我正在Spring Data JPA中尝试JOINED
类型的数据库继承,引用this article 。这很好用。但我必须在我的项目中实现 MappedSuperClass 。我通过以下方式实现:
Base.java
@MappedSuperclass
public abstract class Base {
public abstract Long getId();
public abstract void setId(Long id);
public abstract String getFirstName();
public abstract void setFirstName(String firstName);
}
BaseImpl.java
@Entity
@Inheritance(strategy = InheritanceType.JOINED)
public class BaseImpl extends Base {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
private String firstName;
...
}
Super1.java
@MappedSuperclass
public abstract class Super1 extends BaseImpl {
public abstract String getSuperName();
public abstract void setSuperName(String guideName);
}
Super1Impl.java
@Entity
public class Super1Impl extends Super1 {
private String superName;
...
}
BaseBaseRepository.java
@NoRepositoryBean
public interface BaseBaseRepository<T extends Base> extends JpaRepository<T, Long> { }
BaseRepository.java
@NoRepositoryBean
public interface BaseRepository<T extends Base> extends BaseBaseRepository<Base> { }
BaseRepositoryImpl.java
@Transactional
public interface BaseRepositoryImpl extends BaseRepository<BaseImpl> { }
Super1Repository.java
@NoRepositoryBean
public interface Super1Repository<T extends Super1> extends BaseBaseRepository<Super1> { }
Super1RepositoryImpl.java
@Transactional
public interface Super1RepositoryImpl extends Super1Repository<Super1Impl> { }
我正在尝试在测试用例中保存 Super1
对象:
@Test
public void contextLoads() {
Super1 super1 = new Super1Impl();
super1.setSuperName("guide1");
super1.setFirstName("Mamatha");
super1.setEmail("jhhj");
super1.setLastName("kkjkjhjk");
super1.setPassword("jhjjh");
super1.setPhoneNumber("76876876");
System.out.println(super1Repository.save(super1));
}
但我收到以下错误:
Caused by: org.springframework.beans.factory.BeanCreationException:
Error creating bean with name 'baseRepositoryImpl':
Invocation of init method failed; nested exception is java.lang.IllegalArgumentException:
This class [class com.example.entity.Base] does not define an IdClass
.....
Caused by: java.lang.IllegalArgumentException: This class [class com.example.entity.Base] does not define an IdClass
.......
在Super1Impl
中尝试了@PrimaryKeyJoinColumn(name = "id", referencedColumnName = "id")
,但仍然遇到相同的错误。
最佳答案
该错误是由不正确的存储库接口(interface)声明引起的。
BaseRepository<T extends Base> extends BaseBaseRepository<Base>
应该是
BaseRepository<T extends Base> extends BaseBaseRepository<T>
和
Super1Repository<T extends Super1> extends BaseBaseRepository<Super1>
应该是
Super1Repository<T extends Super1> extends BaseBaseRepository<T>
按照目前的声明,BaseBaseRepository<Base>
表示 Base
的存储库对象和Base
没有@Id
字段,因此出现错误。
关于spring-data-jpa - 如何使用MapperSuperClass在Spring Data JPA中实现数据库继承?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41090908/