【发布时间】：2016-05-16 04:14:12
【问题描述】：
所以，我正在使用 Hibernate 5 和 SpringMVC 4 制作这个 webapp。 由于某种原因，我不能插入具有 OneToMany 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发布的许多解决方案，但都没有为我工作......我不知道我在尝试解决问题时是否做错了。
现在，我最近添加了电话桌。在此之前，我可以添加一个配置文件对象，它将在同一事务中级联地址和用户实体的插入。现在，在代码中添加电话实体后，持久化操作一直在抛出我
列“ProfileId”不能为空
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之间是：
- 更改 Phone 表的 ProfileId 列以使其允许空值，但保留 JoinColumn 注释上的 nullable = false。这样做的结果是它实际上插入了所有实体，但 ProfileId 也被保存为 null。
- 更改了级联类型...仍然是同样的错误。
- 有人说 EntityManager 上下文发生冲突。这不是我的情况。
- 将可以为空的 JoinColumn 注释属性设置为 false。已经这样做了，仍在代码中，也无法正常工作。
- 在关系的两边设置级联类型。也做了，也还在代码上，不工作。
- 将可选的 ManyToOne 关系属性设置为 false。也不行。
和我一样新.. 我的理解是 ProfileId 正在接收 null，因为由于某种原因，在插入 Profile 表之后，生成的 id 尚未在配置文件对象上设置，这导致手机插入失败。但我不知道如何解决。
如果您需要知道我是如何持久化对象的...
sessionFactory.getCurrentSession().persist(profile);
而且 sessionFactory 是自动装配的。 配置文件对象具有以下内容：
{"phones":[{"phoneNumber":"123456789"}],"user":{"username":"Nameeeeee@alksjd.com","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
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"));
}
【问题讨论】：