您的问题在于 JPA 所需的“Discriminator”列。您正在使用@Inheritance
注释,默认情况下将使用InheritanceType.SINGLE_TABLE
战略。这意味着以下内容:
- 您继承的实体
Person
and Company
将进入一个表。
- JPA 将需要一个鉴别器来区分实体类型。
我做了以下操作以使其适合您的用例:
实体:
@Inheritance
@Entity
@Table(name = "user_table")
public abstract class User {
@Id
private long id;
@NotNull
@Column
private String email;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
}
@Entity
public class Company extends User {
@Column(name = "company_name")
private String companyName;
public String getCompanyName() {
return companyName;
}
public void setCompanyName(String companyName) {
this.companyName = companyName;
}
}
@Entity
public class Person extends User {
@Column
private int age;
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
}
数据库架构:
-- user table
create table user_table (
id BIGINT NOT NULL PRIMARY KEY,
email VARCHAR(50) NOT NULL,
age INT,
company_name VARCHAR(50),
dtype VARCHAR(80) -- Discriminator
);
一些测试数据:
insert into user_table(id, dtype, age, email) values
(1,'Person', 25, '[email protected] /cdn-cgi/l/email-protection'),
(2,'Person',22, '[email protected] /cdn-cgi/l/email-protection');
insert into user_table(id, dtype, company_name, email) values
(3,'Company','Acme Consultants', '[email protected] /cdn-cgi/l/email-protection'),
(4,'Company', 'Foo Consultants', '[email protected] /cdn-cgi/l/email-protection');
存储库:
@NoRepositoryBean
public interface UserBaseRepository<T extends User> extends CrudRepository<T, Long> {
T findByEmail(String email);
}
@Transactional
public interface PersonRepository extends UserBaseRepository<Person> {
}
@Transactional
public interface CompanyRepository extends UserBaseRepository<Company> {
}
JUnit 测试:
public class MultiRepositoryTest extends BaseWebAppContextTest {
@Autowired
private PersonRepository personRepository;
@Autowired
private CompanyRepository companyRepository;
@Test
public void testGetPersons() {
List<Person> target = new ArrayList<>();
personRepository.findAll().forEach(target::add);
Assert.assertEquals(2, target.size());
}
@Test
public void testGetCompanies() {
List<Company> target = new ArrayList<>();
companyRepository.findAll().forEach(target::add);
Assert.assertEquals(2, target.size());
}
}
以上测试均通过。这表明 JPA 现在可以正确使用鉴别器来检索所需的记录。
对于您的问题的 JPA 相关理论,请参阅此link https://docs.oracle.com/javaee/6/tutorial/doc/bnbqn.html.