[英]Handling entities inheritance spring boot
我正在使用本教程來處理實體繼承。 我有擴展User實體的個人和公司實體。
@Entity
@Inheritance
public abstract class User {
@Id
private long id;
@NotNull
private String email;
// getters and settres
}
@Entity
public class Person extends User {
private int age;
// getters and settres and other attributs
}
@Entity
public class Company extends User {
private String companyName;
// getters and settres and other attribut
}
然后擴展UserBaseRepository的UserRpository,PersonRepository和Company Repository。
@NoRepositoryBean
public interface UserBaseRepository<T extends User>
extends CrudRepository<T, Long> {
public T findByEmail(String email);
}
@Transactional
public interface UserRepository extends UserBaseRepository<User> { }
@Transactional
public interface PersonRepository extends UserBaseRepository<Person> { }
@Transactional
public interface CompanyRepository extends UserBaseRepository<Company> { }
問題是當調用personRepository.findAll()來獲取所有人時,結果我也得到了公司。
您的問題出在JPA要求的“ Discriminator”列中。 您正在使用@Inheritance
批注,默認情況下將使用InheritanceType.SINGLE_TABLE
策略。 這意味着:
Person
和Company
將進入一個表。 為了使它適用於您的用例,我做了以下工作:
實體:
@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;
}
}
數據庫架構:
-- 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
);
一些測試數據:
insert into user_table(id, dtype, age, email) values
(1,'Person', 25, 'john.doe@email.com'),
(2,'Person',22, 'jane.doe@email.com');
insert into user_table(id, dtype, company_name, email) values
(3,'Company','Acme Consultants', 'acme@company.com'),
(4,'Company', 'Foo Consultants', 'foo@company.com');
倉庫:
@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測試:
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());
}
}
以上測試通過。 這表明JPA現在正確使用了鑒別器來檢索所需的記錄。
有關您的問題的JPA相關理論,請參見此鏈接 。
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