    }

    private Properties hibernateProperties() {
        Properties properties = new Properties();
        properties.put("hibernate.dialect", environment.getRequiredProperty("hibernate.dialect"));
        properties.put("hibernate.show_sql", environment.getRequiredProperty("hibernate.show_sql"));
        properties.put("hibernate.format_sql", environment.getRequiredProperty("hibernate.format_sql"));
        properties.put("hibernate.temp.use_jdbc_metadata_defaults",
                environment.getRequiredProperty("hibernate.temp.use_jdbc_metadata_defaults"));
    }

Answer 1

在代码的某个地方，您应该这样做

Phone phone = new Phone();
//... set phone vars
phone.setProfile(profile);

sessionFactory.getCurrentSession().persist(profile);